problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Example 13. Find $\int \frac{7+x^{2}}{x^{2}\left(x^{2}+2\right)} d x$. | Solution. The fraction can be decomposed into two fractions:
$$
\frac{7}{x^{2}\left(x^{2}+2\right)}+\frac{x^{2}}{x^{2}\left(x^{2}+2\right)}=\frac{7}{x^{2}\left(x^{2}+2\right)}+\frac{1}{x^{2}+2}
$$
What to do with the right fraction, which does not fit the tabulated formulas? We can obtain a fraction with the denomina... | -\frac{7}{2x}-\frac{5}{2\sqrt{2}}\operatorname{arctg}\frac{x}{\sqrt{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,202 |
Example 14. Find $\int(1+2 \sin x)^{2} d x$. | Solution. After squaring, we obtain the non-tabular integral $\int \sin ^{2} x d x$, which can be reduced to tabular ones using the power reduction formula: $\sin ^{2} x=\frac{1}{2}(1-\cos 2 x)$. Thus, we have:
$$
\begin{aligned}
& \int(1+2 \sin x)^{2} d x=\int(1+4 \sin x+2(1-\cos 2 x)) d x= \\
& \quad=3 \int d x+4 \i... | 3x-4\cosx-\sin2x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,203 |
Example 15. Find $\int \frac{x^{3}+2}{x-1} d x$. | Solution. It is possible to convert an improper fraction into the sum of a polynomial and a proper fraction:
$$
\frac{x^{3}+2}{x-1}=\frac{x^{3}-1+3}{x-1}=\frac{x^{3}-1}{x-1}+\frac{3}{x-1}
$$
Therefore,
$$
\int \frac{x^{3}+2}{x-1} d x=\int\left(x^{2}+x+1+\frac{3}{x-1}\right) d x=\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+3 \l... | \frac{x^{3}}{3}+\frac{x^{2}}{2}+x+3\ln|x-1|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,204 |
Example 1. Find $\int e^{x^{2}} \cdot x d x$. | 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 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,205 |
Example 2. Find $\int \frac{\operatorname{arctg}^{4} x}{1+x^{2}} d x$. | Solution. Among the tabular integrals, there are no formulas containing arctangent in the integrand. Let's try to make it the variable of integration: $\operatorname{arctg} x=u$. Then $d u=d(\operatorname{arctg} x)=$ $=(\operatorname{arctg} x)^{\prime} d x=\frac{d x}{1+x^{2}}$. Now it is clear that the arctangent shoul... | \frac{1}{5}\operatorname{arctg}^{5}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,206 |
Example 3. Find $\int \cos ^{3} x \cdot \sin x d x$. | Solution. We notice that $\sin x d x$ is the differential of cosine (up to a sign)
$$
\int \cos ^{3} x \sin x d x=-\int \cos ^{3} x d(\cos x)=-\frac{1}{4} \cos ^{4} x+C
$$
Here, the economy is achieved by not writing out the intermediate integral $\int u^{3} d u$ and its result $\frac{u^{4}}{4}$. | -\frac{1}{4}\cos^{4}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,207 |
Example 5. Find $\int \frac{d x}{1+\sin ^{2} x}$. | Solution. There is no factor that would give the differential of some function (new variable). This means we need to obtain it.
Method 1. The presence of $\sin ^{2} x$ in the denominator intuitively leads to the desire to use the relation $\frac{d x}{\sin ^{2} x}=-d(\operatorname{ctg} x)$. How can we get this fraction... | \ln|\operator | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,208 |
Example 7. Find $\int \sin ^{5} x d x$. | Solution. The factor needed to be placed under the differential sign is obtained by splitting the power:
$$
\begin{aligned}
& \int \sin ^{5} x d x=\int \sin ^{4} x \cdot \sin x d x= \\
& \begin{array}{l}
=\left\{\begin{array}{l}
\text { obtaining the factor } \sin x=-(\cos x)^{\prime}, \\
\text { it is necessary to ex... | -\cosx+\frac{2}{3}\cos^{3}x-\frac{1}{5}\cos^{5}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,209 |
Example 8. Find $\int x^{2} \sin \left(x^{3}+1\right) d x$. | Solution. Let's make the substitution $x^{3}+1=u$. Differentiating this equality, we get $3 x^{2} d x=d u$. Hence, $x^{2} d x=\frac{1}{3} d u$. The integral takes a tabular form:
$$
\int x^{2} \sin \left(x^{3}+1\right) d x=\frac{1}{3} \int \sin u d u=-\frac{1}{3} \cos u+C=-\frac{1}{3} \cos \left(x^{3}+1\right)+C .
$$
... | -\frac{1}{3}\cos(x^{3}+1)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,210 |
Example 9. Find $\int(2 x+3) \sqrt{x+2} d x$. | Solution. It is not obvious what to put under the differential sign, so let's make a substitution that allows us to get rid of the irrationality. Let $\sqrt{x+2}=t$, or $x+2=t^{2}$. From the last equality, we get $x=t^{2}-2, d x=2 t d t$, and $2 x+3=2\left(t^{2}-2\right)+3=$
$=2 t^{2}-1$. Substituting these equalities ... | \frac{4}{5}\sqrt{(x+2)^{5}}-\frac{2}{3}\sqrt{(x+2)^{3}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,211 |
Example 12. Find $\int \frac{\sqrt{4-x^{2}}}{x^{2}} d x$. | Solution. Let $x=2 \sin t, d x=2 \cos t d t, \sqrt{4-x^{2}}=2 \cos t$. We arrive at the integral
$$
\begin{aligned}
\int \frac{\sqrt{4-x^{2}}}{x^{2}} d x=\int \frac{\cos ^{2} t}{\sin ^{2} t} d t=-\operatorname{ctg} t- & t+C= \\
& =-\operatorname{ctg}\left(\arcsin \frac{x}{2}\right)-\arcsin \frac{x}{2}+C
\end{aligned}
... | -\operatorname{ctg}(\arcsin\frac{x}{2})-\arcsin\frac{x}{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,213 |
Example 13. Find $\int \frac{d x}{x^{2} \sqrt{1+x^{2}}}$. | Solution. Let $x=\operatorname{tg} t, \sqrt{1+x^{2}}=\frac{1}{\cos t}, \frac{1}{x^{2}}=\frac{\cos ^{2} t}{\sin ^{2} t}, d x=$ $=\frac{d t}{\cos ^{2} t}, t=\operatorname{arctg} x$. We arrive at an "almost tabular" integral, which after reverse substitutions takes the form $\left(\sin t=\frac{\operatorname{tg} t}{\sqrt{1... | C-\frac{\sqrt{1+x^{2}}}{x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,214 |
Example 1. Find $\int\left(x^{2}-x+1\right) \cos 2 x d x$. | Solution. Here $P(x)=x^{2}-x+1$ is a polynomial of the second degree, which we take as $u$ and integrate by parts twice. The continuity of the solution will be ensured by the following presentation:
$$
\begin{gathered}
\int\left(x^{2}-x+1\right) \cos 2 x d x=\left\{\begin{array}{cc}
x^{2}-x+1=u, d u=(2 x-1) d x \\
\co... | \frac{2x^{2}-2x+1}{4}\sin2x+\frac{2x-1}{4}\cos2x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,216 |
Example 3. Find $\int \ln ^{2} x d x$. | Solution. We integrate by parts twice:
$$
\begin{aligned}
& \int \ln ^{2} x d x=\left\{\begin{aligned}
u & =\ln ^{2} x, & d u & =2 \ln x \cdot \frac{d x}{x} \\
d x & =d v, & v & =x
\end{aligned}\right\}= \\
& =x \ln ^{2} x-2 \int \ln x d x=\left\{\begin{array}{cc}
\ln x=u, & d u=\frac{d x}{x} \\
d x=d v, & v=x
\end{ar... | x\ln^{2}x-2x\lnx+2x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,217 |
Example 4. Find $\int \sqrt{a^{2}+x^{2}} d x$. | Solution. Let the required integral be denoted by $I$ and use the integration by parts formula to form an equation in terms of $I$. We proceed as follows:
$$
\begin{aligned}
I= & \int \sqrt{a^{2}+x^{2}} d x=\left\{\begin{aligned}
\sqrt{a^{2}+x^{2}}=u, & d u=\frac{x}{\sqrt{a^{2}+x^{2}}} d x \\
