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3.17. Calculate
$$
\int_{L}\left(2 x^{2}+3 y^{2} i\right) d z
$$
where $L-$ is the straight line segment connecting the points $z_{1}=1+i, z_{2}=2+3 i$. | Solution. Let's find the equation of the line passing through the points $z_{1}=1+i$ and $z_{2}=2+3 i$. Assume it has the form $\operatorname{Im} z=k \operatorname{Re} z+b$, from which it follows that the system is valid:
$$
\left\{\begin{array} { l }
{ \operatorname { I m } z _ { 1 } = k \operatorname { R e } z _ { ... | \frac{1}{3}(-64+67i) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,766 |
3.18. Calculate
$$
\int_{L}\left(z^{2}+2 z \bar{z}\right) d z
$$
where $L-$ is the arc of the circle $|z|=1, \arg z \in[0, \pi]$. | Solution. It is clear that $L=\left\{z \in \mathbb{C} \mid z=e^{i \varphi}, \varphi \in[0, \pi]\right\}$, then
$$
\begin{gathered}
\int_{L}\left(z^{2}+2 z \bar{z}\right) d z=\left\{\begin{array}{l}
z=e^{i \varphi}, \varphi \in[0, \pi], \\
d z=i e^{i \varphi} d \varphi, \\
\bar{z}=\overline{e^{i \varphi}}=e^{-i \varphi... | -\frac{14}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,767 |
### 3.19. Calculate
$$
\int_{0}^{i} z \sin z d z
$$ | S o l u t i o n. Since the function $f(z)=z \sin z$ is analytic in any domain $D \subset \mathbb{C}$, encompassing any contour $L$ connecting the points 0 and $i$ (the verification of this fact is left to the reader), by the corollary of Cauchy's theorem we have:
$$
\begin{gathered}
\int_{0}^{i} z \sin z d z=-\int_{0}... | -\frac{i}{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,768 |
3.20. Calculate
$$
\int_{1}^{i} \frac{\ln ^{2} z}{z} d z
$$ | Solution. Let's take some region $D$ that does not contain 0. In this region, as it is not difficult to verify, the function
$$
f(z)=\frac{\ln ^{2} z}{z}
$$
is analytic, and therefore, by the corollary of Cauchy's theorem,
$$
\begin{gathered}
\int_{1}^{i} \frac{\ln ^{2} z}{z} d z=\int_{1}^{i} \ln ^{2} z d \ln z=\lef... | \frac{-i\pi^{3}}{24} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,769 |
3.21. Calculate
$$
\int_{L} \frac{60 e^{z}}{z(z+3)(z+4)(z+5)} d z
$$
where $L$ is the unit circle centered at the origin. | Solution. The integrand can be represented in the form
$$
\frac{f(z)}{z-a}
$$
where
$$
f(z)=\frac{60 e^{z}}{(z+3)(z+4)(z+5)}, a=0
$$
The function $f(z)$ is a single-valued analytic function in the domain $D$, such that $D=\{z \in \mathbb{C} \| z \mid \leqslant 2\}$, and $L \subset D$ is a closed continuous line bou... | 2\pii | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,770 |
3.22. Calculate
$$
\int_{L} \frac{\sin 3 z}{(6 z-\pi)^{3}} d z
$$
where $L$ is the unit circle centered at the origin. | Solution. The integrand can be represented in the form
$$
\frac{f(z)}{(z-a)^{3}}
$$
where $f(z)=6^{-3} \sin 3 z, a=\frac{\pi}{6}$. The function $f(z)$ is a single-valued analytic function in the domain $D$, such that
$$
D=\left\{z \in \simeq \| z \left\lvert\, \leqslant \frac{\pi}{2}\right.\right\}
$$
and $L \subse... | -\frac{i\pi}{24} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,771 |
3.23. Calculate
$$
\int_{L} \frac{e^{z}}{z^{2}+9} d z
$$
where $L$ is the circle: a) $|z-3 i|=2 ;$ b) $|z-3 i|=9$. | Solution.
a) It is obvious that:
$$
\begin{aligned}
& \int_{L} \frac{e^{z}}{z^{2}+9} d z=\frac{i}{6} \int\left(\frac{e^{z}}{z+3 i} \cdot \frac{e^{z}}{z-3 i}\right) d z= \\
= & \frac{i}{6} \int_{L} \frac{e^{z}}{z+3 i} d z-\frac{i}{6} \int_{L} \frac{e^{z}}{z-3 i} d z=\frac{i}{6}\left(I_{1}-I_{2}\right) .
\end{aligned}
... | \frac{2\pi\sin3}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,772 |
### 3.24. By computing the integral
$$
\frac{1}{2 \pi i} \int_{L} \frac{d z}{(z-\alpha)\left(z-\frac{1}{a}\right)}
$$
over the contour \( L=\{z \in \overline{S} \mid z \mid=1\} \), prove that for \(\alpha \in (0,1)\) the following equality holds:
$$
\int_{0}^{2 \pi} \frac{d \varphi}{1+\alpha^{2}-2 \alpha \cos \varph... | Solution. The integral
$$
\frac{1}{2 \pi i} \int_{L} \frac{d z}{(z-\alpha)\left(z-\frac{1}{\alpha}\right)}
$$
can be transformed as follows:
$$
\begin{aligned}
& \frac{1}{2 \pi i} \int_{L} \frac{d z}{(z-\alpha)\left(z-\frac{1}{\alpha}\right)}=\frac{\alpha}{2 \pi i\left(1-\alpha^{2}\right)} \int\left(\frac{1}{z-\frac... | \int_{0}^{2\pi}\frac{\varphi}{1+\alpha^{2}-2\alpha\cos\varphi}=\frac{2\pi}{1-\alpha^{2}} | Calculus | proof | Yes | Yes | olympiads | false | 30,773 |
3.28. Expand the function in a Laurent series in powers of $z$
$$
f(z)=\frac{1}{(z-1)(z-2)}
$$
a) in the annulus $1<|z|<2$;
b) in the annulus $1<|z-1|<2$;
c) in the annulus $1<|z-2|<2$. | Solution. The function
$$
f(z)=\frac{1}{(z-1)(z-2)}=\frac{1}{z-2}-\frac{1}{z-1}
$$
has two isolated singular points $z=1$ and $z=2$.
