id string | solution string | answer string | metadata dict | problem string | candidates list | relevance_scores list | relevance_scores_full null |
|---|---|---|---|---|---|---|---|
aops_152764 | Let's use kunny's first choice. Let $u=2-x.$ Then $x=2-u$ so $x-1=1-u$ and $dx=-du.$ When $x-1,\ u=1$ and when $x=2,\ u=0.$
$\int_{1}^{2}(x-1)\sqrt{2-x}\,dx=-\int_{1}^{0}(1-u)\sqrt{u}\,du=\int_{0}^{1}u^{1/2}-u^{3/2}\,du$
$=\left.\frac23u^{3/2}-\frac25u^{5/2}\right|_{0}^{1}=\frac23-\frac25=\frac4{15}.$
Possibil... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Hey guys I need a method to solve this integral using substitution\r\n\r\n\\[\\int_{1}^{2}(x-1) \\sqrt{2-x}\\; dx\\]",
"content_html": "Hey guys I need a method to solve this integral using substitut... | Evaluate the integral
\[
\int_{1}^{2} (x-1)\sqrt{2-x}\,dx.
\] | [
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aops_2433736 | This is also from Virginia Tech 2004 (though I would not be surprised if there were another source before this).
Let $b_n := \tfrac{a_{n+1}}{a_n}$. It clearly suffices to prove the result under the assumption that $\lim_{n\to\infty}b_n = 1$.
We claim that, for all $x$ sufficiently close to $1$,
\[
|1 - x| \geq -\tf... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$\\{a_n\\}_{n=1}^{\\infty}$ is a sequence of positive real numbers such that $\\lim_{n\\rightarrow \\infty}a_n=0$\n\nShow that the series below diverges:\n\n$$\\sum_{n=1}^{\\infty}|1-\\frac{a_{n+1}}{a_n}|$... | Let \(\{a_n\}_{n=1}^{\infty}\) be a sequence of positive real numbers such that \(\lim_{n\to\infty} a_n = 0\). Show that the series
\[
\sum_{n=1}^{\infty}\left|1-\frac{a_{n+1}}{a_n}\right|
\]
diverges. | [
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numina_10031205 | Solution. Since $\int e^{u} d u=e^{u}+C$, in this integral, it is necessary to make $x^{2}$ the variable of integration, i.e., let $x^{2}=u$. In this case, $d u=d x^{2}=2 x d x$. Therefore, we bring the factor $x$ under the differential sign and obtain the differential of $x^{2}$: $x d x=\frac{1}{2} d\left(x^{2}\right)... | \frac{1}{2}e^{x^{2}}+C | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 1. Find $\int e^{x^{2}} \cdot x d x$. | [
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0.... | null |
aops_88628 | Post these questions in the College Playground section next time.
$\int\sin^{4}x\cos x\,dx=\int\sin^{4}x\,d\left(\sin x\right)=\frac{\sin^{5}x}{5}+C$
That step just came from the fact that $d\left(\sin x\right)=\cos x\,dx$.
$\int\frac{5}{x\ln x}\,dx=\int\frac{5}{\ln x}\,d\left(\ln x\right)=5\ln\left(\ln x\righ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Hi, I'm swedish, been a while since I posted here.\r\nI'm not one hundred percent sure of the english terms but I'll try and hopefully you'll understand it anyway.\r\n\r\nCalculate the following Integratio... | Calculate the following integrals using substitution:
1) \(\displaystyle \int \sin^4 x \cos x \, dx.\)
2) \(\displaystyle \int \frac{5}{x\ln x} \, dx.\) | [
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aops_157166 | P = $ \int\frac{1}{\sqrt{(2x-x^{2})^{3}}}\,dx = \int x^{-3/2}(2-x)^{-3/2}\,dx$.
Now change variables according to $ \frac{2}{x}-1 = t^{2}$, and so
$ P =-\frac{1}{2}\int(1+t^{2}) \cdot t^{-2}\,dt = \frac{1-t^{2}}{2t}+C = \frac{1-(\frac{2}{x}-1)}{2\sqrt{\frac{2}{x}-1}}+C$
$ \Rightarrow P = \frac{x-1}{\sqrt{2x-x... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$ \\int \\frac{dx}{\\sqrt{(2x-x^{2})^{3}}}$\r\n\r\nand \r\n\r\n$ \\int \\frac{(x+\\sqrt{1+x^{2}})^{15}dx}{\\sqrt{1+x^{2}}}$",
"content_html": "<img src=\"//latex.artofproblemsolving.com/c/c/c/ccce02f... | 1. \displaystyle \int \frac{dx}{\sqrt{(2x-x^{2})^{3}}}
2. \displaystyle \int \frac{(x+\sqrt{1+x^{2}})^{15}}{\sqrt{1+x^{2}}}\,dx | [
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aops_1309682 | What a mess...
$$H:= \int (\sqrt{\sin 2x}+\sqrt{\cos 2x})\sin x \mathrm d x=\int\sin x\sqrt{\sin 2x}\,\mathrm d x+\int \sin x \sqrt{\cos 2x}\,\mathrm d x$$
[b]The first integral ($I$):[/b]
$$I=\int \sec ^2 x \cdot \frac{\sqrt 2 \tan ^{\frac 32} x}{(\tan ^2 x +1)^2}\,\mathrm d x$$
substitute $u=\tan x$:
$$I=\sqrt 2\int... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$\\int (\\sqrt{\\sin 2x}+\\sqrt{\\cos 2x})\\sin xdx$",
"content_html": "<img src=\"//latex.artofproblemsolving.com/c/f/d/cfd224086930b61c27ac7db04a80c7c75548ce98.png\" class=\"latex\" alt=\"$\\int (\... | \[
\int \bigl(\sqrt{\sin 2x}+\sqrt{\cos 2x}\bigr)\sin x\,dx
\] | [
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aops_1512839 | For $m \ge 1$, just divide by $n$ and take the $m$-th root to see that this is exactly the generalized power means inequality (comparing the $m$-th mean with the arithmetic mean). For $m \le 0$, we do exactly the same, but note that taking the $m$-th root reverses the inequality sign. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": " I need its proof . \nLet $x_1,x_2,...x_n$ be positive real numbers and $m \\in (-\\infty,0] \\cup [1, +\\infty)$ .Prove that : \n$$ x^m_1+x^m_2+...+x^m_n \\geq \\frac{1}{n^{m-1}}.(x_1+x_2+...+x_n)^m$$",
... | Let \(x_1,x_2,\dots,x_n\) be positive real numbers and \(m\in(-\infty,0]\cup[1,\infty)\). Prove that
\[
x_1^m+x_2^m+\cdots+x_n^m \ge \frac{1}{n^{\,m-1}}\bigl(x_1+x_2+\cdots+x_n\bigr)^m.
\]
Proof.
Let \(f(t)=t^m\). For \(m\in(-\infty,0]\cup[1,\infty)\) the function \(f\) is convex on \((0,\infty)\). By Jensen's inequal... | [
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aops_174402 | $ P \equal{} \int \sin(t/2) \tan(t/2)\,dt \equal{} \int \frac {\sin^2(t/2)}{\cos(t/2)}\,dt \equal{}$
$ \equal{} \int \frac {1 \minus{} \cos^2(t/2)}{\cos(t/2)}\,dt \equal{} \int \sec(t/2)\,dt \minus{} \int \cos(t/2)\,dt$. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "I have troubles with this integral:\r\n\r\n\\[ \\int \\sin (t/2)\\tan (t/2)dt\\]\r\nCould you solve it?\r\nthanks.",
"content_html": "I have troubles with this integral:<br>\n<br>\n<img src=\"//latex... | Cleaned and reformatted problem:
Evaluate the integral
\[
\int \sin\!\left(\frac{t}{2}\right)\tan\!\left(\frac{t}{2}\right)\,dt.
\] | [
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aops_1694377 | You must be joking!
\begin{align*}
&\frac{1}{8} \left(2 \sqrt{10-2 \sqrt{5}} \tan ^{-1}\left(\frac{\sqrt{10-2 \sqrt{5}}-4 \sqrt[5]{\tan (x)}}{1+\sqrt{5}}\right)+2 \sqrt{2 \left(5+\sqrt{5}\right)} \tan
^{-1}\left(\frac{\sqrt{2 \left(5+\sqrt{5}\right)}-4 \sqrt[5]{\tan (x)}}{1-\sqrt{5}}\right)\right.\\&\left.+2 \sqrt{... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Calculate $$\\int \\sqrt[5]{\\tan x}\\, dx$$ in terms of elementary functions",
"content_html": "Calculate <img src=\"//latex.artofproblemsolving.com/c/2/3/c23baaccb9086995a36c0a309dcac310e92b826f.p... | Calculate
\[
\int \sqrt[5]{\tan x}\,dx
\]
in terms of elementary functions. | [
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aops_346394 | [quote="kunny"]Rewrite $a_n$ as $a_n=2n\sum_{k=1}^{n} \frac{1}{n^2+2kn}$
Riemman Integral waits for you.[/quote]
How can it be riemann integral if :
$\lim_{n \to \infty} \frac{b-a}{n} \sum_{k=1}^{n}f(a+\frac{k(b-a)}{n}) = \int_{a}^{b}f(x)dx $ and:
$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n}f(\frac{k}{n}) = \int_... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Calculate $ \\lim_{n \\to \\infty} a_{n} $ when:\n$ a_{n} = 2n( \\frac{1}{n^{2}+2n} + \\frac{1}{n^{2}+4n} +\\ldots +\\frac{1}{n^{2}+2n^{2} }) $ \n\n\n\nThe best comparison I've received was:\n$ \\frac{2}{3... | Calculate \(\displaystyle\lim_{n \to \infty} a_{n}\), where
\[
a_{n}=2n\left(\frac{1}{n^{2}+2n}+\frac{1}{n^{2}+4n}+\cdots+\frac{1}{n^{2}+2n^{2}}\right).
\] | [
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numina_10031314 | Solution. Rewriting the given equation in the form $y^{\prime}=(2 x+y)^{2}+2$, we notice the appropriateness of the substitution (change of variable) $2 x+y=v$. Differentiating this equality (with respect to the variable $x$), we get $2+y^{\prime}=v^{\prime}$, or $y^{\prime}=v^{\prime}-2$. The original equation takes t... | 2\operatorname{tg}(2x+C)-2x | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 4. Solve the equation $y^{\prime}-y^{2}=4 x^{2}+4 x y+2$. | [
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numina_10038854 | 27.28. According to problem $27.27 \lim _{x \rightarrow 0}(1+x)^{1 / x}=e$. From this, using the continuity of the logarithm, we obtain $\lim _{x \rightarrow 0} \frac{1}{x} \ln (1+x)=1$. | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 27.28. Prove that $\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}=1$. | [
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aops_1058885 | [quote=Spotsjoy]With particular $\alpha , \beta , \delta$ values, the following inequality holds: $sin\alpha+sin\beta+sin\delta \geq \frac{5}{2}$. Prove, that $cos\alpha + cos\beta + cos\delta \leq \frac{\sqrt{11}}{2}$.[/quote]
Because $\left(\cos\alpha + \cos\beta + \cos\delta \right)^2\leq3\left(\cos^2\alpha + \cos^2... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "With particular $\\alpha , \\beta , \\delta$ values, the following inequality holds: $sin\\alpha+sin\\beta+sin\\delta \\geq \\frac{5}{2}$. Prove, that $cos\\alpha + cos\\beta + cos\\delta \\leq \\frac{\\sq... | Let real numbers \(\alpha,\beta,\delta\) satisfy
\[
\sin\alpha+\sin\beta+\sin\delta \ge \frac{5}{2}.
\]
Prove that
\[
\cos\alpha+\cos\beta+\cos\delta \le \frac{\sqrt{11}}{2}.