d x=d v, & v=x
\end{align... | \frac{1}{2}x\sqrt{^{2}+x^{2}}+\frac{^{2}}{2}\ln|x+\sqrt{^{2}+x^{2}}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,218 |
Example 5. Find $\int \frac{d x}{\left(x^{2}+a^{2}\right)^{n}}$, where $n \geqslant 2-$ an integer. | Solution. Let $I_{n}=\int \frac{d x}{\left(x^{2}+a^{2}\right)^{n}}$. Assuming that $n \geqslant 2$, we will integrate by parts another integral $I_{n-\mathrm{I}}$. We have:
$$
\begin{gathered}
I_{n-1}=\int \frac{d x}{\left(x^{2}+a^{2}\right)^{n-1}}=\left\{\begin{array}{c}
\frac{1}{\left(x^{2}+a^{2}\right)^{n-1}}=u, \q... | \int\frac{}{(x^{2}+^{2})^{2}}=\frac{1}{2^{2}(x^{2}+^{2})}+\frac{1}{2^{3}}\operatorname{arctg}\frac{x}{}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,219 |
Example 6. Find $\int e^{\alpha x} \sin \beta x d x$. | Solution. Let $I=\int e^{\alpha x} \sin \beta x d x$; as above, after applying the integration by parts formula twice, we will form an equation to find $I$. We have:
$$
\begin{gathered}
I=\int e^{\alpha x} \sin \beta x d x=\left\{\begin{array}{cc}
e^{\alpha x}=u, & d u=\alpha e^{\alpha x} d x \\
\sin \beta x d x=d v, ... | \frac{\alpha\sin\betax-\beta\cos\betax}{\alpha^{2}+\beta^{2}}e^{\alphax}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,220 |
Example 7. Find $\int x^{2}(2 x-1)^{31} d x$. | Solution. We will gradually get rid of the first factor:
$$
\begin{aligned}
\int x^{2}(2 x-1)^{31} d x & =\left\{\begin{aligned}
x^{2}=u, & d u=2 x d x \\
(2 x-1)^{31} d x=d u, & v=\frac{1}{64}(2 x-1)^{32}
\end{aligned}\right\}= \\
& =\frac{x^{2}}{64}(2 x-1)^{32}-\frac{1}{32} \int x(2 x-1)^{32} d x=
\end{aligned}
$$
... | \frac{x^{2}}{64}(2x-1)^{32}-\frac{x}{2112}(2x-1)^{33}+\frac{(2x-1)^{34}}{143616}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,221 |
Example 1. Find $\int \frac{7 x^{3}-4 x^{2}-32 x-37}{(x+2)(2 x-1)\left(x^{2}+2 x+3\right)} d x$. | Solution. The integrand is a proper rational fraction $(n=3, m=4)$, which we will represent as the sum of three simpler fractions:
$$
\begin{aligned}
& \frac{7 x^{3}-4 x^{2}-32 x-37}{(x+2)(2 x-1)\left(x^{2}+2 x+3\right)}=\frac{A}{x+2}+\frac{B}{2 x-1}+\frac{C x+D}{x^{2}+2 x+3}= \\
& \quad=\frac{A(2 x-1)\left(x^{2}+2 x+... | 3\ln|x+2|-\frac{5}{2}\ln|2x-1|+\frac{3}{2}\ln(x^{2}+2x+3)-\frac{4}{\sqrt{2}}\operatorname{arctg}\frac{x+1}{\sqrt{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,222 |
Example 3. Find the integral of the improper rational function $I=\int \frac{x^{6}+2 x^{5}-x^{4}+x^{2}+2 x}{x^{4}+1} d x$. | Solution. Dividing the numerator by the denominator (it is recommended to do this "in column"), we single out the integer part of the fraction:
$$
\frac{x^{6}+2 x^{5}-x^{4}+x^{2}+2 x}{x^{4}+1}=x^{2}+2 x-1+\frac{1}{x^{4}+1}
$$
We decompose the proper fraction into the simplest ones, taking into account the equality
$... | \frac{x^{3}}{3}+x^{2}-x+\frac{1}{4\sqrt{2}}\ln\frac{x^{2}+\sqrt{x}+1}{x^{2}-\sqrt{2}x+1}+\frac{1}{2\sqrt{2}}\operatorname{arctg}(\sqrt{2}x+1)+\frac{1}{2\sqrt{2}}\operatorname | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,223 |
Example 4. Find $\int \frac{d x}{x^{4}-3 x^{3}+2 x^{2}}$. | Solution. The denominator can be factored as: $x^{2}\left(x^{2}-3 x+2\right)=x^{2}(x-1)(x-2)$. The integrand fraction is equal to the sum of four type I fractions.
$$
\begin{aligned}
& \frac{1}{x^{2}(x-1)(x-2)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x-1}+\frac{D}{x-2}= \\
& \quad=\frac{A x(x-1)(x-2)+B(x-1)(x-2)+C x^{2}(... | -\frac{1}{2x}+\frac{1}{4}\ln\frac{|x^{3}(x-2)|}{(x-1)^{4}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,224 |
Example 1 (kp. $1^{\circ}$ ). Find $\int \sin 5 x \cos 7 x d x$. | Solution. We have
$\int \sin 5 x \cos 7 x d x=\frac{1}{2} \int(-\sin 2 x+\sin 12 x) d x=\frac{1}{4} \cos 2 x-\frac{1}{24} \cos 12 x+C$. | \frac{1}{4}\cos2x-\frac{1}{24}\cos12x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,225 |
Example 2 (to $2^{\circ}, 3^{\circ}$ ). Find $\int \cos ^{6} 2 x d x$. | Solution. We have:
$$
\begin{aligned}
& \cos ^{6} 2 x=\left(\frac{1+\cos 4 x}{2}\right)^{3}=\frac{1}{8}\left(1+3 \cos 4 x+3 \cos ^{2} 4 x+\cos ^{3} 4 x\right)= \\
& =\frac{1}{8}\left(1+3 \cos 4 x+\frac{3}{2}(1+\cos 8 x)+\left(1-\sin ^{2} 4 x\right) \cos 4 x\right)= \\
& =\frac{1}{8}\left(\frac{5}{2}+4 \cos 4 x+\frac{3... | \frac{1}{8}(\frac{5}{2}x+\sin4x+\frac{3}{16}\sin8x-\frac{1}{12}\sin^{3}4x)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,226 |
Example 3 (to $4^{\circ}$ ). Find $\int \sin ^{4} x \cos ^{3} x d x$. | Solution. We are dealing with case 1); $m=3$ - an odd number. Let $\sin x=u$ and bring $\cos x$ under the differential sign, since $\cos x d x=d(\sin x)$. We have:
$$
\begin{aligned}
\int \sin ^{4} x\left(1-\sin ^{2} x\right) \cos x d x=\int & u^{4}\left(1-u^{2}\right) d u= \\
& =\frac{u^{5}}{5}-\frac{u^{7}}{7}+C=\fra... | \frac{1}{5}\sin^{5}x-\frac{1}{7}\sin^{7}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,227 |
Example 4 (to $4^{\circ}$ ). Find $\int \sin ^{2} x \cos ^{4} x d x$. | Solution. We are dealing with case 2). We reduce the degree of the integrand expression twice, then transform the product into a sum and combine like terms:
$$
\begin{gathered}
\int \sin ^{2} x \cos ^{2} x \cdot \cos ^{2} x d x=\int \frac{1}{4} \sin ^{2} 2 x \cdot \frac{1}{2}(1+\cos 2 x) d x= \\
=\frac{1}{8} \int \fra... | \frac{1}{16}(x-\frac{1}{12}\sin6x+\frac{1}{4}\sin2x-\frac{1}{4}\sin4x)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,228 |
Example 6 (to $5^{\circ}$ ) Find $\int \frac{\sin 2 x d x}{1+\cos 2 x+\sin ^{2} 2 x}$. | Solution. We have $R(\sin 2 x, \cos 2 x)=\frac{\sin 2 x}{1+\cos 2 x+\sin ^{2} 2 x} \quad$ and $R(-\sin 2 x, \cos 2 x)=-R(\sin 2 x, \cos 2 x)$. Case 1$)$ applies. Therefore, let $\cos 2 x=t$. This substitution is equivalent to bringing $\cos 2 x$ under the differential sign. Therefore, we can do without additional formu... | \frac{1}{6}\ln\frac{2-\cos2x}{1+\cos2x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,229 |
Example 7 (to $5^{\circ}$ ). Find $\int \frac{\sin ^{5} 3 x}{\cos ^{5} 3 x} d x$. | Solution. In this case, it is easier to switch to the tangent. We will show all transformations in the course of the solution. We have:
$$
\begin{aligned}
& \int \frac{\sin ^{5} 3 x}{\cos ^{5} 3 x} d x=\int \operatorname{tg}^{5} 3 x d x=\int \operatorname{tg}^{3} 3 x\left(\frac{1}{\cos ^{2} 3 x}-1\right) d x= \\
& =\i... | \frac{\operatorname{tg}^{4}3x}{12}-\frac{1}{6}\operatorname{tg}^{2}3x-\frac{1}{3}\ln|\cos3x|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,230 |
Example 8 (to $5^{\circ}$ ). Find $\int \frac{d x}{\sin ^{2} x+4 \cos ^{2} x}$. | Solution. It seems that the universal substitution $\operatorname{tg} \frac{x}{2}=t$ is theoretically acceptable. Let's try to save on transformations.