a) In the annulus $1<|z|<2$, the function $f(z)$ is analytic, and we can expand it into a Laurent series ${ }^{3}$ :
$$
\begin{aligned}
& f(z)=\frac{1}{(z-1)(z-2)}=\frac{1}{z-2}-\frac... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,775 |
3.29. Expand the function $f(z)=\frac{2 i a}{z^{2}+a^{2}}$ into a Laurent series in the annulus $0<|z-i a|<a$, where $a$ is a positive real number. | Solution. In the ring $0<|z-i a|<a$, the function $f(z)$ is analytic, we can expand it into a Laurent series:
$$
\begin{aligned}
f(z)= & \frac{2 i a}{z^{2}+a^{2}}=\frac{2 i a}{(z+i a)(z-i a)}=\frac{1}{z-i a}-\frac{1}{z+i a}=\frac{1}{z-i a}- \\
& -\frac{1}{2 i a} \frac{1}{1+\frac{z-i a}{2 i a}}=\frac{1}{z-i a}-\frac{1}... | \frac{1}{z-i}-\sum_{k=0}^{+\infty}\frac{i^{k-1}(z-i)^{k}}{2^{k+1}^{k+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,776 |
3.30. Expand the function $f(z)=\frac{z+2}{z^{2}+4 z+3}$ into a Laurent series in the annulus $2<|z+1|<+\infty$. | Solution. The function $f(z)$ has two isolated singular points, namely $z=-1$ and $z=-3$. In the annulus $2<|z+1|<+\infty$, the function $f(z)$ is analytic, and we can expand it into a Laurent series:
$$
\begin{aligned}
f(z)= & \frac{z+2}{z^{2}+4 z+3}=\frac{z+2}{(z+3)(z+1)}=\frac{z+2}{(z+1)\left(1+\frac{2}{z+1}\right)... | (z+2)\sum_{k=0}^{\infty}\frac{2^{k}}{(z+1)^{k+1}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,777 |
3.33. Compute
$$
\int \frac{e^{2 z}}{\left(z+\frac{\pi i}{2}\right)^{2}} d z
$$
where $L$ is the circle $|z|=1$. | Solution. A special point of the integrand function $f(z)=\frac{e^{2 z}}{\left(z+\frac{\pi i}{2}\right)^{2}}$ is the point $z=-\frac{\pi i}{2}$, which represents a pole of second order. There are no other special points of the function $f(z)$ within the region bounded by the circle $|z|=1$. Therefore, by the main theor... | 2\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,778 |
### 3.34. Compute the integral
$$
\int_{L} \frac{\sin z}{z\left(z-\frac{\pi}{2}\right)} d z
$$
where $L-$ is a rectangle bounded by the following lines: $x=2, x=-1, y=2, y=-1$. | Solution. The special points of the integrand function, which in this example has the form
$$
f(z)=\frac{\sin z}{z\left(z-\frac{\pi}{2}\right)}
$$
are the points $z=0$ and $z=\frac{\pi}{2}$. The point $z=0$ is a removable singularity of the integrand function, since at this point
$$
\lim _{z \rightarrow 0} f(z)=\lim... | 4i | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,779 |
3.36. Calculate
$$
\int_{L} \frac{e^{2 z}}{z^{4}+5 z^{2}-9} d z
$$
where $L-$ is the circle of radius $|z|=\frac{5}{2}$. | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,780 | |
3.37. Find the image of the following (functions: a) $f(t)=1$; b) $f(t)=t$; c) $f(t)=e^{-k t}$. | Solution.
a) $\bar{f}(p)=\int_{0}^{-1} e^{p t} d t=\lim _{\alpha \rightarrow+\infty} \int_{0}^{\alpha} e^{-p t} d t=$ $=-\left.\frac{1}{p} \lim _{\alpha \rightarrow+\infty} e^{-p t}\right|_{0} ^{\alpha}=-\frac{1}{p} \lim _{\alpha \rightarrow+\infty}\left(e^{-p \alpha}-1\right)=\frac{1}{p}$
(since $\lim _{\alpha \right... | \frac{1}{p},\frac{1}{p^2},\frac{1}{p+k} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,781 |
3.40. Find the originals of the following functions:
a) $\bar{f}(p)=\frac{p}{p^{2}-2 p+5}$
b) $\bar{f}(p)=\frac{1}{p^{3}-8} ;$
c) $\bar{f}(p)=\frac{p+1}{p(p-1)(p-2)(p-3)}$. | Solution. a) We use elementary techniques to decompose this fraction into the sum of fractions whose originals are known from the table:
$$
\frac{p}{p^{2}-2 p+5}=\frac{p-1+1}{(p-1)^{2}+4}=\frac{p-1}{(p-1)^{2}+4}+\frac{1}{(p-1)^{2}+4}
$$
(completing the square in the denominator).
By formulas 13,12 of Table 3.2, we h... | \begin{aligned}&)\quade^{}(\cos2+\frac{1}{2}\sin2)\\&b)\quad\frac{1}{12}(e^{2}-e^{-}\cos\sqrt{3}-\sqrt{3}e^{-}\sin\sqrt{3})\\&)\quad-\frac{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,782 |
3.41. Using the convolution theorem, find the original function of $\bar{f}(p)=\frac{p}{p^{4}-1}$. | Solution. Let's write $\bar{f}(p)$ as
$$
\bar{f}(p)=\frac{p}{p^{2}-1} \frac{1}{p^{2}+1}
$$
Since $\frac{p}{p^{2}-1} \doteqdot \operatorname{ch} t$ (formula 8 in table 3.2 for $\lambda=1$) and $\frac{1}{p^{2}+1} \doteqdot \sin t$ (formula 5 in table 3.1 for $\omega=1$), by the convolution theorem (formula 14 in table ... | \frac{\operatorname{ch}-\cos}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,783 |
3.42. Solve the differential equation (with initial condition):
\[
\left\{\begin{array}{l}
y^{\prime}(t)-2 y(t)=e^{t} \\
y(0)=0 \\
t>0
\end{array}\right.