\] | [
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aops_1295761 | [hide="Solution"]
Expand the square:
$\lim_{x \rightarrow 0}\frac{\cos^4(x) - 2\cos^2(x) + 1}{x}$
This is of the $\frac{0}{0}$ indeterminate form, so we use L'Hopital's Rule. The above is equivalent to
$\lim_{x \rightarrow 0}\frac{-4\cos^3(x)\sin(x) + 4\sin(x)\cos(x)}{1} = 0$ (by direct substitution).
[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$\\lim_{x \\to 0} \\frac{(1-cos^2(x))^2}{x}$",
"content_html": "<img src=\"//latex.artofproblemsolving.com/f/5/4/f5473d2db2a7871b63ef6a842a5752cf46ccaa6a.png\" class=\"latex\" alt=\"$\\lim_{x \\to 0}... | \[
\lim_{x \to 0} \frac{(1-\cos^2 x)^2}{x}
\] | [
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aops_1950987 | [hide=Solution]Let the limit be $L$.
$$L=\lim_{n\to\infty}\frac{(1^{1^2}2^{2^2}\cdots n^{n^2})^{1/n^3}}{n^{1/3}}$$
Taking the logarithm of both sides,
$$\ln L=\lim_{n\to\infty}\left(\frac1{n^3}(1^2\ln 1+2^2\ln 2+\cdots+n^2\ln n)-\frac13\ln n\right)$$
$$\ln L=\lim_{n\to\infty}\left(\frac1n\left(\left(\frac1n\right)^2\ln... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find the given limit $$\\lim_{ n\\rightarrow \\infty} \\frac{(1^{1^2} . 2^{2^2}..........n^{n^2} )^{1/n^3}}{n^{1/3}}$$",
"content_html": "Find the given limit <img src=\"//latex.artofproblemsolvi... | Find the limit
\[
\lim_{n\to\infty}\frac{\bigl(1^{1^2}\,2^{2^2}\cdots n^{n^2}\bigr)^{1/n^3}}{n^{1/3}}.
\] | [
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aops_609524 | Since $\sqrt{x^{2} +1} + x > 1$ for $x>0$ and $\sqrt{x^{2} +1} - x = \dfrac {1} {\sqrt{x^{2} +1} + x}$, it means $\left \{ \sqrt{x^{2} +1} - x \right \}^{n} = \left (\dfrac {1} {\sqrt{x^{2} +1} + x}\right )^n$.
Now, denoting $u = \sqrt{x^{2} +1} + x$, we have $du = \dfrac {\sqrt{x^2+1} + x} {\sqrt{x^2+1}} dx$; on th... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If $n > 1$, prove that\n\n$\\int_{0}^{\\infty} \\{ \\sqrt{x^{2} +1} - x \\}^{n} \\ dx= \\frac{n}{n^{2}-1}$.",
"content_html": "If <span style=\"white-space:nowrap;\"><img src=\"//latex.artofproblemso... | If \(n>1\), prove that
\[
\int_{0}^{\infty}\bigl(\sqrt{x^{2}+1}-x\bigr)^{n}\,dx=\frac{n}{n^{2}-1}.
\] | [
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aops_1063104 | [hide="sol"] By l'Hopital, $$\lim_{x\to\infty}\dfrac{\ln(1+x)}{\sqrt{x}}=\lim_{x\to\infty}\dfrac{\frac{1}{x+1}}{\frac{1}{2}x^{-1/2}}=\lim_{x\to\infty}\dfrac{2\sqrt{x}}{x+1}=0.$$ [/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find\n\n$\\lim_{x \\to \\infty} \\frac{\\ln(1+x)}{\\sqrt{x}}$.",
"content_html": "Find<br>\n<br>\n<span style=\"white-space:nowrap;\"><img src=\"//latex.artofproblemsolving.com/4/8/2/48284c5f47bea7e4... | Find
\[
\lim_{x \to \infty} \frac{\ln(1+x)}{\sqrt{x}}.
\] | [
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aops_69941 | We make the substitution $x = \tan(t/2)$. We have that $\cos t = \frac{1-x^2}{1+x^2}$ and $dt = \frac{2dx}{1+x^2}$. This turns the integral into
\[ \int_0^\infty \frac{2(1-x^2)dx}{(1+x^2)(x^2(u+1)^2 + (u-1)^2)} \]
Converting to partial fractions we get
\[ \frac{1}{u}\int_0^\infty \frac{u^2+1}{(u+1)^2x^2 + (u-1)^2} ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find\r\n${J = \\int_0^\\pi \\frac{\\cos (t) dt}{1-2u \\cos t + u^2}}$ \r\n $u \\in ]-1;1[$",
"content_html": "Find<br>\n<img src=\"//latex.artofproblemsolving.com/8/7/b/87b3eb96c55902f88b58... | Find
\[
J=\int_{0}^{\pi}\frac{\cos t\,dt}{1-2u\cos t+u^{2}},\qquad u\in(-1,1).
\] | [
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aops_121256 | $\int_{0}^{1}\frac{ln(cos(\frac{\pi x}{2}))}{(x)(x+1)}dx = \int_{0}^{1}\frac{ln(cos(\frac{\pi (1-x)}{2}))}{(x)(x+1)}dx = \int_{0}^{1}\frac{ln(sin(\frac{\pi x}{2}))}{(x)(x+1)}dx$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate \\[\\int_{0}^{1}\\frac{\\log (\\cos \\frac{\\pi x}{2})}{x(x+1)}\\, dx \\]",
"content_html": "Evaluate <img src=\"//latex.artofproblemsolving.com/e/3/a/e3afe83c9b4ecc3c1abc2f2b68f924f8c2c224e... | Evaluate
\[
\int_{0}^{1}\frac{\log\!\left(\cos\frac{\pi x}{2}\right)}{x(x+1)}\,dx.
\] | [
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ours_32665 | To solve the integral \(\int \frac{2x - 9\sqrt{x} + 9}{(x - 3\sqrt{x})^{1/3}} \, \mathrm{dx}\), we use the substitution \(u = \sqrt{x}\), which implies \(x = u^2\) and \(\mathrm{dx} = 2u \, \mathrm{du}\).
Substituting these into the integral, we have:
\[
\int \frac{2x - 9\sqrt{x} + 9}{(x - 3\sqrt{x})^{1/3}} \, \mathr... | \frac{6}{5} (x - 3\sqrt{x})^{5/3} + C | {
"competition": "cmm",
"dataset": "Ours",
"posts": null,
"source": "CMM_2025_Integration_Bee_Hard_Sols.md"
} | Evaluate the integral: \(\int \frac{2x - 9\sqrt{x} + 9}{(x - 3\sqrt{x})^{1/3}} \, \mathrm{dx}\). | [
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aops_2171399 | [quote=jjagmath]
[quote=TuZo][b]No l'Hospital needed: [/b]
\[L=\underset{x\to \alpha }{\mathop{\lim }}\,\frac{\sqrt{1-\cos (f(x))}}{x-\alpha }=\sqrt{2}\underset{x\to \alpha }{\mathop{\lim }}\,\left| \frac{\sin \frac{f(x)}{2}}{\frac{f(x)}{2}} \right|\underset{x\to \alpha }{\mathop{\lim }}\,\frac{\left| \frac{f(x)}{2} \... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $f(x)=ax^2+bx+c$. We have $f(\\alpha)=f(\\beta)=0$. Find the value of\n$$L = \\lim_{x\\rightarrow\\alpha}\\frac{\\sqrt{1-\\cos(f(x))}}{x-\\alpha}.$$\n\n[hide=My Work]I used the fact that $1-\\cos(2\\th... | Let \(f(x)=ax^2+bx+c\). Suppose \(f(\alpha)=f(\beta)=0\). Find
\[
L=\lim_{x\to\alpha}\frac{\sqrt{1-\cos(f(x))}}{x-\alpha}.
\] | [
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... | null |
aops_2091295 | [hide=Contrived solution]
Doing the Weierstrass substitution in reverse, we have
$$\frac{1-x^2}{1+x^2}=\cos(\theta),\qquad\frac{2x}{1+x^2}=\sin(\theta),\qquad\frac{2}{1+x^2}\,dx=d\theta.$$
Therefore, the given integral is equivalent to
$$\int_{0}^{\frac{\pi}{2}}\frac{1}{32}\sin^4(\theta)\cos^5(\theta)\,d\theta=\int_{0}... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate\n$$\n\\int_{0}^{1}{\\frac{x^4(1-x^2)^5}{(1+x^2)^{10}}}dx\n$$\n\nThe answer by WA is [hide=this]$\\frac{1}{1260}$[/hide], but the trig sub solution I initially attempted ended up too messy.",
... | Evaluate
\[
\int_{0}^{1}\frac{x^{4}(1-x^{2})^{5}}{(1+x^{2})^{10}}\,dx.
\] | [
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aops_173281 | [quote="Vinxenz"]Prove that
\[ I\equiv\int_{ \minus{} \infty}^{\infty}\frac {e^{\frac x3}}{1 \plus{} e^x}\ \mathrm {dx} \equal{} \frac {2\pi}{\sqrt {3}} \ .
\]
[/quote]
$ \left\|\begin{array}{c} x: \equal{} 3\ln x \\
\\
\mathrm {dx}: \equal{} \frac 3x\cdot \mathrm {dx}\end{array}\right\|$ $ \implies$ $ I \equal{} 3... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Prove:\r\n\\[ \\int_{\\minus{}\\infty}^{\\plus{}\\infty}\\frac{e^{x/3}}{1\\plus{}e^x}dx\\equal{}\\frac{2\\pi}{\\sqrt{3}}\\]",
"content_html": "Prove:<br>\n<img src=\"//latex.artofproblemsolving.com/6... | Prove
\[
\int_{-\infty}^{\infty}\frac{e^{x/3}}{1+e^{x}}\,dx=\frac{2\pi}{\sqrt{3}}.
\] | [
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aops_3481828 | [hide=Good.Thanks.][quote=User21837561]let $f(x)=\frac{1}{e^x-1}$
By Jensen:
$\sum_{cyc}^{}\frac{1}{2a-1}=\sum_{cyc}^{}f(ln(a)+ln(2))\ge 3f(\frac{1}{3}ln(abc)+ln(2))\ge 3f(3ln(2))=\frac{3}{7}$
$\sum_{cyc}^{}\frac{1}{4a-1}=\sum_{cyc}^{}f(ln(a)+2ln(2))\ge 3f(\frac{1}{3}ln(abc)+2ln(2))\ge 3f(4ln(2))=\frac{1}{5}$
$\sum_{cy... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $ a,b,c> 1 $ and $ abc\\le 64. $ Prove that\n$$ \\frac{1}{2a-1}+\\frac{1}{2b-1}+\\frac{1}{2c-1} \\geq \\frac{3}{7} $$\n$$ \\frac{1}{4a-1}+\\frac{1}{4b-1}+\\frac{1}{4c-1}\\geq \\frac{1}{5} $$\n$$\\frac... | Let \(a,b,c>1\) and \(abc\le 64\). Prove that
\[
\frac{1}{2a-1}+\frac{1}{2b-1}+\frac{1}{2c-1}\ge\frac{3}{7},
\]
\[
\frac{1}{4a-1}+\frac{1}{4b-1}+\frac{1}{4c-1}\ge\frac{1}{5},
\]
\[
\frac{1}{8a-1}+\frac{1}{8b-1}+\frac{1}{8c-1}\ge\frac{3}{31},
\]
\[
\frac{1}{a-1}+\frac{1}{2b-1}+\frac{1}{c-1}\ge\frac{3\bigl(1+4\sqrt[3]{2}... | [
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0.5320094560389104,
0.47518... | null |
aops_2678075 | With $x=t^3$ we find $\int {{t^2 \cdot 3t^2 dt}\over {t^3+1}}=3\int (t-{t\over {t^3+1}})dt=3\int (t+{{1/3}\over {t+1}}-{{1/3(t+1)}\over {t^2-t+1}})dt=3\int (t+{{1/3}\over {t+1}}-{{1/6(2t-1)}\over {t^2-t+1}}-{{1/2}\over {t^2-t+1}})dt$.