$$
\begin{aligned}
& \int \frac{d x}{\sin ^{2} x+4 \cos ^{2} x}=\int \frac{1}{\frac{\sin ^{2} x}{\cos ^{2} x}+4} \cdot \frac{d x}{\cos ^{2} x}=\int \frac{d(\operatorna... | \frac{1}{2}\operatorname{arctg}\frac{\operatorname{tg}x}{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,231 |
Example 1. Find $\int \frac{\operatorname{sh} x d x}{\sqrt{1+\operatorname{ch}^{2} x}}$
Solution It is not hard to notice the possibility of bringing $\operatorname{ch} x$ under the differential sign and applying the tabular formula 13:
$$
\int \frac{\operatorname{sh} x d x}{\sqrt{1+\operatorname{ch}^{2} x}}=\int \fr... | Solution. Consider the given integral as an integral of a rational function of the variable $u=e^{x}$ :
$$
\begin{aligned}
& \int \frac{d x}{\operatorname{ch} x}=2 \int \frac{d x}{e^{x}+e^{-x}}=2 \int \frac{e^{x} d x}{e^{2 x}+1}=\left\{\begin{aligned}
e^{x} & =u \\
e^{x} d x & =d u
\end{aligned}\right\}= \\
&=2 \int \... | 2\operatorname{arctg}e^{x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,232 |
Example 3. Find $\int \frac{x}{\operatorname{sh} \frac{x^{2}}{2}} d x$. | Solution. We have (see formula 4) from point $2^{\circ}$ )
$$
\begin{aligned}
& \int \frac{x d x}{\operatorname{sh} \frac{x^{2}}{2}}=\left\{\begin{array}{c}
\frac{x^{2}}{2}=u \\
x d x=d u
\end{array}\right\}=\int \frac{d u}{\operatorname{sh} u}=\int \frac{d u}{2 \operatorname{sh} \frac{u}{2} \operatorname{ch} \frac{u}... | \ln|\operatorname{}\frac{x^{2}}{4}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,233 |
Example 4. Find $\int \frac{d x}{\operatorname{ch}^{6} x}$. | Solution. We will use formula 1) from point $2^{\circ}$ and its consequence $1-\operatorname{th}^{2} x=\frac{1}{\operatorname{ch}^{2} x}$, obtained by dividing 1) by $\operatorname{ch}^{2} x$. We have:
$$
\begin{aligned}
& \int \frac{d x}{\operatorname{ch}^{6} x}=\int \frac{\left(\operatorname{ch}^{2} x-\operatorname{... | \operatorname{}x-\frac{2}{3}\operatorname{}^{3}x+\frac{1}{5}\operatorname{}^{5}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,234 |
Example 4. Find $\int \frac{d x}{\sqrt{7 x^{2}+9 x+1}}$. | Solution. Let $\sqrt{7 x^{2}+9 x+1}=\sqrt{7} x+t$. Then $7 x^{2}+9 x+$ $+1=7 x^{2}+2 \sqrt{7} x t+t^{2}$
\[
\begin{aligned}
& x=\frac{t^{2}-1}{9-2 \sqrt{7} t}, \quad d x=\frac{-2 \sqrt{7} t^{2}+18 t-2 \sqrt{7}}{9-2 \sqrt{7} t^{2}} d t \\
& \sqrt{7} x+t=\frac{\sqrt{7} t^{2}-\sqrt{7}}{9-2 \sqrt{7} t}+t=\frac{-\sqrt{7} t... | -\frac{1}{\sqrt{7}}\ln|2\sqrt{7}(\sqrt{7x^{2}+9x+1}-\sqrt{7}x)-9|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,235 |
Example 5. Find $\int \frac{d x}{\sqrt{7 x^{2}+9 x+1}}$. | Solution. Let $\sqrt{7 x^{2}+9 x+1}=t x+1$. Then $7 x^{2}+9 x+$ $+1=t^{2} x^{2}+2 x t+1, t=\frac{\sqrt{7 x^{2}+9 x+1}-1}{x}, x=\frac{2 t-9}{7-t^{2}}, t x+1=$ $=\frac{t^{2}-9 t+7}{7-t^{2}}, d x=\frac{2 t^{2}-18 t+14}{\left(7-t^{2}\right)^{2}} d t$
The given integral takes the form
$\int \frac{2 d t}{7-t^{2}}=\frac{1}{... | \frac{1}{\sqrt{7}}\ln|\frac{\sqrt{7x^{2}+9x+1}-1-\sqrt{7}x}{\sqrt{7x^{2}+9x+1}-1+\sqrt{7}x}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,236 |
Example 6. Find $\int \frac{d x}{\sqrt{x^{2}+3 x-4}}$. | Solution. We have $x^{2}+3 x-4=(x-1)(x+4)$. Let $\sqrt{x^{2}+3 x-4}=(x-1) t$, i.e., $(x-1)(x+4)=(x-1)^{2} t^{2}$, $x+4=(x-1) t^{2}$, $x=\frac{t^{2}+4}{t^{2}-1}$, $d x=\frac{-10 t d t}{\left(t^{2}-1\right)^{2}}$, $(x-1) t=\left(\frac{t^{2}+4}{t^{2}-1}-1\right) t=\frac{5 t}{t^{2}-1}$. In this case, $t=\frac{\sqrt{t^{2}+3... | \ln|\frac{\sqrt{x+4}+\sqrt{x-1}}{\sqrt{x+4}-\sqrt{x-1}}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,237 |
Example 7. Find $\int \frac{\sqrt{1+\sqrt[3]{x}}}{x \sqrt{x}} d x$. | Solution. Let's represent the integral in the standard form $\int x^{-3 / 2}\left(1+x^{1 / 3}\right)^{1 / 2} d x . \quad$ Since $\quad m=-\frac{3}{2}, \quad n=\frac{1}{3}, \quad p=\frac{1}{2} \quad$ and $\frac{m+1}{n}+p=-1-$ is an integer, we will perform the third substitution of Chebyshev $(a=1, b=1): x^{-1 / 3}+1=t^... | -2\sqrt{(1+x^{-1/3})^{3}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,238 |
Example 8. Find $\int \sqrt{x}(5-3 \sqrt[3]{x})^{2} d x$. | Solution. We have $m=\frac{1}{2}, n=\frac{1}{3}, p=2$. Let's make the first Chebyshev substitution $x=t^{6}$, or $t=\sqrt[6]{x}$. From here $\sqrt{x}=t^{3}, 5-3 \sqrt[3]{x}=$ $=5-3 t^{2}, d x=6 t^{5}$ Substituting into the desired integral, we get:
$$
\begin{aligned}
& \int \sqrt{x}(5-3 \sqrt[3]{x})^{2} d x=\int t^{3}... | \frac{50}{3}\cdotx\sqrt{x}-\frac{180}{11}x\sqrt[6]{x^{5}}+\frac{54}{13}x^{2}\sqrt[6]{x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,239 |
Example 9. Find $\int \frac{\sqrt[6]{5-3 \sqrt[3]{x}}}{\sqrt[3]{x^{2}}} d x$. | Solution. We have $m=-\frac{2}{3}, n=\frac{1}{3}, p=\frac{1}{6}$ and $\frac{m+1}{n}=-2-$ an integer, so we make the second Chebyshev substitution $5-3 \sqrt[3]{x}=t^{6}$. Therefore, $x^{1 / 3}=\frac{1}{3}\left(5-t^{6}\right), x^{2 / 3}=\frac{1}{9}\left(5-t^{6}\right)^{2}, x=\frac{1}{27}\left(5-t^{6}\right)^{3}$, $d x=-... | -\frac{6}{7}\sqrt[6]{(5-3\sqrt[3]{x})^{7}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,240 |
Example 10. Find $\int \frac{d x}{\sqrt[3]{(2 x-1)^{2}}+\sqrt{2 x-1}}$. | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,241 | |
Example 1. Determine the sign of the integral $I=\int_{-\pi}^{0} \sin x d x$, without calculating it. | Solution. Since the function $y=\sin x$ is negative in the interval $(-\pi ; 0)$, the given integral is negative.