\] | Solution. We proceed to the images:
$$
p \bar{y}(p)-2 \bar{y}(p)=\frac{1}{p-1} \text { or } \bar{y}(p)=\frac{1}{(p-1)(p-2)}
$$
We decompose this rational fraction into partial fractions:
$$
\frac{1}{(p-1)(p-2)}=\frac{A}{p-1}+\frac{B}{p-2} \Rightarrow 1 \equiv A(p-2)+B(p-1)
$$
Setting $p=1$, we get $A=-1$; for $p=2$... | y()=e^{2}-e^{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,784 |
3.43. Solve the Cauchy problem:
$$
\left\{\begin{array}{l}
y^{\prime}(t)+a y(t)=\varphi(t) \\
y(0)=y_{0} \\
t>0
\end{array}\right.
$$ | Solution. Transitioning to images, we get:
$$
p \bar{y}(p)-y_{0}+a \bar{y}(p)=\bar{\varphi}(p)
$$
or
$$
\bar{y}(p)=\frac{\bar{\varphi}(p)+y_{0}}{p+a}
$$
from which
$$
\bar{y}(p)=\frac{y_{0}}{p+a}+\frac{\bar{\varphi}(p)}{p+a}
$$
Transitioning to originals, using Table 3.2 and the convolution theorem, we have:
$$
... | y()=y_{0}e^{-}+\int_{0}^{}\varphi(\tau)e^{-(-\tau)}\tau | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,785 |
3.44. Solve the differential equation $y^{\prime \prime}-2 y^{\prime}-$ $-3 y=e^{3 t}$, if $y(0)=0, y^{\prime}(0)=0$. | Solution. We proceed to the images:
$$
p^{2} \bar{y}(p)-p y(0)-y^{\prime}(0)-2(p \bar{y}(p)-y(0))-3 \bar{y}(p)=\frac{1}{p-3}
$$
or
$$
\bar{y}(p)=\frac{1}{(p-3)\left(p^{2}-2 p-3\right)}=\frac{1}{(p+1)(p-3)^{2}} .
$$
We decompose this rational fraction into partial fractions:
$$
\frac{1}{(p+1)(p-3)^{2}}=\frac{A}{(p-... | y()=\frac{1}{4}e^{3}-\frac{1}{16}e^{3}+\frac{1}{16}e^{-} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,786 |
3.45. Solve the integral equation:
a) $y(\tau)=\int_{0}^{t} y(\tau) d \tau+1$;
b) $\int_{0}^{t} y(\tau) \sin (t-\tau) d \tau=1-\cos t$. | Solution.
a) We construct the image equation:
$$
\bar{y}(p)=\frac{\bar{y}(p)}{p}+\frac{1}{p} \Rightarrow \bar{y}(p)(p-1)=1 \Rightarrow \bar{y}(p)=\frac{1}{p-1}
$$
Therefore, the original function is $y(t)=e^{t}$.
b) The left side of the equation is the convolution of the function $y(t)$ and $\sin t$. Transitioning ... | y()=e^{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,787 |
3.46. Solve the system of differential equations:
a) $\left\{\begin{array}{l}\frac{d y_{1}(t)}{d t}+y_{1}(t)-y_{2}(t)=e^{t}, \\ \frac{d y_{2}(t)}{d t}+3 y_{1}(t)-2 y_{2}(t)=2 e^{t}, \\ y_{1}(0)=1, \quad y_{2}(0)=1, t>0 .\end{array}\right.$
b) $\left\{\begin{array}{l}\frac{d x}{d t}=x+2 y, \\ \frac{d y}{d t}=2 x+y+1 ;\e... | Solution.
a) Transitioning to images, we have the system of linear algebraic equations:
$$
\left\{\begin{array}{l}
p \overline{y_{1}}(p)-1+\overline{y_{1}}(p)-\overline{y_{2}}(p)=\frac{1}{p-1} \\
\overline{y_{2}}(p)-1+3 \overline{y_{1}}(p)-2 \overline{y_{2}}(p)=\frac{2}{p-1}
\end{array}\right.
$$
or
$$
\left\{\begi... | y_{1}()=y_{2}()=e^{},\quadx()=-\frac{2}{3}-2e^{-}+\frac{8}{3}e^{3},\quady()=\frac{1}{3}+2e^{-}+\frac{8}{3}e^{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,788 |
4.1. Find the general solution of the ODE $y^{\prime}=\frac{y^{2}}{x}$. | Solution. Rewrite the equation as $\frac{d y}{d x}=\frac{y^{2}}{x}$. Gather all terms depending on $y$ on the left side, and all terms depending on $x$ on the right side: $\frac{d y}{y^{2}}=\frac{d x}{x}$. Integrate both sides, each with respect to its own variable: $\int \frac{d y}{y^{2}}=\int \frac{d x}{x}$; it is co... | -\frac{1}{\ln(Cx)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,789 |
4.3. Solve the Cauchy problem: $\frac{d y}{d x}=\frac{1}{(x+y)^{2}}, y(0)=1$. | The solution is as follows. $1 \cdot \check{i} s t e p$. Introduce a new unknown function of substitution $t=x+y$, then $\frac{d t}{d x}=1+\frac{1}{t^{2}}$.
$2-\check{u}$ step. Separate the variables: $\frac{t^{2} d t}{1+t^{2}}=d x$, integrate both sides, each with respect to its own variable:
$$
\int \frac{t^{2} d t... | (x+y)-\operatorname{arctg}(x+y)=x+1-\frac{\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,790 |
4.19. Obtain the general solution of the equation $(x+y) d x+$ $+(x-y) d y=0$. | Solution. The functions $P(x, y)=x+y$ and $Q(x, y)=x-y$ are homogeneous functions of the first degree: $P(t x, t y)=$ $=t x+t y=t(x+y)=t P(x, y), Q(t x, t y)=t x-t y=t(x-y)=$ $=t Q(x, y)$. Introduce a new function $u(x)=\frac{y(x)}{x}$, hence $y(x)=x u(x), d y=u d x+x d u$. Substitute these into the original equation $... | x^2+2xy-y^2=C^2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,791 |
4.20. Obtain the general solution of the equation
$$
\frac{d y}{d x}=\frac{x+3 y-5}{4 x-y-7}
$$ | Solution. We form the determinant $\Delta=\left|\begin{array}{cc}1 & 3 \\ 4 & -1\end{array}\right|=-13 \neq 0$.
We have the 2nd case. Introduce new arguments $\alpha=x-h$ and function $\beta=y-k$, where $h$ and $k$ are found from the system of linear equations
$$
\left\{\begin{array}{l}
h+3 k-5=0 \\
4 h-k-7=0
\end{ar... | \frac{4.5\sqrt{0.75}}{0.75}\operatorname{arctg}\frac{((y-1)/(x-2))-0.5}{\sqrt{0.75}}+0.5\ln|\frac{((y-1)/(x-2))-0.5)^{2}}{0.75}+1|=\ln(C|x- | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,792 |
4.21. Obtain the general solution of the equation
$$
\frac{d y}{d x}=\frac{x+y+1}{4 x+4 y-1}
$$ | Solution. Let's form the determinant $\Delta=\left|\begin{array}{ll}1 & 1 \\ 4 & 4\end{array}\right|=0$. We have the third case. Introduce the substitution $z=x+y, 4 x+4 y=4 z, d y=d z-$ $-d x, \frac{d y}{d x}=\frac{d z}{d x}-1$. The equation takes the form $\frac{d z}{d x}-1=\frac{z+1}{4 z-1}$. Separating variables: $... | \frac{4}{5}(x+y)-\frac{1}{5}\ln|x+y|=x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,793 |
4.22. Obtain the general solution of the equation $x y^{3} d x+\left(x^{2} y^{2}-\right.$ $-1) d y=0$. | Solution. Let's make the substitution $y=w^{\beta}$, where the parameter $\beta$ will be determined from the condition of equality of the sums of the degrees of each of the monomials standing in parentheses before $d x$ and $d w$: $x w^{3 \beta} d x+\left(x^{2} w^{3 \beta-1}-w^{\beta-1}\right) \beta d w=0$. We equate t... | \ln|\frac{1}{yx}|+\frac{y^{2}x^{2}}{2}=\ln\frac{C}{|x|} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,794 |
4.38. Integrate the equation $\frac{d y}{d x}+x y=x^{2}$ using two methods: the method of variation of arbitrary constant and the Bernoulli method. | Solution. 1st method. Apply the method of variation of arbitrary constant.