For the final term we can use $(t-1/2)^2+3/4$ and we find a version of the $\arctan$ ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate $$\\int\\frac{x^\\frac{2}{3}}{1+x} dx$$",
"content_html": "Evaluate <img src=\"//latex.artofproblemsolving.com/5/d/4/5d40f6701a7899c41aff6f717df11255c40288d9.png\" class=\"latexcenter\" alt=... | Evaluate
\[
\int \frac{x^{2/3}}{1+x}\,dx.
\] | [
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0.41076... | null |
aops_2180765 | [hide="Sol"]
$$\int arccos\sqrt{\frac{4}{4+x}} dx$$
Integration by parts: $$x\arccos \left(\frac{2}{\sqrt{4+x}}\right)-\int \frac{\sqrt{x}}{x+4}dx$$
$$x\arccos \left(\frac{2}{\sqrt{4+x}}\right)-4\left(-\arctan \left(\frac{1}{2}\sqrt{x}\right)+\frac{1}{2}\sqrt{x}\right)$$
$$x\arccos \left(\frac{2}{\sqrt{4+x}}\right)-4\l... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Please,solve the following integral:\narccos[sqrt{4/(4+x)}].Thank you!",
"content_html": "Please,solve the following integral:<br>\narccos[sqrt{4/(4+x)}].Thank you!",
"post_id": 16309880,
... | Cleaned problem statement (LaTeX):
Evaluate the integral
\[
\int \arccos\!\left(\sqrt{\frac{4}{4+x}}\right)\,dx.
\] | [
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ours_32625 | Solution: \(\arctan \left(e^{x}+1\right)+C\)
First, multiply the integrand by \(1\) in the form of \(\frac{e^{x}}{e^{x}}\):
\[
\int \frac{1}{2+e^{x}+2 e^{-x}} \, \mathrm{d}x = \int \frac{e^{x}}{e^{2x}+2e^{x}+2} \, \mathrm{d}x
\]
The denominator can be rewritten as a sum of squares:
\[
\int \frac{e^{x}}{e^{2x}+2e^{x... | null | {
"competition": "cmm",
"dataset": "Ours",
"posts": null,
"source": "CMM24_Integration_Bee_Solutions.md"
} | Evaluate the integral: \(\int \frac{1}{2+e^{x}+2 e^{-x}} \, \mathrm{d}x\). | [
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aops_118742 | The limit is $\frac{(\ln 2)^{2}}{2}$.
Let $a_{n}$ be the sequence in question. Then
$\frac{1}{n+i}<\ln\frac{n+i}{n+i-1}<\frac{1}{n+i-1}$
It follows that
$\sum_{i=1}^{n}\frac{1}{n+i}\ln \frac{n+i}{n}<a_{n}<\sum_{i=1}^{n}\ln \frac{n+i}{n}\frac{1}{n+i-1}<\frac{n+1}{n}\sum_{i=1}^{n}\ln \frac{n+i}{n}\frac{1}{n}\frac{1... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$\\lim_{n \\to \\infty}{\\sum_{i=1}^{n}ln{\\frac{n+i}{n}}ln{\\frac{n+i}{n+i-1}}}$",
"content_html": "<img src=\"//latex.artofproblemsolving.com/9/2/b/92bf4d09ccd8fefbf52f294f47a8cf05543132a1.png\" cl... | \[
\lim_{n \to \infty} \sum_{i=1}^{n} \ln\!\left(\frac{n+i}{n}\right)\ln\!\left(\frac{n+i}{n+i-1}\right)
\] | [
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aops_2225230 | Notice that the integral is equal to $\Gamma(\frac{3}{2})$. By Legendre's duplication formula :
$\Gamma(1)\Gamma(\frac{3}{2})=2^{1-2}\sqrt{\pi}\Gamma(2)$
And it gives us that $\Gamma(\frac{3}{2})=\frac{\sqrt{\pi}}{2}$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate $\\int_{0}^{\\infty}{e^{-x}\\sqrt{x}~dx}$",
"content_html": "Evaluate <img src=\"//latex.artofproblemsolving.com/0/e/8/0e88231587a2527ca25c2d75d4726fd3d62a1575.png\" class=\"latex\" alt=\"$\... | Evaluate \(\displaystyle \int_{0}^{\infty} e^{-x}\sqrt{x}\,dx.\) | [
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aops_1987246 | $D=\int_{0}^{\pi}\frac{x^2(1-\cos2x)}{2}dx=\int_{0}^{\pi}\frac{x^2}{2}dx-\frac{1}{2}\int_{0}^{\pi}x^2\cos2xdx$
$D=\frac{\pi^3}{6}-\frac{1}{2}[\frac{1}{2}x^2\sin2x|_0^{\pi}-\int_0^{\pi}x\sin2xdx]$
$D=\frac{\pi^3}{6}-\frac{1}{2}[0-(\frac{-1}{2}x\cos2x|_0^{\pi}+\frac{1}{2}\int_0^{\pi}\cos2xdx)]$
$D=\frac{\pi^3}{6}-\frac{1... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "calculate $D=\\int_{0}^{\\pi} (xsinx)^2dx$",
"content_html": "calculate <img src=\"//latex.artofproblemsolving.com/e/d/1/ed1d60f22d1b7992a2d2b8e9b29bfba7046d5d7f.png\" class=\"latex\" alt=\"$D=\\int_... | Calculate
\[
D=\int_{0}^{\pi} (x\sin x)^2\,dx.
\] | [
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aops_406113 | [quote="raoofmirzaei"]please solve this integral:
$ \int\frac{sin^{2}x}{\sqrt{(1+sin^{2}x)}}dx $[/quote]
$\int\frac{sin^{2}x}{\sqrt{(1+sin^{2}x)}}dx $
$=\int\frac{\left (sin^{2}x+1 \right )-1}{\sqrt{(1+sin^{2}x)}}dx $
$=\int\sqrt{1+sin^{2}x}dx -\int\frac{dx}{\sqrt{1+sin^{2}x}}$
There are two types of integrals: The in... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "please solve this integral:\n$ \\int\\frac{sin^{2}x}{\\sqrt{(1+sin^{2}x)}}dx $",
"content_html": "please solve this integral:<br>\n<img src=\"//latex.artofproblemsolving.com/c/4/d/c4db3783caeb02190f6... | Cleaned and reformatted problem (contents unchanged):
Evaluate the integral
\[
\int \frac{\sin^{2}x}{\sqrt{1+\sin^{2}x}}\,dx.
\] | [
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aops_610669 | [quote="lambert"]let $ f: \mathbb{R} - \{0\} \to \mathbb{R} $ defined by $ f(x)=\dfrac{1}{1+ a^{\dfrac{1}{x}}}$ , where $ a>1$ prove that $\lim_{x\to\ 0 ^{+} } f(x) =0 $ and $ \lim_{x\to\ 0^{-}} f(x)=1$[/quote]
Good evening,
$\bullet$ If $x\to 0^+$ then since $a>1$ we have: $\lim_{x\rightarrow 0^+} a^{1/x}=+\infty... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "let $ f: \\mathbb{R} - \\{0\\} \\to \\mathbb{R} $ defined by $ f(x)=\\dfrac{1}{1+ a^{\\dfrac{1}{x}}}$ , where $ a>1$ prove that $\\lim_{x\\to\\ 0 ^{+} } f(x) =0 $ and $ \\lim_{x\\to\\ 0^{-}} f(x)=1$",
... | Let \(f:\mathbb{R}\setminus\{0\}\to\mathbb{R}\) be defined by
\[
f(x)=\frac{1}{1+a^{1/x}},
\]
where \(a>1\). Prove that
\[
\lim_{x\to 0^+} f(x)=0\qquad\text{and}\qquad \lim_{x\to 0^-} f(x)=1.
\] | [
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numina_10030319 | ## Solution.
$$
\begin{gathered}
V=\pi \int_{0}^{\pi} y^{2} d x=\pi \int_{0}^{\pi} \sin ^{2} x d x=\pi \int_{0}^{\pi} \frac{1-\cos 2 x}{2} d x=\frac{\pi}{2} \int_{0}^{\pi}(1-\cos 2 x) d x= \\
=\frac{\pi}{2} x-\left.\frac{1}{2} \sin 2 x\right|_{0} ^{\pi}=\frac{\pi^{2}}{2} .
\end{gathered}
$$ | \frac{\pi^{2}}{2} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 3.18. Calculate the volume of the body formed by rotating the figure bounded by one half-wave of the sine curve $y=\sin x$ around the $O x$ axis. | [
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aops_396246 | $\int_{-1}^1 (1-x^2)^n \,dx = 2 \int_0^1 (1-x^2)^n \,dx = 2 \int_0^{\pi /2} ( \cos t)^{2n+1}\,dt$. Set $I_n = \int_0^{\pi /2} ( \cos t)^{2n+1}\,dt$. Then integrating by parts we get the recurrence relation
$I_n = \frac{2n}{2n+1} I_{n-1}$,
with $I_0 = 1$. Therefore,
$I_n = \prod_{k=1}^n \frac{2k}{2k+1} = \prod_{k=1}^... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Show that:\n\n$\\int_{-1}^{1}(1-x^{2})^{n}dx=\\frac{2^{2n+1}(n!)^{2}}{(2n+1)!}$\n\nwhere n is a positive integer.",
"content_html": "Show that:<br>\n<br>\n<img src=\"//latex.artofproblemsolving.com/e... | Show that
\[
\int_{-1}^{1}(1-x^{2})^{n}\,dx=\frac{2^{2n+1}(n!)^{2}}{(2n+1)!},
\]
where \(n\) is a positive integer. | [
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aops_194073 | [hide="Number 1"]$ P \equal{} \int \frac{x}{(7x \minus{} 10 \minus{} x^2)^{3/2}}\,dx \equal{} \int \frac{x}{\left[\frac{9}{4} \minus{} \left(x \minus{} \frac{7}{2}\right)^2\right]^{3/2}}\,dx$,
and now we are to make the substitution $ x \minus{} \frac{7}{2} \equal{} \frac{3}{2} \sin t$: the result is
$ P \equal{} \in... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Solve these without using Euler Substitution\r\n\r\n1. $ \\int \\frac {xdx}{({\\sqrt {7x \\minus{} 10 \\minus{} x^2})}^3} dx$\r\n\r\n2. $ \\int \\frac {dx}{x \\plus{} \\sqrt {x^2 \\minus{} x \\plus{} 1}} d... | Cleaned and reformatted problem (no comments, indices, or hints; content unchanged):
Solve these without using Euler substitution.
1. \[
\int \frac{x\,dx}{\bigl(\sqrt{7x-10-x^{2}}\bigr)^{3}}
\]
2. \[
\int \frac{dx}{x+\sqrt{x^{2}-x+1}}
\] | [
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aops_280832 | Consider I = $ \int \frac {x^{x}\left( x^{2x} \plus{} 1\right) \left( \ln x \plus{} 1\right) }{x^{4x} \plus{} 1}dx$
$ x^x \equal{} y \ \Rightarrow \ x^x(lnx\plus{}1)dx \equal{} dy$
I = $ \int \frac{y^2\plus{}1}{y^4\plus{}1} dy \equal{} \int \frac{1\plus{}\frac{1}{y^2}}{y^2\plus{}\frac{1}{y^2}}dy$
= $ \int \... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate $ \\int_{0^{\\plus{}}}^{1}\\frac{x^{x}\\left( x^{2x}\\plus{}1\\right) \\left( \\ln x\\plus{}1\\right) }{x^{4x}\\plus{}1}dx$\r\n\t\r\nAnswer [hide]$ \\equal{}0$[/hide]",
"content_html": "Eval... | Evaluate
\[
\int_{0^{+}}^{1}\frac{x^{x}\bigl(x^{2x}+1\bigr)\bigl(\ln x+1\bigr)}{x^{4x}+1}\,dx.