Answer. $I<0$. | I<0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,242 |
Example 2. Compare the integrals $I_{1}=\int_{0}^{1} x d x$ and $I_{2}=\int_{0}^{1} x^{3} d x$, without calculating them | Solution. Since in the interval $(0 ; 1)$ we have $x>x^{3}$, then $\int_{0}^{1} x d x>$ $>\int_{0}^{1} x^{3} d x$
Answer. $I_{1}>I_{2}$ | I_{1}>I_{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,243 |
Example 3. Evaluate the integral $I=\int_{-\pi / 2}^{\pi / 2} \frac{2 d x}{7-3 \cos 2 x}$. | Solution. If $x \in\left(-\frac{\pi}{2} ; \frac{\pi}{2}\right)$, then $-1<\cos 2 x<1$, and therefore $4<7-3 \cos 2 x<10$. Consequently, $\frac{1}{5}<\frac{2}{7-3 \cos 2 x}<\frac{1}{2}$. Applying property 6.3) of the definite integral with $m=\frac{1}{5}, M=\frac{1}{2}, b-a=\pi$, we obtain the estimate $\frac{\pi}{5}<I<... | \frac{\pi}{5}<I<\frac{\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,244 |
Example 4. Calculate the definite integral $\int_{0}^{1} \frac{d x}{1+x^{2}}$. | Solution. We are dealing with a tabular antiderivative.
$$
\int_{0}^{1} \frac{d x}{1+x^{2}}=\left.\operatorname{arctg} x\right|_{0} ^{1}=\operatorname{arctg} 1-\operatorname{arctg} 0=\frac{\pi}{4}-0=\frac{\pi}{4}
$$
Answer. $\frac{\pi}{4}$. | \frac{\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,245 |
Example 5. Calculate the integral $\int_{0}^{2}\left(x^{2}+x-1\right) e^{x / 2} d x$. | Solution. Integrate by parts twice.
$$
\begin{gathered}
\int_{0}^{2}\left(x^{2}+x-1\right) e^{x / 2} d x=\left\{\begin{array}{cc}
x^{2}+x-1=u, & d u=(2 x+1) d x \\
e^{x / 2} d x=d v, & v=2 e^{x / 2}
\end{array}\right\}= \\
=\left.2\left(x^{2}+x-1\right) e^{x / 2}\right|_{0} ^{2}-2 \int_{0}^{2}(2 x+1) e^{x / 2} d x=
\e... | 2(3e-5) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,246 |
Example 6 Calculate the integral $\int_{5}^{12} \frac{\sqrt{x+4}}{x} d x$ and find the integral mean value of the integrand $f(x)=\frac{\sqrt{x+4}}{x}$ over the interval of integration $[5 ; 12]$. | Solution. We integrate by substitution. In this process, we transition to another definite integral with a different integrand and different limits.
Let $\sqrt{x+4}=u$. Then $x+4=u^{2}, x=u^{2}-4, d x=2 u d u$. Therefore, if $x=5$, then $u=3$, and if $x=12$, then $u=4$. We obtain
$$
\begin{aligned}
\int_{5}^{12} \fra... | 2\ln\frac{5e}{3};\frac{2}{7}\ln\frac{5e}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,247 |
Example 7. Calculate the integral $\int_{0}^{\pi / 2} \frac{\sin ^{3} x}{2+\cos x} d x$ | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,248 | |
Example 4. Calculate the area of the figure bounded by the astroid $x=4 \cos ^{3} t, y=4 \sin ^{3} t$. | Solution. Due to the symmetry of the astroid relative to the coordinate axes (Fig. 2.8), it is sufficient to compute one quarter of the area and then multiply the result by 4.
We use the formula $S=4 \int_{0}^{4} y d x$ (see
. We have
$$
y^{\prime}=\frac{x}{2}-\frac{1}{2 x}=\frac{x^{2}-1}{2 x}, 1+y^{\prime 2}=1+\left(\frac{x^{2}-1}{2 x}\right)^{2}=\frac{\left(1+x^{2}\right)^{2}}{4 x^{2}}
$$
and
$$
\begin{aligned}
& L=\int_{1}^{e} \sqrt{1+y^{\prime 2}} d x=\int_{1}^{e} \frac{1+x^{2}}{2 x} d x=\left.\frac{1}{2}... | \frac{1}{4}(e^{2}+1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,250 |
Example 6. Find the length of the arc of the curve $y=\arcsin \sqrt{x}-\sqrt{x-x^{2}}$ from $x_{1}=0$ to $x_{2}=1$. | Solution. We have $y^{\prime}=\frac{1}{2 \sqrt{x} \sqrt{1-x}}-\frac{1-2 x}{2 \sqrt{x-x^{2}}}=\frac{x}{\sqrt{x-x^{2}}}=$ $=\sqrt{\frac{x}{1-x}}$ $L=\int_{0}^{1} \sqrt{1+y^{\prime 2}} d x=\int_{0}^{1} \sqrt{1+\frac{x}{1-x}} d x=\int_{0}^{1} \frac{1}{\sqrt{1-x}} d x=-\left.2 \sqrt{1-x}\right|_{0} ^{1}=2$.