1st step. Solve the corresponding homogeneous equation $\frac{d y}{d x}+x y=0$ by the method of separation of variables:
$$
\frac{d y}{y}=-x d x ; \int \frac{d y}{y}=-\int x d x ; \ln |y|=-\ln |x|+\ln C ; \quad y(x)=\frac{C}{x}
$$
2nd step. R... | y(x)=\frac{x^{3}}{4}+\frac{C}{x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,795 |
4.39. Integrate the Bernoulli equation
$$
\frac{d y}{d x}+\frac{y}{x}=\frac{1}{x^{2}} \cdot \frac{1}{y^{2}}
$$
using two methods: the Bernoulli method and the method of variation of arbitrary constants, after first reducing it to a linear equation. | The problem is solved. 1st method. First, we will use the Bernoulli method.
1st step. Represent the unknown function $y(x)$ as a product of two new unknown functions $u(x)$ and $v(x)$: $y(x)=u(x) \cdot v(x)$, where $\frac{d y}{d x}=v(x) \frac{d u}{d x}+u(x) \frac{d v}{d x}$.
2nd step. Substitute the expressions for $... | y(x)=\sqrt[3]{\frac{3}{2x}+\frac{C_{1}}{x^{3}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,796 |
4.59. $(2 y-3) d x+\left(2 x+3 y^{2}\right) d y=0, y(0)=1$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
4.59. $(2 y-3) d x+\left(2 x+3 y^{2}\right) d y=0, y(0)=1$. | Solution. Step 1. Check condition (4.5):
$$
\begin{gathered}
P(x, y)=2 y-3, Q(x, y)=2 x+3 y^{2}, \\
\frac{\partial P}{\partial y}=\frac{\partial(2 y-3)}{\partial y}=2 ; \quad \frac{\partial Q}{\partial x}=\frac{\partial\left(2 x+3 y^{2}\right)}{\partial x}=2 ; \quad \frac{\partial P}{\partial y}=\frac{\partial Q}{\par... | 2xy-3x+y^{2}-1=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,798 |
4.60. $\left(3 x^{2} y^{2}+7\right) d x+2 x^{3} y d y=0, y(0)=1$. | Solution. $1-\check{i}$ step. Check condition (4.5):
$$
P(x, y)=3 x^{2} y^{2}+7, Q(x, y)=2 x^{3} y
$$
$2-\bar{i}$ step. Use formula (4.6):
$$
\begin{aligned}
& U(x, y)=\int_{x_{0}}^{x}\left(3 x^{2} y^{2}+7\right) d x+\int_{y_{0}}^{y} 2 x_{0}^{3} y d y=C, \\
& U(x, y)=y^{2} \int_{x_{0}}^{x}\left(3 x^{2} d x+7 \int_{x... | y^{2}x^{3}+7C_{1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,799 |
4.61. Integrate the equation
$$
\left(x^{2} y^{2}-1\right) d x+2 x^{3} y d y=0
$$
by finding the integrating factor. | Solution.
$$
P(x, y)=x^{2} y^{2}-1, Q(x, y)=2 x^{3} y, \frac{\partial P}{\partial y}=2 x^{2} y, \frac{\partial Q}{\partial x}=6 x^{2} y
$$
Thus, the equality $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$ does not hold, and the equation is not an exact differential equation. First, we will try to find ... | xy^{2}+\frac{1}{x}=C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,800 |
4.78. Find the singular solutions of the equation
$$
F\left(x, y, \frac{d y}{d x}\right)=\left(\frac{d y}{d x}\right)^{2}-(6 x+y) \frac{d y}{d x}+6 x y=0
$$ | Solving this equation with respect to $\frac{d y}{d x}$, we obtain two equations: $\frac{d y_{1}}{d x}=y_{1}$ and $\frac{d y_{2}}{d x}=6 x$, the right-hand sides of which satisfy the condition of existence and uniqueness of the solution to the Cauchy problem at any point in the plane $(x, y)$. The general solutions of ... | -6x-12 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,801 |
4.79. Integrate the Lagrange equation
$$
y=x\left(\frac{d y}{d x}\right)^{2}-\frac{d y}{d x}
$$ | The problem is solved. Let's move to the parametric representation $p=\frac{d y}{d x}$, then according to formula (4.7) we get:
$$
\begin{aligned}
& a(p)=p^{2}, \quad b(p)=-p \\
& \frac{d x}{x p}+\frac{2}{p-1} x=\frac{1}{p^{2}-p} \\
& x=u(p) v(p) \\
& v \frac{d u}{d p}+u \frac{d v}{d p} \frac{2}{p-1} u v=\frac{1}{p^{2... | {\begin{pmatrix}\frac{p-\ln|p|+C}{(p-1)^{2}}\\xp^{2}-p\end{pmatrix}.} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,802 |
4.80. Integrate the Clairaut's equation
$$
y=x \frac{d y}{d x}+\frac{1}{2(d y / d x)}
$$ | Solution. Let $p=\frac{d y}{d x}$, then from the equation we get $y=x p+\frac{1}{2 p}$. Differentiate both sides of the last equation and replace $d y$ with $p d x: p d x=p d x+x d p-\frac{d p}{2 p^{2}}$. Transform the obtained equation: $d p\left(x-\frac{1}{2 p^{2}}\right)=0$. From this it follows: a) $d p=0, p=C=$ co... | Cx+\frac{1}{2C}ory^{2}=2x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,803 |
4.92. Find the solution to the Cauchy problem for the equation of the form
$y^{(n)}(x)=F(x): y^{\prime \prime \prime}=24 x, y(0)=3, y^{\prime}(0)=2, y^{\prime \prime}(0)=4$. | Solution. After the first integration, we get $y^{\prime \prime}(x)=\int 24 x d x+C_{1}=24 \cdot \frac{x^{2}}{2}+C_{1}$. After the second integration, we find $y^{\prime}(x)=\int\left(12 x^{2}+C_{1}\right) d x=12 \cdot \frac{x^{3}}{3}+C_{1} x+C_{2}$. Finally, after the third integration, we write the general solution:
... | y(x)=x^{4}+2x^{2}+2x+3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,804 |
4.93. Solve the Cauchy problem with the differential equation \(3 y^{\prime \prime}=4 \frac{x^{3}}{\left(y^{\prime}\right)^{2}}\) and the initial condition \(y(1)=0, y^{\prime}(1)=2\). After the substitution \(y^{\prime}=p(x)\), we obtain the equation \(3 p^{\prime}=4 \frac{x^{3}}{p^{2}}\). | Solution. Let's find its general solution, for this we rewrite the equation in the form: $3 \frac{d p}{d x}=4 \frac{x^{3}}{p^{2}}$. We separate the variables: $3 p^{2} d p=4 x^{3} d x$. We integrate both sides with respect to their variables:
$$
\int 3 p^{2} d p=\int 4 x^{3} d x, 3 \frac{p^{3}}{3}=4 \frac{x^{4}}{4}+C,... | y(x)=\int_{1}^{x}\sqrt[3]{^{4}+7} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,805 |
4.94. Solve the Cauchy problem: $y^{\prime \prime}-8 y^{3}=0 ; y(0)=-1 ; y^{\prime}(0)=2$. | Solution. Multiply both sides of the equation by $y^{\prime}$ and notice that the left side becomes a complete derivative:
$$
y^{\prime} \cdot y^{\prime \prime}-8 y^{3} \cdot y^{\prime}=\frac{d}{d x}\left(\frac{y^{\prime 2}}{2}-2 y^{4}\right)=0
$$
From this, considering the initial conditions, we find
$$
y^{\prime 2... | -\frac{1}{1+2x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,806 |
4.109. Find the solution to the Cauchy problem: $y^{\prime \prime}-3 y^{\prime}+2 y=0$, $y(0)=1, y^{\prime}(0)=0$. | Solution. We form the characteristic equation $r^{2}-3 r+2=0$. Its roots are real and distinct: $r_{1}=1, r_{2}=2$. The general solution is:
$$
y(x)=C_{1} e^{\eta x}+C_{2} e^{r_{2} x}=C_{1} e^{x}+C_{2} e^{2 x}
$$
We find $y^{\prime}(x)=C_{1} e^{x}+2 C_{2} e^{2 x}$. Using the initial conditions:
$$
\begin{gathered}
y... | y(x)=2e^{x}-e^{2x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,807 |
4.110. Find the solution to the Cauchy problem: $y^{\prime \prime}-4 y^{\prime}+4 y=0$, $y(0)=2, y^{\prime}(0)=1$. | Solution. We form the characteristic equation: $r^{2}-4 r+4=0$. Its roots are real and equal to each other: $r_{1}=r_{2}=r=2$. The general solution has the form:
$$
\begin{aligned}
& y(x)\left(C_{1}+x C_{2}\right) e^{r_{x}}=\left(C_{1}+x C_{2}\right) e^{2 x} \\
& y(x)=C_{1} e^{\eta x}+C_{2} e^{r_{2} x}=C_{1} e^{x}+C_{... | y(x)=(1-3x)e^{2x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,808 |
4.111. Find the solution to the Cauchy problem: $y^{\prime \prime}-16 y^{\prime}+20 y=0$, $y(0)=3, y^{\prime}(0)=0$. | Solution. We form the characteristic equation $r^{2}-16 r+20=0$. Its roots are complex conjugates:
$$
\begin{aligned}
r_{1,2} & =\frac{4 \pm \sqrt{(-4)^{2}-4 \cdot 1 \cdot 5}}{2 \cdot 1}=\frac{4 \pm \sqrt{-4}}{2}= \\
& =\frac{4 \pm \sqrt{4(-1)}}{2}=\frac{4 \pm 2 \cdot \sqrt{-1}}{2}=2 \pm i
\end{aligned}
$$
The genera... | y(x)=(3\cosx-6\sinx)e^{2x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,809 |
4.176. $\frac{d x}{d t}=-2 x, \frac{d y}{d t}=-y$. | Solution. We form the characteristic equation
$$
\left|\begin{array}{cc}
(-2-\lambda) & 0 \\
0 & (-1-\lambda)
\end{array}\right|=(\lambda+1)(\lambda+2)=0
$$
The roots of the characteristic equation are real, distinct, and negative:
$$
\lambda_{1}=-1<0, \lambda_{2}=-2<0
$$
(case $1 \mathrm{a})-$ stable node. | \lambda_{1}=-1<0,\lambda_{2}=-2<0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,812 |
4.177. $\frac{d x}{d t}=x-2 y, \frac{d y}{d t}=x+4 y$. | Solution. We form the characteristic equation
$$
\left|\begin{array}{cc}
(1-\lambda) & -2 \\
1 & (4-\lambda)
\end{array}\right|=\lambda^{2}-5 \lambda+6=0
$$
The roots of the characteristic equation are real, distinct, and positive:
$$
\lambda_{1}=2>0, \lambda_{2}=3>0
$$
(case $1 \mathrm{~b}$) - unstable node. | \lambda_{1}=2,\lambda_{2}=3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,813 |
4.179. $\frac{d x}{d t}=y, \frac{d y}{d t}=-4 x-2 y$. | Solution. We form the characteristic equation
$$
\left|\begin{array}{cc}
(0-\lambda) & 1 \\
-4 & (-2-\lambda)
\end{array}\right|=2 \lambda^{2}+2 \lambda+4=0
$$
The roots of the characteristic equation are complex conjugates, the real part is negative, and the imaginary part is non-zero:
$$
\left.\lambda_{1,2}=-\frac... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,814 |
4.180. $\frac{d x}{d t}=x-y, \frac{d y}{d t}=x+y$. | Solution. We form the characteristic equation
$$
\left|\begin{array}{cc}
(1-\lambda) & -1 \\
1 & (1-\lambda)
\end{array}\right|=\lambda^{2}-2 \lambda+2=0
$$
The roots of the characteristic equation are complex conjugates, with a positive real part and a non-zero imaginary part:
$$
\lambda_{1,2}=1 \pm i (\text{case }... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,815 |
4.184. $\frac{d x}{d t}=y, \frac{d y}{d t}=0$. | The solution is as follows. We form the characteristic equation
$$
\left|\begin{array}{cc}
(0-\lambda) & 1 \\
0 & (0-\lambda)
\end{array}\right|=\lambda^{2}=0
$$
The roots of the characteristic equation are real, coincident, and equal to zero: $\lambda_{1}=\lambda_{2}=0$. All points on the $O X$ axis are equilibrium ... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,818 |
4.185. $\frac{d x}{d t}=x, \frac{d y}{d t}=0$. | Solution. We form the characteristic equation
$$
\left|\begin{array}{cc}
(1-\lambda) & 1 \\
0 & (0-\lambda)
\end{array}\right|=\lambda^{2}=0
$$
The roots of the characteristic equation are real: $\lambda_{1}=1, \lambda_{2}=0$. All points on the Y-axis are equilibrium points (case $3 \mathrm{~d}$).