\] | [
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ours_31035 | Solution: We start by simplifying the integrand. Notice that for any \( x > 0 \), the identity \(\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}\) holds. This is because:
\[
\tan^{-1} x + \cot^{-1} x = \tan^{-1} x + \left(\frac{\pi}{2} - \tan^{-1} x\right) = \frac{\pi}{2}
\]
Thus, the integrand \(\tan^{-1} x + \cot^{-1} x\... | \frac{\pi}{2} | {
"competition": "bmt",
"dataset": "Ours",
"posts": null,
"source": "CASP18TBA.md"
} | Evaluate the integral:
\[
\int_{1}^{2} \left(\tan^{-1} x + \cot^{-1} x\right) \, dx
\] | [
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0.... | null |
aops_180619 | A way of thinking:
$ \int\frac{dx}{1\plus{}x^2}\equal{}\arctan x\plus{}C.$
Based on that, $ \int\frac{dx}{a^2\plus{}x^2}$ should involve an arctangent.
Treat this as a physical problem. Assume that $ x$ and $ a$ are both measured with the same units of measurement - maybe they're both distances. General princi... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find th integral of dx / (9+x^2).\r\n\r\na) 3arctan(x/3) +c\r\nb) 1/3arctan(x/3) + c\r\nc) 1/9arctan(x/3) + c\r\nd) 1/3arctan(x) + c\r\ne) 1/9arctan(x) + c",
"content_html": "Find th integral of dx /... | Find the integral
\[
\int \frac{dx}{9+x^2}.
\]
Choices:
a) \(3\arctan\!\left(\frac{x}{3}\right)+C\)
b) \(\tfrac{1}{3}\arctan\!\left(\frac{x}{3}\right)+C\)
c) \(\tfrac{1}{9}\arctan\!\left(\frac{x}{3}\right)+C\)
d) \(\tfrac{1}{3}\arctan(x)+C\)
e) \(\tfrac{1}{9}\arctan(x)+C\) | [
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aops_1628862 | [quote=Yimself]the standard substitution $tan(x/2)=u$ after splitting the integral from 0 to pi and from pi to 2pi (because tangent is not continous there) Will this work for you?[/quote]
Yes but first write $a\sin(u)+b\cos(u)=\sqrt{a^2+b^2}\sin(u+\sin^{-1}(\frac{b}{\sqrt{a^2+b^2}}))$
and peform the substiution $v=u+... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Can you help me out with this? How to find it? $\\int_{0}^{2\\pi }\\frac{sinu}{acosu+bsinu+c}du$",
"content_html": "Can you help me out with this? How to find it? <img src=\"//latex.artofproblemsolvi... | Compute the integral
\[
\int_{0}^{2\pi} \frac{\sin u}{a\cos u + b\sin u + c}\,du.
\] | [
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aops_1990199 | Both methods work, I think here I prefer #3.
- If $x=\cosh t$ then $x^2-1=(\sin ht)^2$ and we get $\int {{\sinh t dt}\over {\sinh t+1}}=t-\int{{ 2d(e^t)}\over {e^{2t}+2e^t-1}}$.
- If $x={1\over {\sin t}}$ then we get $\int_{\pi/4}^{\pi/2} {{\cos t dt}\over {\sin t(\sin t+\cos t)}}=\int_{\pi/4}^{\pi/2} ({1\over {\sin t... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate:\n$$\\int_{1}^{\\sqrt{2}} \\dfrac{1}{1+\\sqrt{x^2-1}} dx$$\nIs it possible to use trigonometric substitution? I tried substituting $x=\\sec\\theta$ but failed halfway......What is the fastest way?... | Evaluate:
\[
\int_{1}^{\sqrt{2}} \frac{1}{1+\sqrt{x^{2}-1}}\,dx.
\]
Is it possible to use trigonometric substitution? If so, finish it using the substitution \(x=\sec\theta\). | [
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aops_221033 | $ \lim_{n\to\infty}\frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}2}\plus{}\cdots\plus{}\frac{1}{kn}$
$ \equal{}\lim_{n\to\infty}\frac{1}{n}\cdot\frac{n}{n\plus{}1}\plus{}\frac{1}{n}\cdot\frac{n}{n\plus{}2}\plus{}\cdots\plus{}\frac{1}{n}\cdot\frac{1}{k}$
$ \equal{}\int_1^k\frac{1}{x}dx$
$ \equal{}\ln k$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How do I prove that:\r\n$ \\lim_{n\\to\\infty} \\frac{1}{n\\plus{}1} \\plus{} \\cdots \\plus{} \\frac{1}{kn} \\equal{} ln(k)$\r\n\r\n(I think it is true :P)\r\nI've tried something with $ f(x) \\equal{} \\... | Prove that
\[
\lim_{n\to\infty}\sum_{i=n+1}^{kn}\frac{1}{i}=\ln k.
\] | [
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aops_3447203 | [quote name="MetaphysicalWukong" url="/community/p33254533"]
why is the last step after the implication true? ^^
[/quote]
the substitution $u=x-1$ is used, which gives $\int_{-1}^0 f(u^2-1)\,du=\int_0^1 f(u^2-1)\,du$, and then you finish as @2above said | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $f \\in C(\\mathbb R)$. Is it true that $2\\times \\int_0^1 f(x^2-2x)=\\int_0^2 f(x^2-2x)$ ?",
"content_html": "Let <span style=\"white-space:pre;\"><img src=\"//latex.artofproblemsolving.com/d/2... | Let \(f\in C(\mathbb{R})\). Is it true that
\[
2\int_0^1 f(x^2-2x)\,dx=\int_0^2 f(x^2-2x)\,dx\;?
\] | [
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aops_3532183 | There is an elementary way to find integrals like that but a faster way is to do it using properties of the digamma function. Let
$$I_n:=\int_0^{\infty}\frac{dx}{1+x+ \cdots + x^n}, \ \ \ \ \ n \ge 2.$$
Then the substitution $x=t^r, \ r=\frac{1}{n+1}$ gives
$$I_n=\int_0^{\infty}\frac{1-x}{1-x^{n+1}} \ dx=r\int_0^{\inft... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Determine the exact value of $$\\int_{0}^{\\infty} \\frac{1}{\\sum_{n=0}^{10} x^n} \\,dx$$",
"content_html": "Determine the exact value of <img src=\"//latex.artofproblemsolving.com/8/d/7/8d75dd8942b... | Determine the exact value of
\[
\int_{0}^{\infty} \frac{1}{\sum_{n=0}^{10} x^n}\,dx.
\] | [
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0.721156... | null |
aops_453378 | thanks for your hints, but it can't be zero..
how do i should count this line integral:
$\int_S |y|ds$
where S is the line like Lemniscate of Bernoulli given by $(x^2+y^2)^2=a^2(x^2-y^2)$?
I know the parametric equation:
$x=\frac{a \cos{t}}{1+\sin^2{t}}$ and $y=\frac{a \sin{t} \cos{t}}{1+\sin^2{t}}$
thus
$x'_t=a \cdot... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find the value of\n$\\int_{0}^{2\\pi} \\frac{\\sin{t} \\cos{t}}{(1+\\sin^2{t})^{\\frac{3}{2}} } dt$\nThanks for help.",
"content_html": "Find the value of<br>\n<img src=\"//latex.artofproblemsolving.... | Find the value of
\[
\int_{0}^{2\pi} \frac{\sin t\cos t}{\bigl(1+\sin^{2}t\bigr)^{3/2}}\,dt.
\] | [
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aops_124679 | Ok. If no one else will do it I will show you how to do it.
Note that $\left(1+\frac{1}{n}\right)^{n+\frac12}= e \cdot e^{\left(n+\frac{1}{2}\right)\ln\left(1+\frac{1}{n}\right)-1}$
We bring out the $e$ so that the thing remaining in the exponent tends to 0 for big $n$, so that we can taylor expand $\exp$ around... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How can this limit be solved\r\n $\\lim_{n\\to \\infty}n^{2}\\; ((1+\\frac{1}{n})^{n+\\frac{1}{2}}-e)$ ?",
"content_html": "How can this limit be solved<br>\n<img src=\"//latex.artofproblemsolving.co... | Find the value of the limit
\[
\lim_{n\to\infty} n^{2}\Bigl(\Bigl(1+\frac{1}{n}\Bigr)^{n+\tfrac{1}{2}}-e\Bigr).
\] | [
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0.415033... | null |
aops_2933208 | 1) We have $\underset{x\to 0}{\mathop{\lim }}\,{{\left( 1+3x \right)}^{\frac{1}{x}}}={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\frac{3x}{x}}}={{e}^{3}}$
2) $\underset{x\to \infty }{\mathop{\lim }}\,x{{\sin }^{2}}\frac{1}{x}=\underset{x\to \infty }{\mathop{\lim }}\,x\frac{{{\sin }^{2}}\frac{1}{x}}{{{x}^{2}}\frac{1}{{{x}... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Compute $\\lim_{x\\to 0^+} (1+3x)^{1\\over x}$ and $\\lim_{x\\to \\infty}x\\sin^2{1 \\over x}$ without using L'Hopital's rule. ",
"content_html": "Compute <img src=\"//latex.artofproblemsolving.com/e... | Compute the following limits without using L'Hôpital's rule:
1. \(\displaystyle \lim_{x\to 0^+} (1+3x)^{1/x}\).
2. \(\displaystyle \lim_{x\to\infty} x\sin^2\!\left(\frac{1}{x}\right).\) | [
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aops_1239663 | Here is the direct output from Mathematica, you should be able to figure out how to get to the final answer by manipulating and substituting on the integrand.
$$\frac{\sqrt{2} \cos (x) \sqrt{\tan ^2(x)-2016} \left(\sqrt{2017} \sin ^{-1}\left(\frac{1}{12} \sqrt{\frac{2017}{14}} \sin (x)\right)-\tan ^{-1}\left(\frac{\sq... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluation of $\\int \\sqrt{\\tan^2 x-2016}dx$",
"content_html": "Evaluation of <img src=\"//latex.artofproblemsolving.com/2/f/c/2fc4b66c19b16a87c0f299df0d65b0395d0d4d07.png\" class=\"latex\" alt=\"$... | Evaluate the integral
\[
\int \sqrt{\tan^2 x - 2016}\,dx.
\] | [
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0.388667... | null |
aops_1522204 | $$ \int \ln(1+x^2)\arctan{x}dx $$
$$ x = \tan t $$
$$ 1 + \tan^2 t = \frac{1}{\cos^2 t} $$
$$ -2 \int \frac{t \cdot \ln \cos t}{\cos^2 t}dt = -2\ln \cos t \cdot \int \frac{t }{\cos^2 t}dt -2 \int \tan t \left ( \int \frac{\tau }{\cos^2 \tau}d\tau \right ) dt $$
$$ \int \frac{t }{\cos^2 t}dt = t \tan t - \int \tan t... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Compute: $\\int_{0}^{1}\\ln(1+x^2)\\arctan{x}dx$",
"content_html": "Compute: <img src=\"//latex.artofproblemsolving.com/6/4/3/643333d61a18940e95bfae760323b9bbaa1bcf49.png\" class=\"latex\" alt=\"$\\i... | Compute:
\[
\int_{0}^{1} \ln\bigl(1+x^{2}\bigr)\arctan x\,dx.