Answer. $L=2$. | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,251 |
Example 7. Find the length of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. | Solution. We will use the parametric equations of the ellipse and the corresponding formula (4) for the length: $x=a \cos t, y=b \sin t$, $d x=-a \sin t d t, d y=b \cos t d t, \sqrt{d x^{2}+d y^{2}}=\sqrt{a^{2} \sin ^{2} t+b^{2} \cos ^{2} t} d t=$ $=\sqrt{\left(b^{2}-a^{2}\right) \cos ^{2} t+a^{2}} d t=\sqrt{a^{2}+c^{2... | \int_{0}^{2\pi}\sqrt{1+\varepsilon^{2}\cos^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,252 |
Example 8. Find the length of the arc of the curve $r=a \cos ^{3} \frac{\varphi}{3}(a>0)$. | Solution. Since $r \geqslant 0$, then $\cos ^{3} \frac{\varphi}{3} \geqslant 0$, which means $\frac{\varphi}{3} \in \left[-\frac{\pi}{2} ; \frac{\pi}{2}\right]$, or $\varphi \in \left[-\frac{3 \pi}{2} ; \frac{3 \pi}{2}\right]$. The cosine function is an even function, so we will limit ourselves to calculating the lengt... | \frac{3\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,253 |
Example 9. Calculate the volume of the solid of revolution around the $O x$ axis of the figure bounded by the lines $2 y=x^{2}$ and $2 x+2 y=3$. | Solution. The desired volume $V$ is the difference $V_{1}-V_{2}$ of the volumes of two bodies. The first is obtained by rotating the segment $A B$ around the $O x$ axis, the second by rotating the arc of the parabola $A O B$ (Fig. 2.9). The limits of integration are the projections of points $A$ and $B$ onto the $O x$ ... | \frac{272}{15}\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,254 |
Example 10. Calculate the volume of the body formed by rotating the astroid $x=a \cos ^{3} t, y=a \sin ^{3} t$ around the line $y=-a$, where $a>0$. | Solution. We will use the symmetry of the astroid relative to the coordinate axes (Fig. 2.10). The desired volume $V$ is the difference between the volumes $V_{1}-V_{2}$ of two bodies. The first is obtained by rotating the arc $A B C$ of the astroid, and the second by rotating the arc $A D C$ around the line $y=-a$. In... | \frac{3}{4}\pi^{2}^{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,255 |
Example 11. Find the volume of the body bounded by two cylinders $x^{2}+y^{2}=R^{2}$ and $x^{2}+z^{2}=R^{2}$. | Solution. In Fig. 2.11, one-eighth of the body located in the first octant $(x \geqslant 0, y \geqslant 0, z \geqslant 0)$ is shown. The cross-section by a plane perpendicular to the $O x$ axis is a square. If the section is made through the point with abscissa $(x, 0,0)$, then the side of the square is $a=y=z=\sqrt{R^... | \frac{16}{3}R^{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,256 |
Example 12. The arc of the cubic parabola $y=\frac{1}{3} x^{3}$, enclosed between the points $O(0,0)$ and $A(1,1 / 3)$, rotates around the $O x$ axis. Find the area of the surface of revolution. | Solution. According to formula (9), we get:
$$
\begin{aligned}
& \qquad S=2 \pi \int_{0}^{1} \frac{1}{3} x^{3} \sqrt{1+x^{4}} d x=\left.\frac{\pi}{9}\left(1+x^{4}\right)^{3 / 2}\right|_{0} ^{1}=\frac{\pi}{9}(2 \sqrt{2}-1) \\
& \text { Answer. } S=\frac{\pi}{9}(2 \sqrt{2}-1)
\end{aligned}
$$
## Exercises
(below, all ... | \frac{\pi}{9}(2\sqrt{2}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,257 |
Example 1. The velocity of a body changes according to the law $v=\frac{1}{2} t^{2}$ (m/s). What distance will the body travel in 12 s? What is the speed of motion? | Solution. The path is equal to the integral of the speed:
$$
S=\int_{0}^{12} \frac{1}{2} t^{2} d t=\left.\frac{1}{6} t^{3}\right|_{0} ^{12}=288(M)
$$
The average speed is $v_{\mathrm{cp}}=\frac{S}{t}=\frac{288}{12}=24(\mathrm{m} / \mathrm{c})$.
Answer. $S=288 \mathrm{~m} ; v_{\text {avg }}=24 \mathrm{~m} / \mathrm{c... | S=288\mathrm{~};v_{\text{avg}}=24\mathrm{~}/\mathrm{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,258 |
Example 2. What work is required to stretch a spring by $18 \mathrm{~cm}$, if a force of $24 \mathrm{H}$ stretches the spring by $3 \mathrm{~cm}$? | Solution. According to Hooke's Law, the elastic force stretching the spring is proportional to this extension, i.e., $F(x)=k x$. According to the condition: $F(0.03 \text{m})=24 \text{N}$. From the equation $24=0.03 k$ we find $k=800$, so $F(x)=800 x$ is the force stretching the spring. The work done by this force is
... | 25.92 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,259 |
Example 3. Calculate the mass $m$ and the moment of inertia of a uniform flat rod of length $l$ relative to its end (density is $\rho$). | Solution. Let's align the rod with the segment $[0 ; l]$ of the $O x$ axis (the rod is defined by the graph of the function $y=0, x \in[0 ; l]$. Then $\left(\sqrt{1+y^{\prime 2}}=1\right)$
$$
m=\int_{0}^{l} \rho d x=\rho l
$$
The moment of inertia of the rod relative to its end is
$$
M_{y}=\int_{0}^{l} \rho x^{2} d ... | \rho;\frac{\rho^{3}}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,260 |
Example 4. For a homogeneous material curve consisting of one half of a cycloid arch $x=a(t-\sin t), y=a(1-\cos t)$, $t \in[0 ; \pi]$, find (assuming $\rho=1$ ):
1) the mass of the curve;
2) the static moments with respect to the coordinate axes;
3) the coordinates of the center of mass;
4) the moments of inertia with... | Solution. We will use the formulas for calculating the desired quantities in parametric form. First, we find the element of arc length of the curve. We have $d x=a(1-\cos t) d t, d y=a \sin t d t, d l=\sqrt{d x^{2}+d y^{2}}=$ $=\sqrt{a^{2}(1-\cos t)^{2}+a^{2} \sin ^{2} t} d t=\sqrt{a^{2}(2-2 \cos t)} d t=2 a \sin \frac... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,261 |
Example 7. Investigate the convergence of the improper integrals:
$$
\text { a) } \int_{1}^{+\infty} \frac{x^{2}+2 x+\sin 2 x}{x^{3} \ln x+3 x+5} d x ; \quad \text { b) } \quad \int_{1}^{+\infty} \frac{x^{2}+2 x+\sin 2 x}{x^{3} \ln ^{2} x+3 x+5} d x
$$ | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,262 | |
Example 1. Express the double integral $I=\int_{D} \int f(x, y) d x d y$ as iterated integrals, if $D-$ is the region depicted in Fig. 3.4, and the boundary is given by the equations
$$
y=\cos \frac{x}{2}, \quad x=-\frac{\pi}{2}, \quad y=\frac{x}{2}-\frac{\pi}{4}, \quad x=\frac{\pi}{2}
$$ | Solution. Let's write $D$ as a system of inequalities and form repeated integrals.
The analysis of Fig. 3.4 shows that the region $D$ is bounded from below by the graph of the function $y=\frac{x}{2}-\frac{\pi}{4}$, and from above by the graph of the function $\cos \frac{x}{2}$, and these functions are defined on the ... | \begin{aligned}I=&\int_{-\pi/2}^{\pi/2}\int_{x/2}^{\cos(x/2)}f(x,y)=\int_{-\pi/4}^{0}\int_{-\pi/2}^{2y+\pi/2}f(x,y)+\\&+\int_{0} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,263 |
Example 2. Change the order of integration:
$$
I=\int_{0}^{1} d x \int_{0}^{x^{2}} f(x, y) d y+\int_{1}^{\sqrt{2}} d x \int_{0}^{\sqrt{2-x^{2}}} f(x, y) d y
$$ | Solution. We will construct the region of integration (Fig. 3.5), express it as a system of inequalities, and transition to another iterated integral:
$$
\begin{aligned}
D:\left\{\begin{arr... | \int_{0}^{1}\int_{\sqrt{y}}^{\sqrt{2-y^{2}}}f(x,y) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,264 |
Example 3. Change the order of integration in the double integral $I=\int_{0}^{1} d x \int_{\sqrt{x}}^{2-x} f(x, y) d y$. Sketch the region of integration. | Solution. First, we need to construct the region of integration $D$. The integration in the inner integral is performed over $y$ from the parabola $y=\sqrt{x}$ to the line $y=2-x$, and in the outer integral - over $x$ from the point $x=0$ to the point $x=1$ (Fig. 3.6).