### 4.5.1.2.
PROB... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,819 |
4.201. $\frac{d x}{d t}=-x+4 y-4 x y^{3}, \frac{d y}{d t}=-2 y-x^{2} y^{2}$. | Solution. There is no general method for constructing a Lyapunov function. In most cases, it is sought in the form \( v(x, y) = a x^{2 n} + b y^{2 n} \). In this case, we will look for \( v(x, y) \) in the form: \( v(x, y) = a x^{2} + b y^{2} \). We have:
\[
\begin{gathered}
\frac{d v}{d t} = 2 a x \left( -x + 4 y - 4... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,820 |
4.202. $\frac{d x}{d t}=y-2 x^{3}, \frac{d y}{d t}=-x-5 y^{3}$. | Let's construct the function $v(x, y)=x^{2}+y^{2}$. We will show that it satisfies all the conditions of Theorem 4.14, i.e., it is a Lyapunov function.
a) It is differentiable everywhere, $v(x, y)>0$ for $x \neq 0$, $y \neq 0$, $v(0,0)=0$, and the origin is a point of strict minimum.
b) $\frac{d v}{d t}=2 x\left(y-2 x... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,821 |
4.203. $\frac{d x}{d t}=y-\frac{x}{2}-\frac{x y^{3}}{2}, \frac{d y}{d t}=-y-2 x+x^{2} y^{2}$. | Solution. We construct the function $v(x, y)=2 x^{2}+y^{2}$ and show that it is a Lyapunov function.
a) It is differentiable everywhere and $v(x, y)>0$ for $x \neq 0$, $y \neq 0$, $v(0,0)=0$ and the origin is a point of strict minimum.
b) $\frac{d v}{d t}=4 x\left(y-\frac{x}{2}-\frac{x y^{3}}{2}\right)+2 y\left(-y-2 ... | proof | Calculus | other | Yes | Yes | olympiads | false | 30,822 |
4.205. Investigate the stability by the first approximation
$$
\frac{d x}{d t}=2-e^{y}-3 x-\cos x, \frac{d y}{d t}=2 y+8 \sin x
$$ | The problem is solved. Let's write down the first three terms of the Maclaurin series expansion for the exponential, sine, and cosine functions.
$$
\begin{gathered}
e^{y}=1+y+\frac{y^{2}}{2}+\ldots ; \cos y=1-\frac{y^{2}}{2}+\frac{y^{4}}{24}-\ldots \\
\sin x=x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\ldots
\end{gathered}
$$... | \lambda_{1,2}=-\frac{1}{2}\\frac{\sqrt{7}i}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,823 |
Example 1.1. Find the integrals:
a) $\int \sin (7 x+3) d x$,
b) $\int \frac{d x}{x+5}$.
c) $\int(3 x+5)^{99} d x$,
d) $\int e^{x^{3}} x^{2} d x$
e) $\int \sin ^{3} x \cos x d x$. | Solution. a) Here the substitution $t=7 x+3$ is made, as in this case the indefinite integral is reduced to a tabular integral of sine
$$
\begin{aligned}
t=7 x+3 \Rightarrow d t & =d(7 x+3)=(7 x+3)^{\prime} d x=7 d x, \text { i.e. } \\
d t & =7 d x \Rightarrow d x=\frac{1}{7} d t
\end{aligned}
$$
Taking into account ... | \begin{aligned})&\quad-\frac{1}{7}\cos(7x+3)+C\\b)&\quad\ln|x+5|+C\\)&\quad\frac{(3x+5)^{100}}{300}+C\\)&\quad\frac{1}{3}e^{x^{3}}+C\\e)&\quad\frac | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,824 |
Example 1.2. Find $\int \frac{x d x}{1+x^{2}}$. | Solution. This integral could be found using the same technique as in Example 1.1. However, we will show how the integrand is formally transformed after introducing a new variable of integration. Introduce a new variable of integration $t$, related to the old variable by the equation $t=1+x^{2}$. Next, we transform the... | \frac{1}{2}\ln(1+x^{2})+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,825 |
Example 1.3. Compute $\int x e^{x} d x$. | Solution. Let $u=x, e^{x} d x=d v$. Then $d u=d x$, and $v=\int d v=\int e^{x} d x=e^{x}$ (it is not necessary to write the constant of integration here). By formula (1.10) we have
$$
\int x e^{x} d x=x e^{x}-\int e^{x} d x=x e^{x}-e^{x}+C=e^{x}(x-1)+C
$$ | e^{x}(x-1)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,826 |
Example 1.4. Calculate $\int e^{x} \cos x d x$. | Solution. Let $u=\cos x, d v=e^{x} d x$. Then $d u=-\sin x d x, v=e^{x}$, and by formula (1.10) we have:
$$
\int e^{x} \cos x d x=e^{x} \cos x+\int e^{x} \sin x d x
$$
The second integral on the right side of the obtained equality is integrated by parts again, and the necessary actions for integration by parts are co... | \frac{e^{x}}{2}(\cosx+\sinx)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,827 |
Example 1.5. Given the improper fraction $\frac{6 x^{3}+5 x^{2}+3 x-4}{x^{2}+4}$. It is necessary to represent it as the sum of a polynomial and a proper fraction. | Solution.