\] | [
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0.45050... | null |
aops_239149 | \[ \left. {\left. {\int\limits_1^6 {\frac{2}
{{\sqrt {x \plus{} 3} }}dx \equal{} } 2\int\limits_1^6 {\frac{{d\left( {x \plus{} 3} \right)}}
{{\sqrt {x \plus{} 3} }}} \equal{} 2\int\limits_1^6 {\left( {x \plus{} 3} \right)^{ \minus{} \frac{1}
{2}} d\left( {x \plus{} 3} \right)} \equal{} 2\left( {2\left( {x \plus{} ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Compute $ \\displaystyle{\\int\\limits_1^6 {\\frac{2}{\\sqrt{x\\plus{}3}}}dx}$",
"content_html": "Compute <img src=\"//latex.artofproblemsolving.com/d/8/5/d85c736b83fd75a2bd6a4025dbad4180b66e18ff.png... | Compute
\[
\int_{1}^{6} \frac{2}{\sqrt{x+3}}\,dx.
\] | [
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aops_289678 | $ \equal{}\int_0^1 \frac{1}{\sqrt{t\plus{}t^2}} \, dt\equal{}2 \text{ArcSinh}[1]$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Compute\r\n\\[ \\int_1^e\\frac {dx}{\\sqrt {x^2\\ln x \\plus{} (x\\ln x)^2}}.\\]",
"content_html": "Compute<br>\n<img src=\"//latex.artofproblemsolving.com/e/0/c/e0ce8aad2597bd9493d324f62c3e08b64ca9b... | Compute
\[
\int_1^e \frac{dx}{\sqrt{x^2\ln x + (x\ln x)^2}}.
\] | [
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numina_10032749 | Solution. Based on the geometric meaning of the definite integral, we conclude that the desired area is expressed by the integral
$$
S=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x d x
$$
Evaluating this integral, we get
$$
S=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x d x=\left.\sin x\right|_{-\frac{\pi}{2}} ^{\fra... | 2 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 8. Determine the area bounded by the arc of the cosine curve from $x=-\frac{\pi}{2}$ to $x=\frac{\pi}{2}$ and the $O x$ axis. | [
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aops_198396 | 1)
$ I_{n \plus{} 1} \equal{} \int_0^1 x^{n \plus{} 1} \sqrt {1 \minus{} x} \, \mbox{d}x$
By integration by parts we get
$ u \equal{} x^{n \plus{} 1} , \quad dv \equal{} \sqrt {1 \minus{} x} dx$
$ I_{n \plus{} 1} \equal{} \left[ x^{n \plus{} 1} \cdot \left( \minus{} \frac {2}{3} \right) (1 \minus{} x)^{3/2} \right]... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "[b]1)[/b] Let $ I_n\\equal{}\\int_{0}^{1} x^n\\sqrt{1\\minus{}x}dx$\r\nProve that \r\n$ I_{n\\plus{}1}\\equal{}\\frac{2n\\plus{}2}{2n\\plus{}5}I_n$\r\n[b]2)[/b] Let $ f: R\\to [0,1]$ be a continuous functi... | 1) Let
\[I_n=\int_{0}^{1} x^n\sqrt{1-x}\,dx.\]
Prove that
\[I_{n+1}=\frac{2n+2}{2n+5}\,I_n.\]
2) Let \(f:\mathbb{R}\to[0,1]\) be a continuous function. Prove that
\[\int_0^{\pi} x\,f(\sin x)\,dx=\frac{\pi}{2}\int_0^{\pi} f(\sin x)\,dx.\] | [
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aops_174471 | [quote="Carcul"]Compute $ \int \frac {x^{17}}{x^{24} \plus{} 1}\ \mathrm {dx}$.[/quote]
[b][u]An easy extension.[/u][/b] For $ \alpha\in \mathcal R^*_\plus{}$ , $ \int\frac {x^{3\alpha \minus{}1}}{x^{4\alpha}\plus{}1}\ \mathrm {dx}\equal{}$ $ \frac {1}{\alpha}\cdot\int\frac {\left(x^{\alpha}\right)^2}{\left(x^{\alpha}... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Compute $ \\int \\frac{x^{17}}{x^{24} \\plus{} 1}\\,dx$.",
"content_html": "Compute <span style=\"white-space:nowrap;\"><img src=\"//latex.artofproblemsolving.com/2/6/6/266e9beea39826f27ae71853b92fd5... | Compute \(\displaystyle \int \frac{x^{17}}{x^{24} + 1}\,dx.\) | [
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aops_545342 | Do the sub : $x\mapsto -t$ to get : \[I= \int\limits_{\frac{-\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \frac{\left(\arcsin x\right)^2}{1+2^x}\ \mathrm{d}x= \int\limits_{\frac{-\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \frac{2^t \left(\arcsin t\right)^2}{1+2^t}\ \mathrm{d}t.\] Since $I$ is finite, we get :\[I= \frac{1}{2} \int\limits... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Calculate: $\\int_{-\\frac{\\sqrt{2}}{2}}^{\\frac{\\sqrt{2}}{2}}\\frac{(\\arcsin x)^2}{1+2^x}dx$.",
"content_html": "Calculate: <span style=\"white-space:nowrap;\"><img src=\"//latex.artofproblemsolv... | Calculate:
\[
\int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \frac{(\arcsin x)^2}{1+2^x}\,dx.
\] | [
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aops_562275 | [quote="PhamKhacLinh"]Find the limit of the sequence:
a, $a_n = \frac{1}{1^3} + \frac{1}{2^3} +...+ \frac{1}{n^3}$
b, $b_n = \frac{1}{1^2} + \frac{1}{2^2} +...+ \frac{1}{n^2}$
( this sequence was solved by Euler , but i can't find this solution of this, may you help me? )
Extension : Can solve with k power ? (k is any ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find the limit of the sequence:\na, $a_n = \\frac{1}{1^3} + \\frac{1}{2^3} +...+ \\frac{1}{n^3}$\nb, $b_n = \\frac{1}{1^2} + \\frac{1}{2^2} +...+ \\frac{1}{n^2}$\n( this sequence was solved by Euler , but ... | Find the limits of the sequences:
a) \(a_n=\displaystyle\sum_{m=1}^n \frac{1}{m^3}\).
b) \(b_n=\displaystyle\sum_{m=1}^n \frac{1}{m^2}\).
Extension: determine the limit of \( \displaystyle\sum_{m=1}^n \frac{1}{m^k}\) for an integer \(k\). | [
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aops_2070154 | let $xg(x)=f(x)$ so we have $g(x)=g(x^x)$ and $g$ is continuous now we let $x^x=y$ and call $x=T(y)$ so $T(y)^{T(y)} =y$ for sure if we have $y>1$ then we have $y>T(y)$ bu doing this infinity times $T(T(T(...(T(Y))...)))$ leads to $1$ since $g$ is continuous we have $g(x)=g(1)$ for $x>1$ and if $x<1$ its pretty much th... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find all continuous functions $f : \\mathbb{R}^+\\to\\mathbb{R}$\n$f(x)=x^{x-1}f(x^x)$\nfor all positive number x",
"content_html": "Find all continuous functions <img src=\"//latex.artofproblemsolvi... | Find all continuous functions \(f:\mathbb{R}^+\to\mathbb{R}\) such that
\[
f(x)=x^{x-1}f(x^x)\quad\text{for all }x>0.
\] | [
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aops_1081822 | (b) $\int_{0}^{\pi} \frac{1}{5 + 3 \cos ^{2} x}\ dx=\int_0^\pi\dfrac{1+tan^2x}{8+5tan^2x}dx=2\int_0^{+\infty}\dfrac{du}{8+5u^2}=\frac{1}{\sqrt{10}}\Biggl[arctan\Bigl(u\sqrt{\frac{5}{8}}\Bigr)\Biggr]_0^{+\infty}=\dfrac{\pi}{2\sqrt{10}}$. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "(a) Evaluate $\\int_{0}^{a} \\frac{1}{5 + 3 \\cos x}\\ dx$ when (i) $a= \\pi$ and (ii) when $a = 2 \\pi$.\n\n(b) Evaluate $\\int_{0}^{\\pi} \\frac{1}{5 + 3 \\cos ^{2} x}\\ dx$.",
"content_html": "(a)... | (a) Evaluate \(\displaystyle \int_{0}^{a} \frac{1}{5 + 3\cos x}\,dx\) when (i) \(a=\pi\) and (ii) \(a=2\pi\).
(b) Evaluate \(\displaystyle \int_{0}^{\pi} \frac{1}{5 + 3\cos^{2}x}\,dx\). | [
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aops_2688460 | $$\lim\limits_{x\rightarrow 0}\frac{x^2-\sin^2x}{x^2\sin^2 x}=\lim_{x\rightarrow 0}\frac{2x-2\cos x\sin x}{2x^2\cos x\sin x+2x\sin^2 x}=\lim_{x\rightarrow 0}\frac{x-\sin x\cos x}{x\sin x\left ( x\cos x\left ( \frac{\sin x}{x\cos x}+1 \right ) \right )}=$$
$$=\underbrace{\lim_{x\rightarrow 0}\frac{1}{\cos x}}_{1}\lim_{x... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Using the I'Hospital's rule find $$\\lim\\limits_{x\\to 0}\\frac{x^2-\\sin^2x}{x^2\\sin^2x}=\\frac13.$$",
"content_html": "Using the I'Hospital's rule find <img src=\"//latex.artofproblemsolving.com/... | Using L'Hôpital's rule, find
\[
\lim_{x\to 0}\frac{x^2-\sin^2 x}{x^2\sin^2 x}.
\] | [
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numina_10206974 | 1. Denote \( x_n = \sqrt{n + a_1} + \sqrt{n + a_2} + \cdots + \sqrt{n + a_n} - n\sqrt{n + a_0} \).
2. Take \( a_0 = -1 \). This simplifies the expression for \( x_n \) as follows:
\[
x_n = \sum_{i=1}^n \sqrt{n + a_i} - n\sqrt{n - 1}
\]
3. Let \( k_n \) be the number of terms among \( a_1, a_2, \ldots, a_n \) w... | null | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "aops_forum"
} | Prove that exists sequences $ (a_n)_{n\ge 0}$ with $ a_n\in \{\minus{}1,\plus{}1\}$, for any $ n\in \mathbb{N}$, such that:
\[ \lim_{n\rightarrow \infty}\left(\sqrt{n\plus{}a_1}\plus{}\sqrt{n\plus{}a_2}\plus{}...\plus{}\sqrt{n\plus{}a_n}\minus{}n\sqrt{n\plus{}a_0}\right)\equal{}\frac{1}{2}\] | [
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aops_480283 | 1.
[hide]Let's say \[W = \lim_{x \to 0} \left( \frac{1 + 2^x + 3^x}{3}\right)^{\frac{1}{x}}\] then by giving $\ln$s we'll get \[\ln W = \lim_{x \to 0}\frac{\ln\left( 1 + 2^x + 3^x\right) - \ln 3}{x}\] where the limit on the right side can be computed by L'Hospital's Rule. That is, $\displaystyle \lim_{x\to 0}\frac{2^x\... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate\n$ \\lim_{x \\to 0} ( \\frac{1+2^x+3^x}{3} )^{1/x} $",
"content_html": "Evaluate<br>\n<img src=\"//latex.artofproblemsolving.com/9/a/e/9ae06b98e012205cdd110d55defa7025e46d76e8.png\" class=\"... | Evaluate
\[
\lim_{x \to 0}\left(\frac{1+2^{x}+3^{x}}{3}\right)^{1/x}.