If we change the order of integration, then the i... | \int_{0}^{1}\int_{0}^{y^{2}}f(x,y)+\int_{1}^{2}\int_{0}^{2-y}f(x,y) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,265 |
Example 4. Calculate the double integral $I=\iint_{D} y^{2} \sin x d x d y$, where $D$ is the region bounded by the lines $x=0, x=\pi, y=0, y=1+\cos x$. | Solution. The region $D$ is described by the system of inequalities
$$
\left\{\begin{array}{l}
0 \leqslant x \leqslant \pi \\
0 \leqslant y \leqslant 1+\cos x
\end{array}\right.
$$
Therefore, the given double integral can be written as an iterated integral:
$$
\begin{aligned}
I=\int_{0}^{\pi} \sin x d x \int_{0}^{1+... | \frac{4}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,266 |
Example 5. Calculate the double integral $I=\int_{D} \int(x+y) d x d y$, where $D-$ is the region bounded by the lines $x=0, y=x^{2}+x-3$, $2 y=3 x(x \geqslant 0)$. | Solution. The region $D$ is shown in Fig. 3.7. It is bounded from above by the segment of the line $y=\frac{3}{2} x$, from below by the parabola $y=x^{2}+x-3$, and from the left by the $O y$ axis. Let's determine the abscissa of the point $A$ of intersection of the line and the parabola:
$$
\left\{\begin{array}{l}
y=\... | \frac{14}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,267 |
Example 6. Calculate the areas of figures bounded by the curves:
a) $\left\{\begin{array}{l}y=x \sqrt{9-x^{2}} \\ y=0(0 \leqslant x \leqslant 3)\end{array}\right.$
b) $\left\{\begin{array}{l}y=2 x-x^{2}+3 \\ y=x^{2}-4 x+3\end{array}\right.$ | Solution. We will construct the corresponding regions (Fig. 3.8) and determine the appropriate limits of integration from them, omitting the systems of inequalities. Thus:
a) $S=\int_{D} \int d x d y=\int_{0}^{3} d x \int_{0}^{x \sqrt{9-x^{2}}} d y=$
$=\left.\int_{0}^{3} d x \cdot y\right|_{0} ^{x \sqrt{9-x^{2}}}=\in... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,268 |
Example 7. Calculate the area of the figure bounded above by the lines $x-y+2=0, y=2-\frac{1}{2} x$, and below by the parabola $y=x^{2}-2 x-$ $-8$. | Solution. Referring to Fig. 3.9. The equation of $AC$ is $y=x+2$, and $CB$ is described by the equation $y=2-\frac{1}{2} x$. Then,
$$
S=\int_{-2}^{0} d x \int_{x^{2}-2 x-8}^{x+2} d y+\int_{0}^{2} d x \int_{x^{2}-2 x-8}^{2-x / 2} d y=42
$$
Answer. $S=42$. | 42 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,269 |
Example 8. Calculate the volume of the body bounded above by the surface $z=x y^{2}$, and below by the rectangle $0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 2$. | Solution. The required volume can be calculated using the formula
$$
V=\int_{D} \int_{2} z d x d y, \text { where } D-\text { is the given rectangle, and } z=x y^{2} .
$$
The double integral can be reduced to the product of definite integrals, whose limits of integration are numbers:
$$
V=\int_{0}^{1} x d x \cdot \i... | \frac{4}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,270 |
Example 1. Calculate the area of the figure bounded by the curve $(2 x+3 y+5)^{2}+(x+2 y-3)^{2}=64$. | Solution. Let $\left\{\begin{array}{l}2 x+3 y+5=u, \\ x+2 y-3=v,\end{array}\right.$ or $\left\{\begin{array}{l}x=2 u-3 v-19, \\ y=-u+2 v+11 .\end{array}\right.$
We have: $\frac{\partial x}{\partial u}=2, \frac{\partial x}{\partial v}=-3, \frac{\partial y}{\partial u}=-1, \frac{\partial y}{\partial v}=2, I(u, v)=\left|... | 64\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,271 |
Example 2. Calculate the area of the figure bounded by the closed curve $\left(x^{2}+y^{2}\right)^{2}=4 a y^{3}, a>0$. | Solution. We transition to polar coordinates:
$$
\left\{\begin{array}{l}
x=r \cos \varphi \\
y=r \sin \varphi
\end{array}\right.
$$
Then the equation of the given curve takes the form
$$
\left(r^{2}\left(\cos ^{2} \varphi+\sin ^{2} \varphi\right)\right)^{2}=4 a \cdot r^{3} \sin ^{3} \varphi, \quad \text { or } \quad... | \frac{5}{2}\pi^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,272 |
Example 3. Calculate the volume of the body bounded by the surfaces $a z=a^{2}-x^{2}-y^{2}, z=0, x^{2}+y^{2}+a x=0(a>0)$. | Solution. The given body is a part of a paraboloid of revolution $a z=a^{2}-x^{2}-y^{2}$, enclosed within the cylinder $x^{2}+y^{2}+a x=0(a>0)$.
The integration domain $D$ is the circle $x^{2}+y^{2}+a x \leqslant 0$, or $\left(x+\frac{a}{2}\right)^{2}+y^{2} \leqslant \frac{a^{2}}{4}$, located in the II and III quadran... | \frac{5\pi^{3}}{32} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,273 |
Example 1. Calculate the surface area of $z=\frac{x y}{a}$, located inside the cylinder $x^{2}+y^{2}=a^{2}, a>0$. | Solution. We have $z_{x}^{\prime}=\frac{y}{a}, \quad z_{y}^{\prime}=\frac{x}{a}, \quad \sqrt{1+z_{z}^{\prime 2}+z_{y}^{\prime 2}}=$ $=\frac{1}{a} \sqrt{a^{2}+y^{2}+x^{2}}$. The domain of integration is the circle $x^{2}+y^{2} \leqslant a^{2}$. We will transition to polar coordinates. We have:
$$
\begin{gathered}
S=\in... | \frac{2\pi}{3}^{2}(2\sqrt{2}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,274 |
Example 3. Calculate the mass of the surface $z=x y$, located inside the cylinder $x^{2}+\frac{y^{2}}{4}=1$, if the density is $\rho=\frac{|z|}{\sqrt{1+x^{2}+y^{2}}}$. | Solution. Given the symmetries of the integration region $\sigma$: $x^{2}+\frac{y^{2}}{4} \leqslant 1$, the equation of the surface, and the density function, it is sufficient to compute the integral over one quarter of the region and multiply the result by 4: $m=4 \int_{D} \rho(x, y, z) d \sigma$, where $D$ is the qua... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,275 |
Example 4. Calculate the moments of inertia of the circle $x^{2}+y^{2} \leqslant-x$ with respect to the coordinate axes and the origin. Assume the density of the circle is equal to one. | Solution. We will transition to polar coordinates. The circle $x^{2}+y^{2} \leqslant -x$, i.e., $\left(x+\frac{1}{2}\right)^{2}+y^{2} \leqslant \frac{1}{4}$, is located in the second and third quadrants, so $\frac{\pi}{2} \leqslant \varphi \leqslant \frac{3 \pi}{2}$. In this case, $0 \leqslant r \leqslant -\cos \varphi... | I_{Ox}=\frac{\pi}{64},I_{Oy}=\frac{5\pi}{64},I_{O}=\frac{3\pi}{32} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,276 |
Example 5. Calculate the coordinates of the center of gravity of the square $0 \leqslant x \leqslant 2,0 \leqslant y \leqslant 2$ with density $\rho=x+y$. | Solution. First, we calculate the mass and static moments of the square. We have:
$$
m=\iint_{D} \rho d x d y=\int_{0}^{2} d x \int_{0}^{2}(x+y) d y=8
$$
$$
\begin{gathered}
M_{x}=\int_{D} x \rho d x d y=\int_{0}^{2} x d x \int_{0}^{2}(x+y) d y=\left.\int_{0}^{2}\left(x^{2} y+\frac{x y^{2}}{2}\right)\right|_{0} ^{2} ... | x_{}=\frac{7}{6},y_{}=\frac{7}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,277 |
Example 6. Find the coordinates of the center of gravity of a homogeneous plane figure bounded by the lines $y=\frac{1}{2} x^{2} ; y=2$. | Solution. Since the figure is homogeneous (we assume $\rho(x, y)=1$), then due to its symmetry relative to the $O y$ axis (Fig. 3.15), the abscissa of the center of gravity $\bar{x}=0$. For the ordinate, we have:
 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,278 |
Example 1. Calculate the mass of the tetrahedron bounded by the planes $x=0, y=0, z=0$ and $x / 10+y / 8+z / 3=1$, if the density distribution of mass at each point is given by the function $\rho=(1+x / 10+y / 8+z / 3)^{-6}$. | Solution. We have $m=\iint_{W} \int \rho d V$. The triple integral is reduced to a double and a definite one (see point $4^{\circ}$):
$$
m=\iint_{D} d x d y \int_{0}^{z} \frac{d z}{\left(1+\frac{x}{10}+\frac{y}{8}+\frac{z}{3}\right)^{6}}
$$
The upper limit, or the exit point from the region, is the ordinate of the po... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,279 |
Example 2. Calculate the static moments of the tetrahedron bounded by the coordinate planes and the plane $x+y+z=1$, if the density of the tetrahedron is $\rho=x y$. | Solution.