$$
\frac{6 x^{3}+5 x^{2}+3 x-4}{x^{2}+4}=\frac{6 x\left(x^{2}+4\right)+5 x^{2}-24 x+3 x-4}{x^{2}+4}=
$$
we added and subtracted $24 x$ in the numerator and grouped $6 x^{3}+$ $+24 x$, then, after grouping, we factored out the common factor $6 x$. To divide a sum by any expression, it is necessary to divide ... | \frac{6x^{3}+5x^{2}+3x-4}{x^{2}+4}=6x+5-\frac{21x+24}{x^{2}+4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,828 |
Example 1.6. Find: $\int \frac{x_{3}+2 x^{2}-4 x-1}{(x-2)^{2}\left(x^{2}+x+1\right)} d x$. | Solution. According to the theorem of decomposition into partial fractions, the decomposition of the integrand, representing a proper rational fraction, has the form:
$$
\frac{x^{3}+2 x^{2}-4 x-1}{(x-2)^{2}\left(x^{2}+x+1\right)}=\frac{A}{(x-2)^{2}}+\frac{B}{(x-2)}+\frac{M x+N}{x^{2}+x+1}
$$
We bring the expression o... | -\frac{1}{(x-2)}+\frac{11}{7}|\lnx-2|-\frac{2}{7}\ln(x^{2}+x+1)+\frac{8}{7\sqrt{3}}\operatorname{arctg}\frac{2x+1}{\sqrt{3}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,829 |
Example 1.7. Find $\int \frac{6 d x}{x(x-1)(x-2)(x-3)}$. | Solution. The decomposition of the integrand into a sum of the simplest fractions has the form:
$$
\frac{6}{x(x-1)(x-2)(x-3)}=\frac{A}{x}+\frac{B}{(x-1)}+\frac{C}{(x-2)}+\frac{-D}{(x-3)}=
$$
The coefficients of the decomposition $A, B, C, D$ are found by equating the numerators in the right and left parts of the equa... | -\ln|x|+3\ln|x-1|-3\ln|x-2|+\ln|x-3|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,830 |
Example 1.8. Find $\int\left(\frac{1-\cos x}{1+\cos x}\right)^{2} \frac{d x}{3 \sin x+4 \cos x+5}$. | Solution. Applying the universal trigonometric substitution, we get
$$
\begin{aligned}
& \int\left(\frac{1-\cos x}{1+\cos x}\right)^{2} \frac{d x}{3 \sin x+4 \cos x+5}= \\
= & \int \frac{\left(1-\frac{1-t^{2}}{1+t^{2}}\right)^{2}}{\left(1+\frac{1-t^{2}}{1+t^{2}}\right)^{2}} \frac{2 d t}{\left(1+t^{2}\right)\left(\frac... | 2(\frac{1}{3}\operatorname{tg}^{3}\frac{x}{2}-3\operatorname{tg}^{2}\frac{x}{2}+27\operatorname{tg}\frac{x}{2}-108\ln|\operatorname{tg}\frac{x}{2}+3|-\frac{81}{\operatorname{tg}\frac{x}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,831 |
Example 1.9. Find $\int \frac{\sin 2 x d x}{\sin ^{4} x+\cos ^{4} x}$. | Solution. Since the integrand is an even function of $\sin x$ and $\cos x$:
$$
\begin{aligned}
R(\sin x, \cos x) & =\frac{2}{\sin ^{4} x+\cos ^{4} x} \\
R(-\sin x,-\cos x) & =\frac{2(-\sin x)(-\cos x)}{(-\sin x)^{4}+(-\cos x)^{4}}= \\
& =\frac{2 \sin x \cos x}{\sin ^{4} x+\cos ^{4} x}=R(\sin x, \cos x)
\end{aligned}
$... | \operatorname{arctg}(\tan^{2}x)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,832 |
Example 1.10. Find $\int \sin ^{2} x \cos ^{3} x d x$. | Solution. Since the integrand is odd with respect to $\cos x$:
$$
\begin{gathered}
R(\sin x, \cos x)=\sin ^{2} x \cos ^{3} x \\
R(\sin x,-\cos x)=-\sin ^{2} x \cos ^{3} x=-R(\sin x, \cos x)
\end{gathered}
$$
we apply the substitution $t=\sin x$. In this case, transforming the integrand according to the general substi... | \frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,833 |
Example 1.11. Find $\int \frac{\sin ^{3} x}{\cos ^{5} x} d x$. | Solution. Since the integrand is odd with respect to $\sin x$, applying the substitution $t=\cos x$ we get, considering that $\sin ^{2} x=1-\cos ^{2} x, d t=-\sin x d x$ :
$$
\begin{gathered}
\int \frac{\sin ^{3} x}{\cos ^{5} x} d x=\int \frac{1-\cos ^{2} x}{\cos ^{5} x} \sin x d x=-\int \frac{1-t^{2}}{t^{5}}= \\
=-\i... | \frac{1}{4\cos^{4}x}-\frac{1}{2\cos^{2}x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,834 |
Example 1.12. Find $\int \sin ^{4} x \cos ^{4} x d x$. | Solution.
$$
\begin{gathered}
\int \sin ^{4} x \cos ^{4} x d x=\frac{1}{16} \int(2 \sin x \cos x)^{4} d x= \\
=\frac{1}{16} \int \sin ^{4} 2 x d x=\frac{1}{16} \int\left(\frac{1-\cos 4 x}{2}\right)^{2} d x= \\
=\frac{1}{64} \int\left(1-2 \cos 4 x+\cos ^{2} 4 x\right) d x= \\
=\frac{1}{64}\left(\int d x-2 \int \cos 4 x... | \frac{1}{128}(3x-4\sin4x+\frac{1}{8}\sin8x)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,835 |
Example 1.13. Find $\int \sin 7 x \cos 3 x d x$. | Solution.
$$
\begin{gathered}
\int \sin 7 x \cos 3 x d x=\frac{1}{2} \int(\sin 10 x+\sin 4 x) d x= \\
=\frac{1}{20} \int \sin 10 x d(10 x)+\frac{1}{8} \int \sin 4 x d(4 x)= \\
=-\frac{1}{20} \cos 10 x-\frac{1}{8} \cos 4 x+C
\end{gathered}
$$
Using formulas (1.21), indefinite integrals of functions that are the produc... | -\frac{1}{20}\cos10x-\frac{1}{8}\cos4x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,836 |
Example 1.14. Find $\int \sin x \cos 2 x \sin 5 x d x$. | Solution.