\] | [
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aops_3397693 | [quote=happyhippos] So the integral is 0. $\blacksquare$[/quote]
Making the substitution $t=\frac\pi2-x$ we get $I=-I\,\rightarrow\,\boxed{\,I=0\,}$
| null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate the integral:\n$\\displaystyle \\int_{0}^{\\frac{\\pi}{2}} \\frac{\\cos x-\\sin x}{1+\\cos x\\sin x}dx$.",
"content_html": "Evaluate the integral:<br>\n<span style=\"white-space:pre;\"><img ... | Evaluate the integral:
\[
\int_{0}^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\cos x\sin x}\,dx.
\] | [
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numina_10032603 | Solution. By adding and subtracting 1 from $\cos x$ and applying the corresponding formula, we get
$$
\begin{aligned}
& \lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}=\lim _{x \rightarrow 0}[1-(1-\cos x)]^{\frac{1}{x}}=\lim _{x \rightarrow 0}\left(1-2 \sin ^{2} \frac{x}{2}\right)^{\frac{1}{x}}= \\
& =\lim _{x \rightarr... | 1 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 13. Find $\lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}$. | [
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aops_201447 | Is this the integral?
$ \int\frac{100x\pi}{1\plus{}x}\ dx$
I left out the limits of integration because you can do that. Since $ 100\pi$ is a constant, you can pull this out and the integral becomes
$ 100\pi\int\frac{x\ dx}{1\plus{}x}$
Now think substitutions. Can you figure out which one? | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "fnInt(100xpi/1+x,x,0,x) = fnInt(100xpi/1+x,x,x,10000)?",
"content_html": "fnInt(100xpi/1+x,x,0,x) = fnInt(100xpi/1+x,x,x,10000)?",
"post_id": 1107992,
"post_number": 1,
"post_time_u... | \[
\int_{0}^{x} \frac{100x\pi}{1+x}\,dx = \int_{x}^{10000} \frac{100x\pi}{1+x}\,dx
\] | [
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aops_424252 | hello, your integrand is equivalent to
${\frac {\sin \left( x \right) +\cos \left( x \right) }{2\,\sin \left(
x \right) \cos \left( x \right) +2\, \left( \cos \left( x \right)
\right) ^{2}-1}}
$
plugging
$\cos(x)=\frac{1-z^2}{1+z^2}$
$\sin(x)=\frac{2z}{1+z^2}$
we get
with $dx=\frac{2dz}{1+z^2}$ the following integra... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$\\int \\frac{\\sin{x}+\\cos{x}}{\\sin{2x}+\\cos{2x}}dx$\n\nI can't find the solution..",
"content_html": "<img src=\"//latex.artofproblemsolving.com/c/f/4/cf456a7ccb7969a043709ee8050099e552be9e1e.pn... | Cleaned-up problem statement (LaTeX):
\[
\int \frac{\sin x + \cos x}{\sin 2x + \cos 2x}\,dx
\] | [
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aops_380346 | Sketch: For each $n$ there exists a unique $m=m_n$ such that $m\pi \le a_n <(m+1)\pi.$ Relative to $n^3$, the diffference between $m\pi$ and $ a_n$ is negligible; let's pretend $m\pi = a_n$. Then $\int_{0}^{m\pi}(1+|\sin x|)dx = \int_{0}^{a_{n}}(1+|\sin x|)dx$. The integral on the left equals $m(2+\pi)$ by $\pi$-p... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "A sequence $a_n$ is defined by $\\int_{a_n}^{a_{n+1}} (1+|\\sin x|)dx=(n+1)^2\\ (n=1,\\ 2,\\ \\cdots),\\ a_1=0$.\n\nFind $\\lim_{n\\to\\infty} \\frac{a_n}{n^3}$.",
"content_html": "A sequence <img s... | A sequence \(a_n\) is defined by
\[
\int_{a_n}^{a_{n+1}} \bigl(1+|\sin x|\bigr)\,dx=(n+1)^2\qquad (n=1,2,\dots),\qquad a_1=0.
\]
Find \(\displaystyle\lim_{n\to\infty}\frac{a_n}{n^3}\). | [
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aops_538724 | set $x=e^{-t}$, the integral would be : \[\int\limits_0^{+\infty} e^{-a t }\cdot (-t)^n \ \mathrm{d}t \overset{at=y}{=}a^{-1}(-1)^n \int\limits_0^{+\infty}e^{-y} \frac{y^n}{a^n} \ \mathrm{d}y= \frac{(-1)^n\Gamma(n+1)}{a^{n+1}}= \frac{(-1)^n (n!)}{a^{n+1}}.\] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$\\int_{0}^{1} x^{a-1} \\cdot (\\ln x)^n dx$ \n\n where $a \\in { 2,3,4,....} $ and $ n \\in{ N}$\n\n[hide=\"answer\"] $\\frac{(-1)^n (n!)}{a^{n+1}}$[/hide]",
"content_html": "<img src=\"//latex.... | Compute the integral
\[
\int_{0}^{1} x^{a-1}(\ln x)^n\,dx,
\]
where \(a\in\{2,3,4,\dots\}\) and \(n\in\mathbb{N}\). | [
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numina_10032609 | Solution. When $x \rightarrow 2$, we have an indeterminate form of $0 \cdot \infty$. Setting $x=2-\alpha$ and taking the limit, we find
$$
\begin{aligned}
& \lim _{x \rightarrow 2}(2-x) \operatorname{tg} \frac{\pi}{4} x=\lim _{\alpha \rightarrow 0} \alpha \operatorname{tg} \frac{\pi}{4}(2-\alpha)=\lim _{\alpha \righta... | \frac{4}{\pi} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 5. Find $\lim _{x \rightarrow 2}(2-x) \tan \frac{\pi}{4} x$. | [
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numina_10031983 | Solution. Here $u=\sin x$. Then we get
$$
y_{x}^{\prime}=\frac{u_{x}^{\prime}}{u}=\frac{(\sin x)^{\prime}}{\sin x}=\frac{\cos x}{\sin x}=\operatorname{ctg} x
$$ | \operatorname{ctg}x | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 253. $y=\ln \sin x$. | [
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aops_511548 | Integration by parts:
\[I_{n}=\int\frac{1}{(x^2+1)^{n}}\ dx = x \cdot \frac{1}{(x^2+1)^{n}}- \int x \cdot (-n)\frac{1}{(x^2+1)^{n+1}}(2x)\ dx\]
\[I_{n}= \frac{x}{(x^2+1)^{n}}+2n \int \frac{x^{2}}{(x^2+1)^{n+1}}\ dx = \frac{x}{(x^2+1)^{n}}+2n \int \frac{x^{2}+1-1}{(x^2+1)^{n+1}}\ dx \]
\[I_{n}= \frac{x}{(x^2+1)^{n}} ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find a formula expressing $\\int \\dfrac{dx}{(x^2+1)^{n+1}}$ in terms of $\\int \\dfrac{dx}{(x^2+1)^{n}}$ for $n \\geq 1$.",
"content_html": "Find a formula expressing <img src=\"//latex.artofproblem... | Find a formula expressing
\[
\int \frac{dx}{(x^2+1)^{n+1}}
\]
in terms of
\[
\int \frac{dx}{(x^2+1)^{n}}
\]
for \(n\ge 1\). | [
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aops_232333 | [b]@ Poincare:[/b] Why don't I write out the first few terms in this form.
$ \left(\frac {1}{1!} \minus{} \frac {1}{2!} \right) \plus{} \left(\frac {1}{2!} \minus{} \frac {1}{3!} \right) \plus{} \left(\frac {1}{3!} \minus{} \frac {1}{4!} \right) \plus{} \left(\frac {1}{4!} \minus{} \frac {1}{5!} \right) \plus{} \ldo... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "find the value of the sum of n/(n+1)! as n uns from 1 to infinity.",
"content_html": "find the value of the sum of n/(n+1)! as n uns from 1 to infinity.",
"post_id": 1283456,
"post_number... | Find the value of the infinite series
\[
\sum_{n=1}^{\infty} \frac{n}{(n+1)!}.
\] | [
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numina_10030354 | Solution. Let's compare the given series with the harmonic series
$$
1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}+\ldots
$$
Each term \(a_{n}=\frac{1}{\sqrt[3]{n}}\) of the given series, starting from the second, is greater than the corresponding term \(b_{n}=\frac{1}{n}\) of the harmonic series. Since the harmonic s... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 5.5. Investigate the convergence of the series
$$
1+\frac{1}{\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}}+\ldots+\frac{1}{\sqrt[3]{n}}+\ldots
$$ | [
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aops_262831 | That is to say, $ \Gamma(x) \Gamma(1 - x) = \frac {\pi}{\sin(\pi x)}$.
Which, together with what Kent Merryfield did before gives:
${ \int_0^1\frac {dx}{\sqrt [4]{1 - x^4}} = \frac14\Gamma\left(\frac14\right)\Gamma\left(\frac34\right) = \frac14 \frac {\pi}{\sin(\frac14\pi)} = \frac {\pi }{4 \cdot\frac12 \sqrt2} = \... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "\\[ \\int {\\frac {{dx}}{{\\sqrt [4]{{1 \\minus{} x^4 }}}}}\r\n\\]\r\n\r\n\\[ \\int {\\sqrt [3]{{x^3 \\minus{} 6x}}} dx\r\n\\]\r\n\r\n\\[ \\int\\limits_{ \\minus{} 1}^1 {\\frac {{dx}}{{\\sqrt {1 \\plus{} x... | \[
\int \frac{dx}{\sqrt[4]{1-x^4}}
\]
\[
\int \sqrt[3]{x^3-6x}\,dx
\]
\[
\int_{-1}^1 \frac{dx}{\sqrt{1+x^2}+\sqrt{1-x^2}}
\]
\[
\int \frac{\sin x-\cos x}{x+\sin 2x}\,dx
\]
\[
\int_0^\pi \frac{a-b\cos x}{a^2+b^2-2ab\cos x}\,dx
\] | [
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aops_112296 | $2.\blacktriangleright$ Prove that $x\in\left(0,\frac{\pi}{2}\right)\Longrightarrow \ln (x+1)>x+\cos x-1\ .$
$\left\{\begin{array}{c}\underline{f(x)=\ln (x+1)-x-\cos x+1}\\\\ f'(x)=\frac{1}{x+1}-1+\sin x\\\\ f''(x)=-\frac{1}{(x+1)^{2}}+\cos x\\\\ f'''(x)=\frac{2}{(x+1)^{3}}-\sin x\\\\ f''''(x)=-\frac{6}{(x+1)^{4}}-\... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$1.\\blacktriangleright$ Ascertain $x>-2$ for which $2^{x}+6^{x}+3^{x}=2x^{2}+6x+3\\ .$\r\n$2.\\blacktriangleright$ Prove that $x\\in(0,\\frac{\\pi}{2})\\Longrightarrow x+\\cos x-1<\\ln (x+1)<x\\ .$\r\n$3.... | 1. Ascertain \(x>-2\) for which
\[
2^{x}+6^{x}+3^{x}=2x^{2}+6x+3.
\]
2. Prove that for \(x\in\left(0,\frac{\pi}{2}\right)\),
\[
x+\cos x-1<\ln(x+1)<x.
\]
3. Solve the system
\[
\begin{cases}
e^{x}+e^{y}=2,\\[4pt]
xy=x+y.
\end{cases}
\] | [
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aops_1460367 | I will elaborate on what I think jg123 is trying to say.
Your first solution is perfectly correct, but when you solved
$\frac{d}{dt}(t+\frac{1}{t})^a$ you have to use the chain rule. So you can let $v=t+\frac{1}{t}$.