$$
\begin{gathered}
M_{x y}=\iiint_{W} \rho d V=\iiint_{W} x y z d x d y d z=\int_{0}^{1} x d x \int_{0}^{1-x} y d y \int_{0}^{1-x-y} z d z= \\
=\int_{0}^{1} x d x \int_{0}^{1-x} y \cdot \frac{(1-x-y)^{2}}{2} d y= \\
=\frac{1}{2} \int_{0}^{1} x d x \int_{0}^{1-x}\left(y(1-x)^{2}-2(1-x) y^{2}+y^{3}\right) d y... | M_{xy}=\frac{1}{720},M_{yz}=\frac{1}{360},M_{xz}=\frac{1}{360} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,280 |
Example 3. Calculate the mass of a part of the cone $36\left(x^{2}+y^{2}\right)=z^{2}$, located inside the cylinder $x^{2}+y^{2}=1(x \geqslant 0, z \geqslant 0)$, if the density is $\rho=\frac{5\left(x^{2}+y^{2}\right)}{6}$. | Solution. The limits of the variable $z: z=0$ and $z=6 \sqrt{x^{2}+y^{2}}$, since a vertical line will intersect the body at points on the base $(z=0)$ and the cone. The region of integration for the double integral is the circle $x^{2}+y^{2} \leqslant 1$, so it is convenient to switch to polar coordinates. We have:
$... | 2\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,281 |
Example 4. Calculate the volume of the body bounded by the surfaces: $x+z=6 ; y=\sqrt{x} ; y=2 \sqrt{x} ; z=0$. | Solution. The volume $V$ of a spatial body is equal to the triple integral over the spatial domain $W$ occupied by the body:
$$
V=\iint_{H} \int d x d y d z
$$
The first of the equations defining the boundary of the body in the problem statement is the equation of a plane parallel to $O y$ and intersecting the axes $... | \frac{48}{5}\sqrt{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,282 |
Example 5. Calculate the moment of inertia of a right circular cylinder of radius $R$ and height $H$ about a diameter of its middle section, if the density is constant and equal to $k$. | Solution. Let's choose a coordinate system such that the $O z$ axis is directed along the axis of the cylinder, and the plane $O x y$ passes through its center. We need to compute $I_{x}=\iint_{W} \int\left(y^{2}+z\right) \cdot k d x d y d z$.
We switch to cylindrical coordinates (omitting some details):
$$
\begin{al... | I_{x}=k\piHR^{2}(\frac{2}{3}H^{2}+\frac{R^{2}}{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,283 |
Example 6. Calculate the coordinates of the center of gravity of the upper half of the sphere $x^{2}+y^{2}+z^{2}=R$, assuming the density is $k$. | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,284 | |
Example 1. Calculate the line integral of the first kind $\int_{L} \sqrt{x^{3} y} d l$, where $L-$ is the arc of the cubic parabola $y=x^{3}$, connecting the points $O(0,0)$ and $A(1,1)$. | Solution. We have $d l=\sqrt{1+y^{\prime 2}} d x=\sqrt{1+9 x^{4}} d x$. Therefore,
$$
\begin{aligned}
& \int_{L} \sqrt{x^{3} y} d l=\int_{0}^{1} \sqrt{x^{3} \cdot x^{3}} \cdot \sqrt{1+9 x^{4}} d x=\int_{0}^{1}\left(1+9 x^{4}\right)^{1 / 2} \frac{d\left(1+9 x^{4}\right)}{36}= \\
&=\left.\frac{1}{36} \cdot \frac{2}{3}\l... | \frac{1}{54}(10\sqrt{10}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,285 |
Example 2. Find the mass of the curve $y=\frac{1}{2} x^{2}+1$, if its linear density is $\rho(x)=x+1, x \in[0 ; \sqrt{3}]$. | Solution. We have $y=\frac{1}{2} x^{2}+1, \quad y^{\prime}=x, \quad d l=\sqrt{1+y^{\prime 2}} d x=$ $=\sqrt{1+x^{2}} d x$. Further,
$$
\begin{aligned}
m=\int_{l} \rho(x) d l= & \int_{0}^{\sqrt{3}}(x+1) \sqrt{1+x^{2}} d x= \\
& =\int_{0}^{\sqrt{3}} x \sqrt{1+x^{2}} d x+\int_{0}^{\sqrt{3}} \sqrt{1+x^{2}} d x=
\end{align... | \frac{7}{3}+\frac{1}{2}(\ln(2+\sqrt{3})+2\sqrt{3}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,286 |
Example 3. Find the mass and coordinates of the center of gravity of the arc of the astroid $x=10 \cos ^{3} t, y=10 \sin ^{3} t, 0 \leqslant t \leqslant \frac{\pi}{2}$, if its density $\rho=1$. | Solution. First, we determine the element of arc length $d l=\sqrt{x^{\prime 2}+y^{\prime 2}} d t . \quad$ We have $\quad x^{\prime}=-30 \cos ^{2} t \sin t, \quad y^{\prime}=30 \sin ^{2} t \cos t$, $d l=30 \sqrt{\cos ^{4} t \sin ^{2} t+\sin ^{4} t \cos ^{2} t} d t=30 \sin t \cos t d t=15 \sin 2 t d t$.
Next, we sequen... | 15,M_{}(4,4) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,287 |
Example 4. Calculate the mass of a quarter of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=$ $=1, x \geqslant 0, y \geqslant 0$, if the density is $\rho=x y$. | Solution. The mass of the arc of the ellipse will be calculated using the formula $m=$ $=\int_{L} x y d l$. We will use the parametric equations of the ellipse: $x=a \cos t, y=b \sin t, t \in\left[0 ; \frac{\pi}{2}\right]$. We have:
$x^{\prime}=-a \sin t, y^{\prime}=b \cos t, d l=\sqrt{x^{\prime 2}+{y^{\prime}}^{2}} d... | \frac{(^{2}++b^{2})}{3(+b)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,288 |
Example 5. Calculate the mass of the material segment $AB$, if $A(-2,1,0), B(-1,3,5)$, and the density at each point $M$ is proportional to the distance from $M$ to $A$ with a proportionality coefficient $k$. | Solution. Let's write down the equations of the segment $A B$. As a directing vector of the segment $A B$, we can take the vector $\overrightarrow{A B}=(1,2,5)$. Then the equations of $A B$ are $x=-2+t, y=1+2 t, z=5 t, t \in[0 ; 1]$. Now let's compute the differential $d l$ of the length of the segment:
$$
d l=\sqrt{(... | 15k | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,289 |
Example 6. Given points $A(4,5), B(4,0)$ and $C(0,5)$. Calculate the second type integral $\int_{L}(4 x+8 y+5) d x+(9 x+8) d y$, where $L:$ a) segment $O A$; b) broken line $O C A$; c) parabola $y=k x^{2}$, passing through points $O$ and $A$. | Solution. The calculation boils down to formulating equations for $L$ and reducing the line integral to a definite integral. For clarity, we will use a diagram (Fig. 3.21).
a) $O A: y=\frac{5}{4} x, x \in[0 ; 4], d y=\frac{5}{4} d x$.
## $I(O A)=$
$=\int_{0}^{4}(4 x+10 x+5) d x+\left(9 x \cdot \frac{5}{4}+10\right) d... | 262;252;\frac{796}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,290 |
Example 7. Calculate the integral $I=\int_{\overparen{A B}}\left(x^{2}-2 x y\right) d x+\left(y^{2}-\right.$ $-2 x y) d y$, where $\overparen{A B}$ is the arc of a parabola connecting points $A(-1,1)$ and $B(1,1)$ and passing through the origin. | Solution. The arc $\overparen{A B}$ can be given by the function $y=x^{2}$, $x \in[-1,1]$. Then $d y=2 x d x$, and the line integral reduces to a definite integral:
$$
\begin{aligned}
I= & \overbrace{\overparen{A B}}\left(x^{2}-2 x y\right) d x+\left(y^{2}-2 x y\right) d y= \\
& =\int_{-1}^{1}\left[\left(x^{2}-2 x^{3}... | -\frac{14}{15} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,291 |
Example 8. Check the conditions of Green's theorem for the line integral $\int_{L} 2 x y d x + x^{2} d y$ and compute this integral along the parabola $y=\frac{x^{2}}{4}$ from the origin to the point $A(2,1)$. | Solution. We have $P(x, y)=2 x y, Q(x, y)=x^{2}$. These functions are defined, continuous, and differentiable at any point $(x, y)$ in the plane. We have $\frac{\partial P}{\partial y}=2 x, \frac{\partial Q}{\partial x}=2 x$. The conditions of Green's theorem are satisfied. Therefore, the given integral is independent ... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,292 |
Example 9. Calculate the work of the force $\vec{F}=(x-y, 1)$ along the semicircle $x^{2}+y^{2}=4(y \geqslant 0)$ when a material point moves from $A(2,0)$ to $B(-2,0)$. | Solution. The work of force $F$ is calculated by the formula (see section 5.2) $A=\int_{L} \vec{F} \cdot \overrightarrow{d s}$, where $\vec{F}=(x-y, 1), \overrightarrow{d s}=(d x, d y)$. We will use the parametric equations of $L$ (Fig. 3.22): $x=2 \cos t, y=2 \sin t, t \in[0 ; \pi]$. In this case,
 d x+2 x y d y$, where $L-$ is the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. | Solution. We apply Green's formula and compute the double integral, transitioning to "generalized" polar coordinates. We have: $P=x^{2}-y^{2}, Q=2 x y, \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=2 y+2 y=4 y$. Therefore,
$$
\begin{aligned}
& \int\left(x^{2}-y^{2}\right) d x+2 x y d y=4 \iint_{D} y d x ... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,294 |
Example 11. Calculate the integral $I=\oint_{L} \frac{y}{x^{2}+y^{2}} d x-\frac{x}{x^{2}+y^{2}} d y$, where $L$ is the circle:
a) $x^{2}+y^{2}=1$, b) $(x-1)^{2}+y^{2}=1$, c) $(x-1)^{2}+(y-1)^{2}=1$. | Solution. Let's check the conditions of Green's theorem.
$$
P=\frac{y}{x^{2}+y^{2}}, Q=-\frac{x}{x^{2}+y^{2}}, \frac{\partial P}{\partial y}=\frac{x^{2}-y^{2}}{\left(x^{2}+y^{2}\right)^{2}}, \frac{\partial Q}{\partial x}=\frac{x^{2}-y^{2}}{\left(x^{2}+y^{2}\right)^{2}}
$$
If $(x, y) \neq(0,0)$, then the functions $P$... | -2\pi;doesnotexist;0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,295 |
Example 1. Calculate the integral $I=\iint_{\sigma} \frac{d \sigma}{(1+x+z)^{2}}$, where $\sigma-$ is the part of the plane $x+y+z=1$, located in the first octant. | Solution. The surface $\sigma$ can be written explicitly: $z=1-x-y$. From this, $z_{x}^{\prime}=-1, z_{y}^{\prime}=-1, d \sigma=\sqrt{1+{z_{x}^{\prime}}^{2}+{z_{y}^{\prime}}^{2}} d x d y=\sqrt{3} d x d y$. The projection of the surface $\sigma$ onto the plane $O x y$ is a triangle $0 \leqslant x \leqslant 1,0 \leqslant... | \frac{\sqrt{3}}{2}(2\ln2-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,297 |
Example 2. Calculate the area of the part of the paraboloid of revolution $3 y=x^{2}+z^{2}$, which is located in the first octant and bounded by the plane $y=6$. | Solution. Apply the formula $S=\int_{\sigma} \int d \sigma$. It is convenient to project $\sigma$ onto the plane $O x z$, in which case the surface $\sigma$ is explicitly given by the equation $y=\frac{1}{3}\left(x^{2}+z^{2}\right)$, and the projection of $\sigma$ onto the plane $O x z$ is a quarter circle $x^{2}+z^{2}... | \frac{39\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,298 |
Example 3. Calculate the area of the part of the surface of the paraboloid of revolution $2z = x^2 + y^2$, enclosed within the cylinder $x^2 + y^2 = R^2$. | Solution. We apply the formula $S=\iint d \sigma$, where $\sigma-$ is the surface of the paraboloid $z=\frac{x^{2}+y^{2}}{2}$ above the circle $x^{2}+y^{2} \leqslant R^{2}$, and we will perform the calculations using a different scheme, unlike in Example 2. We transition to the double integral $S=\iint_{x^{2}+y^{2} \le... | \frac{2\pi}{3}(\sqrt{(1+R^{2})^{3}}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,299 |
Example 4. Calculate the integral $I=\iint_{\sigma}(x \cos \alpha+\cos \beta+$ $\left.+x z^{2} \cos \gamma\right) d \sigma$, where $\sigma$ is the part of the sphere $x^{2}+y^{2}+z^{2}=1$ located in the first octant. | Solution. Let $D_{1}, D_{2}, D_{3}$ be the projections of the surface of the unit sphere onto the coordinate planes $O y z, O x z, O x y$ respectively. We will compute each of the component integrals separately:
$$
\begin{aligned}
& I_{\mathrm{I}}=\iint_{\sigma} x \cos \alpha d \sigma=\iint_{D_{1}} x d y d z=\iint_{D_... | \frac{2}{15}+\frac{5\pi}{12} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,300 |
Example 5. Calculate the flux P of the vector field $\vec{F}=y z \vec{i}+$ $+x z \vec{j}+x y \vec{k}$ through the portion of the plane $x+y+z=1$, located in the first octant, along the normal vector to this plane. | Solution. The flux of the vector field $\vec{F}$ is calculated using the formula $\Pi=\int_{\sigma} \int(\vec{F} \cdot \vec{n}) d \sigma$. For the segment of the plane $z=1-x-y$, we have $\cos \gamma=\frac{1}{\sqrt{3}}$ (the normal vector of the plane $x+y+z=1$ has coordinates $\vec{N}=\{1,1,1\},|\vec{N}|=\sqrt{3}$, th... | \frac{1}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,301 |
Example 6. Find the static moments with respect to the coordinate planes and the coordinates of the center of gravity of a homogeneous triangular plate $x+y+z=a, x \geqslant 0, y \geqslant 0, z \geqslant 0$. | Solution. The mass of the plate can be easily found geometrically. It coincides with the area of an equilateral triangle with side $\sqrt{2} a: m=s=\frac{a^{2} \sqrt{3}}{2}$. Since this triangle is equilateral, the static moments relative to the coordinate planes are equal:
$$
\begin{aligned}
M_{x y}=M_{y z}= & M_{x z... | M_{xy}=M_{yz}=M_{xz}=\frac{\sqrt{3}}{6}^{3},x_{}=y_{}=z_{}=\frac{}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,302 |
Example 7. Calculate the moment of inertia about the $O y$ axis of the hemisphere $x^{2}+y^{2}+z^{2}=R^{2}, y \geqslant 0$. | Solution. We will use the formula $I_{y}=\iint\left(x^{2}+z^{2}\right) d \sigma$, where $\sigma$ can be explicitly defined as: $y=\sqrt{R^{2}-x^{2}-z^{2}}$. In this case, $d \sigma=\sqrt{1+y_{x}^{\prime 2}+{y_{z}^{\prime}}^{\prime 2}} d y d z=\frac{R}{\sqrt{R^{2}-x^{2}-y^{2}}} d y d z$, and $I_{y}=$ $=\iint_{x^{2}+z^{2... | \frac{4}{3}\piR^{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,303 |
Example 8. Calculate the second type integral
$$
\iint_{\sigma} x d y d z + y d x d z + z d x d y
$$
where $\sigma-$ is the outer side of the sphere $x^{2} + y^{2} + z^{2} = R^{2}$. | 4\piR^3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,304 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.