$\frac{d}{dt}(t+\frac{1}{t})^a =\frac{d}{dt}(v)^a = a\cdot v^{a-1} \cdot \frac{dv}{dt}$, $\frac{dv}{dt... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If $x=\\left(t+\\frac1t\\right)^a$ find $\\frac{dx}{dt}$\n\nTaking log both sides\n1... $\\log x = a \\log \\left(\\frac{t^2+1}{t}\\right)$\nDifferentiating w.r.t. t\n2... $\\frac1x \\times \\frac{dx}{dt}=... | If \(x=\left(t+\dfrac{1}{t}\right)^a\), find \(\dfrac{dx}{dt}\).
Taking logarithms,
\[
\log x = a\log\!\left(\frac{t^2+1}{t}\right).
\]
Differentiate with respect to \(t\):
\[
\frac{1}{x}\frac{dx}{dt}
= a\cdot\frac{d}{dt}\log\!\left(\frac{t^2+1}{t}\right).
\]
Hence
\[
\frac{dx}{dt}
= a x\cdot\frac{t^2-1}{t^2(t^2+1)}
=... | [
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aops_1719745 | Since we have $\int \sqrt{1-x^2}dx = \int \frac{1-x^2}{\sqrt{1-x^2}}dx = \int \frac{dx}{\sqrt{1-x^2}} + \int x(\sqrt{1-x^2})' dx = \arcsin x +x\sqrt{1-x^2}-\int \sqrt{1-x^2}dx,$ this means that $\int \sqrt{1-x^2}dx = \frac{1}{2}\left(\arcsin x +x\sqrt{1-x^2} \right)$ and so the required integral $$\frac{1}{2\pi}\left[ ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Prove the rationality of the number $ \\frac{1}{\\pi }\\int_{\\sin\\frac{\\pi }{13}}^{\\cos\\frac{\\pi }{13}} \\sqrt{1-x^2} dx. $",
"content_html": "Prove the rationality of the number <img src=\"//l... | Prove the rationality of the number
\[
\frac{1}{\pi}\int_{\sin\frac{\pi}{13}}^{\cos\frac{\pi}{13}} \sqrt{1-x^2}\,dx.
\] | [
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aops_1208598 | $I=\int \cos(\ln x) dx \\
=x\cos(\ln x) +\int \sin(\ln x) dx +\text{ constant} \\
=x\cos(\ln x) +x\sin(\ln x) -I +\text{ constant} \\
\therefore I=\frac{1}{2} \{x\cos(\ln x) +x\sin(\ln x)\} +c$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate two of the following integrals using the method of integration by parts.\n\n$\\int cos(lnx)\\; \\; dx$ and $\\int sin^2x\\cdot e^{2x}\\; \\; dx$\n",
"content_html": "Evaluate two of the foll... | Evaluate two of the following integrals using the method of integration by parts:
1. \(\displaystyle \int \cos(\ln x)\,dx\).
2. \(\displaystyle \int \sin^2 x\,e^{2x}\,dx\). | [
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0.5... | null |
aops_257200 | This is a first-order linear ODE; we can standardize our methods for dealing with such problems. First step: get a standard form by making the coefficient of $ y'$ $ 1.$
$ y'\plus{}\frac1xy\equal{}\frac1{x^2}$
The integrating factor is $ e^{\int \frac1x\,dx}\equal{}e^{\ln x}\equal{}x.$
$ xy'\plus{}y\equal{}\fr... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Given that: $ x^2y'$+xy=1 and $ y(1)\\equal{}2$. Find the function y",
"content_html": "Given that: <span style=\"white-space:nowrap;\"><img src=\"//latex.artofproblemsolving.com/8/b/0/8b0b99a8981ebc... | Given that \(x^2 y' + x y = 1\) and \(y(1)=2\). Find the function \(y(x)\). | [
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aops_1718536 | $ \displaystyle 2I= \int \frac{2 \sin x}{\sin x \cos x+1} \ dx = \int \frac{ \sin x + \cos x}{\sin x \cos x+1} \ dx -\int \frac{ \cos x - \sin x}{\sin x \cos x+1} \ dx $
Let $u = \sin x - \cos x$. Then $du = (\sin x +\cos x) \ dx$ and $1+\sin x \cos x = \dfrac{3-u^2}{2}$ which can be used for the first integrand. For ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$\\displaystyle \\int\\frac{1}{\\cos x+\\csc x}dx$",
"content_html": "<img src=\"//latex.artofproblemsolving.com/5/c/0/5c0c9f7b79d8858d6240e67e60e0b6fa7750cd85.png\" class=\"latex\" alt=\"$\\displays... | Evaluate the integral
\[
\int \frac{1}{\cos x + \csc x}\,dx.
\] | [
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aops_1749189 | Taking the substitution $x=\sec(a)$ the integral becomes
$\int \frac{ \cos{a}}{1-\cos^3{a}}da$
\\
$=\frac{1}{3}\int \frac{da}{1-\cos{a}}-\frac{1}{3} \int\frac{1-\cos{a}}{1+\cos{a}+\cos^2{a}}da$
$\int \frac{da}{1-\cos{a}}=\int \frac{1+\cos{a}}{\sin^2{a}}da=\int (\csc^2{a}+\cot{a}\csc{a})da$
$=-\cot{a}-\csc{a}+C=\fra... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "I've been struggling with this integration. Can any one help?\n$\\int\\frac{x}{(x^3-1)\\sqrt{x^2-1}}dx$",
"content_html": "I've been struggling with this integration. Can any one help?<br>\n<img src=... | Compute the integral
\[
\int \frac{x}{(x^3-1)\sqrt{x^2-1}}\,dx.
\] | [
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aops_1543549 | i got the solve
write 1/(3n+1) as integral from 0 to 1 x^(3n)
so it means ((-1)^n)/(3n+1)= integral from 0 to 1 (-x)^3n
so sum of ((-1)^n)/(3n+1) = integral from 0 to 1 sum(-x)^(3n) dx = integral from 0 to 1 (dx/(1+x^3) | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "evaluate sum from n=0 to infinity \n((-1)^n)/(3n+1)",
"content_html": "evaluate sum from n=0 to infinity<br>\n((-1)^n)/(3n+1)",
"post_id": 9352909,
"post_number": 1,
"post_time_unix... | Evaluate the sum
\[
\sum_{n=0}^{\infty} \frac{(-1)^n}{3n+1}.
\] | [
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aops_22465 | Of course, Peter meant to say that $f(x)=-x\ln x$ is concave. (Better, say that $f(x)=-x\log_{2004}x$ is concave.)
I'm using "concave" to mean the negative of a convex function. In first-year calculus terms, that's "concave downward."
This is Jensen's inequality. I'm moving it to "solved." | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Suppose that $x_1>0,x_2>0,\\cdots\\cdots,\\ x_{2005}>0$ and $x_1+x_2+\\cdots+x_{2005}=1$.\r\nProve the following inequality.\r\n\r\n\\[-x_1\\log _{2004}{x_1}-x_2\\log _{2004}{x_2}-\\cdots\\cdots -x_{2004}\... | Suppose that \(x_1>0,x_2>0,\ldots,x_{2005}>0\) and \(x_1+x_2+\cdots+x_{2005}=1\). Prove the inequality
\[
- x_1\log_{2004} x_1 - x_2\log_{2004} x_2 - \cdots - x_{2005}\log_{2004} x_{2005}
\le \log_{2004} 2005.
\] | [
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aops_13979 | [tex]
\begin{eqnarray*}
(\sin x)(2 + \cos x) = 2\sin x + \sin x \cos x\\
&=&2\sin x + \frac{\sin 2x}{2}
\end
[/tex]
There are two things that might make this as large as possible: a large value for sin x and a large value for sin 2x. The x that makes sin x the larges is pi/2. Thus if x = pi/2, then
\[
2\s... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "find the maximum value of $f(x) = \\sin x (2+\\cos x)$, can it be solved without calculus?",
"content_html": "find the maximum value of <span style=\"white-space:nowrap;\"><img src=\"//latex.artofpro... | Find the maximum value of
\[
f(x)=\sin x\,(2+\cos x).
\]
Can this be solved without calculus? | [
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numina_10031506 | Solution.
1. Since the integrand has the form $R(\operatorname{tg} x)$, we make the substitution $\operatorname{tg} x=t$.
Substituting into the integrand
$$
\operatorname{tg} x=t, \quad d x=\frac{d t}{1+t^{2}}
$$
we get
$$
\frac{3 \operatorname{tg}^{2} x-1}{\operatorname{tg}^{2} x+5} d x=\frac{3 t^{2}-1}{t^{2}+5} ... | -x+\frac{4}{\sqrt{5}}\operatorname{arctg}(\frac{\operatorname{tg}x}{\sqrt{5}})+C | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 2. Find the indefinite integral
$$
\int \frac{3 \operatorname{tg}^{2} x-1}{\operatorname{tg}^{2} x+5} d x
$$ | [
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aops_272462 | $ \frac {dV}{dt} = 9\sin\sqrt {t + 1} \\
\\
dV = 9\sin\sqrt {t + 1}dt \\
\\
\int\limits_{V_i}^{V_f} dV = \int\limits_0^{t_f} 9\sin\sqrt {t + 1}dt \\
\\
V_i = 81.637,\;\;t_f = 6 \\
\\
\text{Let, }t + 1 = x^2 \\
dt = 2xdx \\
\\
\therefore \int 9\sin\sqrt {t + 1}dt = \int 2x\sin x dx \\
\\
= 2x\{\cos x\... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "A spherical tank contains $ 81.637$ gallons of water at time $ t\\equal{}0$ minutes. For the next $ 6$ minutes, water flows out of the tank at the rate of $ 9\\sin{\\sqrt{t\\plus{}1}}$ gallons per minute. ... | A spherical tank contains \(81.637\) gallons of water at time \(t=0\) minutes. For the next \(6\) minutes, water flows out of the tank at the rate of \(9\sin\sqrt{t+1}\) gallons per minute. How many gallons of water are in the tank at the end of the \(6\) minutes? | [
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numina_10030318 | Solution. To get rid of the irrationality, we make the substitution $1+x=t^{2}$. Then
$$
\begin{aligned}
& \int_{3}^{8} \frac{x d x}{\sqrt{1+x}}=\left|\begin{array}{l}
1+x=t^{2} \\
d x=2 t d t \\
x=3 \rightarrow t=2 \\
x=8 \rightarrow t=3
\end{array}\right|=\int_{2}^{3} \frac{\left(t^{2}-1\right) 2 t d t}{t}=2 \int_{2... | \frac{32}{3} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 3.13. Using the Newton-Leibniz formula, compute the definite integral
$$
\int_{3}^{8} \frac{x}{\sqrt{1+x}} d x
$$ | [
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aops_345089 | Its just the idea of Riemann Definition ,
$ \int_{a}^{b}f(x).dx\equal{} \lim_{n \to \infty}\frac{b\minus{}a}{n}\sum_{r\equal{}1}^{n}f(a\plus{}r\Delta x)$ :) | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find: \r\n\r\n$ \\lim_{n \\to \\infty} \\frac {1}{n}[(\\frac {1}{n})^2 \\plus{} (\\frac {2}{n})^2 \\plus{} ... \\plus{} (\\frac {n \\minus{} 1}{n})^2]$",
"content_html": "Find:<br>\n<br>\n<img src=\"... | Find
\[
\lim_{n \to \infty}\frac{1}{n}\left[\left(\frac{1}{n}\right)^2+\left(\frac{2}{n}\right)^2+\cdots+\left(\frac{n-1}{n}\right)^2\right].
\] | [
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ours_4474 | For \( b_{n} = \frac{a_{n}}{n} \), the condition takes the form
\[
n\left(b_{n+1} - b_{n}\right) \geq (n+2)\left(b_{n+2} - b_{n+1}\right).
\]
If \( b_{k+1} \leq b_{k} \) for some \( k \), it follows that \( b_{n+1} \leq b_{n} \) for every \( n \geq k \). Thus, the sequence \(\left(b_{n}\right)_{n \geq k}\) is decreas... | null | {
"competition": "bulgarian_comps",
"dataset": "Ours",
"posts": null,
"source": "EMT-All-2010-12 кл-sol.md"
} | Let \( a_{1}, a_{2}, \ldots \) be a sequence of positive numbers such that \( 2 a_{n+1} \geq a_{n} + a_{n+2} \) for every \( n \in \mathbb{N} \). Prove that the sequence with general term \( \frac{a_{n}}{n} \) is convergent. | [
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aops_1799013 | [quote=Tan][hide="Alternate Solution"]
Note $\int_{0}^{\infty}e^{-x} /\sqrt{x} dx = \int_{0}^{\infty} x^{-1/2}e^{-x} dx = \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$.
[/hide][/quote]
Yes but the derivation of $\Gamma(\frac{1}{2})=\sqrt{\pi}$ actually uses the Gaussian integral, so that's pretty much a circular argume... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Integrate $\\int_{0}^{\\infty}e^{-x} /\\sqrt{x} dx$. Is it valid to let $x= y^2$ such that $y$ is positive and conclude it is equal to the gaussian integral? (Reason why I'm asking is because the book I'm... | Integrate
\[
\int_{0}^{\infty}\frac{e^{-x}}{\sqrt{x}}\,dx.
\]
Is it valid to let \(x=y^{2}\) with \(y>0\) and conclude the value by relating it to the Gaussian integral? | [
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aops_1267462 | [quote=jmenks]$$\int_{0}^{\infty} \frac{\ln(1+x^2)}{1+x^2} \, \mathrm{d}x$$[/quote]
\begin{align*}
\int_{0}^{\infty} \frac{\ln \left ( 1+x^2 \right )}{1+x^2} \, {\rm d}x &\overset{x=\tan u}{=\! =\! =\! =\!} \int_{0}^{\pi/2} \frac{\ln \left ( 1+\tan^2 x \right )}{1+\tan^2 x} \sec^2 x \, {\rm d}x \\
&= \int_{0}^{\pi/2}... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$$\\int_{0}^{\\infty} \\frac{\\ln(1+x^2)}{1+x^2} \\, \\mathrm{d}x$$",
"content_html": "<img src=\"//latex.artofproblemsolving.com/4/f/7/4f7e7e49a7c8a3d507b4173dcb413aadd01b3fab.png\" class=\"latexcen... | \[
\int_{0}^{\infty} \frac{\ln(1+x^{2})}{1+x^{2}}\,dx
\] | [
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aops_520799 | I guess you mean 'without using' (not the same as useless) Stirling.
$\log \left(\frac{(n!)^\frac{1}{n}}{n}\right)=\frac{1}{n}\sum_{k=1}^n \log \frac{k}{n}\to \int_0^1 \log xdx=-1$, so our original limit is $e^{-1}$. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Compute this limit useless Stirling formula ( $\\sqrt[n]{n!}\\sim \\frac{n}{e}$) :\n\n\\[\\underset{n\\to \\infty }{\\mathop{\\lim }}\\,\\frac{\\sqrt[n]{n!}}{n}\\]",
"content_html": "Compute this lim... | Compute the following limit using Stirling's formula \(\sqrt[n]{n!}\sim \dfrac{n}{e}\):
\[
\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}.
\] | [
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ours_34689 | To prove the inequality, we consider the series:
\[
S = \frac{1}{3^3} + \frac{1}{4^3} + \cdots + \frac{1}{n^3}
\]
We need to show that \( S < \frac{1}{12} \).
First, note that for \( k \geq 3 \), we have:
\[
\frac{1}{k^3} < \frac{1}{k(k-1)(k-2)}
\]
This is because:
\[
k^3 = k \cdot k \cdot k > k \cdot (k-1) \cdot... | null | {
"competition": "irish_mo",
"dataset": "Ours",
"posts": null,
"source": "1990.md"
} | Let \( n \geq 3 \) be a natural number. Prove that
\[
\frac{1}{3^{3}}+\frac{1}{4^{3}}+\cdots+\frac{1}{n^{3}}<\frac{1}{12}
\] | [
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aops_522630 | ${ I }_{ 2 }=\int _{ 0 }^{ 1 }{ \frac { { Tan }^{ -1 }x }{ x } } $
Substitute $x=tan(t)$
${ I }_{ 2 }=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { t({ sec }^{ 2 }t) }{ tan(t) } dt } $
${ I }_{ 2 }=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ (2t)csc(2t)dt } $
Now let $2t=k$
Then the integral becomes
${ I }_{ 2 }=\frac { 1 }... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$ I_1 = \\int_0^{\\pi/2} xcosec(x) dx $\n$ I_2= \\int_0^1 \\frac{tan^{-1}(x)}{x} dx $\n \nFind $ I_1/I_2 $\n\n[hide=\"Try\"]I put $ x = sin\\theta $ in $I_1 $\nand Im stuck with $ I_1/I_2 = \\int_0^1 \\fra... | Let
\(I_1=\displaystyle\int_{0}^{\pi/2} x\csc x\,dx,\)
\qquad
\(I_2=\displaystyle\int_{0}^{1}\dfrac{\tan^{-1}x}{x}\,dx.\)
Find \(\dfrac{I_1}{I_2}.\) | [
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numina_10046892 | ## Solution
$$
\begin{aligned}
& \int_{0}^{\frac{1}{\sqrt{2}}} \frac{\arccos ^{3} x-1}{\sqrt{1-x^{2}}} d x=\int_{0}^{\frac{1}{\sqrt{2}}} \frac{\arccos ^{3} x}{\sqrt{1-x^{2}}} d x+\int_{0}^{\frac{1}{\sqrt{2}}} \frac{-d x}{\sqrt{1-x^{2}}}= \\
& =\int_{0}^{\frac{1}{\sqrt{2}}}-(\arccos x)^{3} d(\arccos x)+\left.\arccos x\... | \frac{15\pi^{4}}{2^{10}}-\frac{\pi}{4} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | ## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{1}{\sqrt{2}}} \frac{(\arccos x)^{3}-1}{\sqrt{1-x^{2}}} d x
$$ | [
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numina_10033411 | Solution.
$$
\lim _{x \rightarrow 0}(1-\cos x)^{x}=\left(0^{0}\right)=e^{\lim _{x \rightarrow 0} x \ln (1-\cos x)}
$$
## Find separately
$$
\begin{aligned}
& \lim _{x \rightarrow 0} x \cdot \ln (1-\cos x)=(0 \cdot(-\infty))=\lim _{x \rightarrow 0} \frac{(\ln (1-\cos x))^{\prime}}{(1 / x)^{\prime}}=\lim _{x \rightarr... | 1 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 6.59. $\lim _{x \rightarrow 0}(1-\cos x)^{x}$. | [
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numina_10102623 | [Proof] Consider mathematician $A$, who takes two naps, denoted as $t_{1}$ and $t_{2}$. The other 4 mathematicians each have a nap that coincides with one of $A$'s naps, meaning each of them naps at either $t_{1}$ or $t_{2}$. By the pigeonhole principle, among these 4 people, there must be two who nap at the same time,... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | $5 \cdot 45$ In a speech, there are 5 mathematicians, each of whom dozes off twice, and every two of them have a moment when they are dozing off simultaneously. Prove that there must be a moment when 3 of them are dozing off simultaneously. | [
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numina_10120194 | 【Analysis】The total number of possibilities when each of the two people draws one card is $7 \times 7=49$. Among these, the combinations where the sum of the numbers is 8 are: $1+7,2$ $+6,3+5,4+4,5+3,6+2,7+1$, totaling 7 possible combinations. Therefore, the probability that the sum of the numbers on the two cards is 8... | \frac{7}{49}=\frac{1}{7} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | $3 、$ A and B each hold 7 cards, on which the numbers $1 、 2 、 3 、 4 、 5 、 6 、 7$ are written respectively. If both draw one card, what is the probability that the sum of the numbers on the two cards is 8? | [
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ours_22358 | We refer to Pascal's triangle. To solve the problem, we consider the sum of all possible values of \(\binom{i}{j}\) subject to the restrictions that \(i, j \geq 0\) and \(i+j \leq n\). This sum is \(2^{n}-1\) (the sum of Pascal's triangle rows below row \(n\)). We need to subtract this from our problem's sum.
By arran... | 27633 | {
"competition": "pumac",
"dataset": "Ours",
"posts": null,
"source": "Team_Round_2023-2.md"
} | What is the sum of all possible \(\binom{i}{j}\) subject to the restrictions that \(i \geq 10\), \(j \geq 0\), and \(i+j \leq 20\)? Count different \(i, j\) that yield the same value separately - for example, count both \(\binom{10}{1}\) and \(\binom{10}{9}\). | [
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0.4... | null |
numina_10266137 | 1. Let \( F \) be the set of students who play football, \( B \) be the set of students who play basketball, and \( S \) be the set of students who play baseball.
2. We are given:
\[
|F| = 128, \quad |B| = 291, \quad |S| = 318, \quad |F \cap B \cap S| = 36
\]
3. Let \( |F \cap B| \) be the number of students w... | 274 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "aops_forum"
} | At Mallard High School there are three intermural sports leagues: football, basketball, and baseball. There are 427 students participating in these sports: 128 play on football teams, 291 play on basketball teams, and 318 play on baseball teams. If exactly 36 students participate in all three of the sports, how many ... | [
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8663,
49586,
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15611,
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3331,
41250,
47035,
1083,
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30880,
15647,
3449,
109... | [
0.43095984920916985,
0.5085065825141035,
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0.46003378227969893,
0.512790110655388,
0.3073370997801875,
0.5397314423061106,
0.405... | null |
numina_10079682 | Attention, here there are two indices: $n$ and $k$, we need to be meticulous about how we will perform the induction. We will prove by induction on $n$ the following property:
$$
\mathcal{P}(n)=" \forall 0 \leq k \leq n,\binom{n}{k}=\binom{n}{k}=\frac{n!}{k!(n-k)!} "
$$
The initialization is obvious $(0!=1)$. Let's n... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Let us define for all $n \geq 0$ and all $0 \leq k \leq n$ the binomial coefficients $\binom{n}{k}$ (pronounced " $k$ out of $n$ ") as follows:
$$
\binom{n}{0}=\binom{n}{n}=1 \text { and }\binom{n+1}{k+1}=\binom{n}{k}+\binom{n}{k+1}
$$
Prove that
$$
\binom{n}{k}=\frac{n!}{k!(n-k)!}
$$ | [
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28038,
376... | [
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0.2436941294389474,
0.2821124277590068,
0.4852184306951922,
0.43... | null |
numina_10169754 | There are 9 single-digit numbers, 90 two-digit numbers, and 900 three-digit numbers. The total number of digits is $9 + 2 \cdot 90 + 3 \cdot 900 = 2889 > 1999$.
This means that the 1999th digit must be found among the three-digit numbers. For the three-digit numbers, 1999 - $189 = 1810$ digits remain. $1810 = 603 \cdo... | 7 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Someone wrote down the integers from 1 to 1999 next to each other. What digit stands at the 1999th position? | [
20141,
14711,
1975,
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20005,
31278,
31011,
61452,
3178,
3708,
62319,
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8021,
26797,
5217,
4440,
422,
3540,
40823,
30848,
43955,
7206,
18842,
40233,
5190,... | [
0.3905878740259248,
0.33600772078803987,
0.5954534506157404,
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0.5710277733838831,
0.5077768269873968,
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0.4100676081276041,
0.47359361199492683,
0.5065035606742075,
0.36369091683406135,
0.4292887713931907,
0.4077831924776799,
0.39... | null |
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