$$
\begin{aligned}
& \int \sin x \cos 2 x \sin 5 x d x=\frac{1}{2} \int(\sin 3 x-\sin x) \sin 5 x d x= \\
& =\frac{1}{4} \int \cos 2 x d x-\frac{1}{4} \int \cos 8 x d x- \\
& \quad-\frac{1}{4} \int \cos 4 x d x+\frac{1}{4} \int \cos 6 x d x= \\
& =\frac{1}{8} \sin 2 x-\frac{1}{32} \sin 8 x-\frac{1}{16} \sin ... | \frac{1}{8}\sin2x-\frac{1}{32}\sin8x-\frac{1}{16}\sin4x+\frac{1}{24}\sin6x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,837 |
Example 1.15. Find $\int \frac{d x}{(x+3)^{1 / 2}+(x+3)^{2 / 3}}$. | Solution. According to the given, the rationalizing substitution for the integral is $x+3=t^{6}$, since $\operatorname{LCM}(2,3)=6$. Then
$$
x=t^{6}-3, (x+3)^{1 / 2}=t^{3}, (x+3)^{2 / 3}=t^{4}, d x=6 t^{5} d t
$$
## We have
$$
\begin{gathered}
\int \frac{d x}{(x+3)^{1 / 2}+(x+3)^{2 / 3}}=6 \int \frac{t^{5} d t}{t^{3... | 3\sqrt[3]{x+3}-6\sqrt[6]{x+3}+6\ln|\sqrt[6]{x+3}+1|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,838 |
Example 1.17. Find $\int \frac{d x}{x+\sqrt{x^{2}-x+2}}$. | Solution. Since $a=1>0$, we apply the first Euler substitution. We have $\sqrt{x^{2}-x+2}=t-x$. Further,
$$
\begin{gathered}
x=\frac{t^{2}-2}{2 t-1} \cdot d x=\frac{2\left(t^{2}-t+2\right)}{(2 t-1)^{2}} d t \\
\int \frac{d x}{x+\sqrt{x^{2}-x+2}}=2 \int \frac{t^{2}-t+2}{(2 t-1)^{2} t} d t=
\end{gathered}
$$
$$
\begin{... | \begin{gathered}4\ln|x+\sqrt{x^{2}-x+2}|-\frac{7}{2}\ln|2x+2\sqrt{x^{2}-x+2}-1|-\frac{2}{2(2x+2\sqrt{x^{2}-x+2}-1)}+C\end{gathered} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,840 |
Example 1.18. Find the integral $\int x^{1 / 3}\left(2+3 x^{1 / 2}\right)^{3} d x$. | Solution. Here $m=1 / 3, n=1 / 2, p=3$. The least common multiple of the denominators of $m$ and $n$, i.e., the numbers 2 and 3, is 6. Therefore, we set $t=\sqrt[6]{x}$. Then $x=t^{6}, d x=6 t^{5} d t$, $x^{1 / 3}=t^{2}, x^{1 / 2}=t^{3}$. As a result of applying the specified substitution, the integral is transformed a... | 6\sqrt[3]{x^{4}}+\frac{216}{11}\sqrt[6]{x^{11}}+\frac{162}{7}\sqrt[3]{x^{7}}+\frac{162}{17}\sqrt[6]{x^{17}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,841 |
Example 1.19. Find $\int x^{1 / 2}\left(3+2 x^{3 / 4}\right)^{1 / 2} d x$. | Solution. In the considered case $m=1 / 2, n=3 / 4$, $p=1 / 2, \frac{m+1}{n}=2$ - an integer.
Let $t=\sqrt{3+2 x^{3 / 4}}$. Then $t^{2}=3+2 x^{3 / 4}, x=$ $=\left(\frac{t^{2}-3}{2}\right)^{4 / 3}, d x=\left(\frac{1}{2}\right)^{4 / 3} \frac{8}{3}\left(t^{2}-3\right)^{1 / 3} t d t$. Expressing the integrand in terms of ... | \frac{2}{15}\sqrt{(3+2x^{3/4})^{5}}-\frac{2}{3}\sqrt{(3+2x^{3/4})^{3}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,842 |
Example 1.20. Find $\int x^{1 / 4}\left(1+x^{1 / 2}\right)^{-7 / 2} d x$. | Solution. In the given case, $m=\frac{1}{4}, n=\frac{1}{2}$, $p=-\frac{7}{2}, \frac{m+1}{n}+p=-1-$ is an integer. Let $t=\sqrt{1+\frac{1}{x^{1 / 2}}}$.
Then $t^{2}=1+\frac{1}{x^{1 / 2}}, \quad x^{1 / 2}=\frac{1}{t^{2}-1}, \quad x=\frac{1}{\left(t^{2}-1\right)^{2}}, \quad d x=$ $=-4 t\left(t^{2}-1\right)^{-3} d t, \qua... | \frac{4}{5}\frac{\sqrt[4]{5}}{\sqrt{(1+x^{1/2})^{5}}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,843 |
Example 2.1. $I=\int_{0}^{2} x^{3} d x$ | Solution. $I=\int_{0}^{2} x^{3} d x=\left.\frac{x^{4}}{4}\right|_{0} ^{2}=\frac{2^{4}}{4}-\frac{0^{4}}{4}=4$. | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,844 |
Example 2.2. $I=\int_{0}^{2}(x-1)^{3} d x$.
| Solution.
$$
\begin{aligned}
I= & \int_{0}^{2}(x-1)^{3} d x=\int_{0}^{2}\left(x^{3}-3 x^{2}+3 x-1\right) d x= \\
& =\int_{0}^{2} x^{3} d x-3 \int_{0}^{2} x^{2} d x+3 \int_{0}^{2} x d x-\int_{0}^{2} d x= \\
& =\left.\frac{x^{4}}{4}\right|_{0} ^{2}-\left.3 \frac{x^{3}}{3}\right|_{0} ^{2}+\left.3 \frac{x^{2}}{2}\right|_{... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,845 |
Example 2.3. $I=\int_{0}^{\pi / 4}\left(\frac{4}{\cos ^{2} x}-\frac{5 x^{2}}{1+x^{2}}\right) d x$. | Solution.
$$
\begin{gathered}
I=4 \int_{0}^{\pi / 4} \frac{d x}{\cos ^{2} x}-5 \int_{0}^{\pi / 4} \frac{\left(x^{2}+1-1\right) d x}{1+x^{2}}= \\
=\left.4 \operatorname{tg} x\right|_{0} ^{\pi / 4}-\left.5 x\right|_{0} ^{\pi / 4}+5 \int_{0}^{\pi / 4} \frac{d x}{1+x^{2}}= \\
=4\left(\operatorname{tg}\left(\frac{\pi}{4}\r... | 4-5\frac{\pi}{4}+5\operatorname{arctg}(\frac{\pi}{4}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,846 |
Example 2.5. $I=\int_{-1}^{1} x|x| d x$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
However, since the text provided is already in a form that is commonly used in English for mathematical expressions, the translation is e... | Solution. Since the integrand is an odd function and the limits of integration are symmetric about zero, then
$$
\int_{-1}^{1} x|x| d x=0
$$
In the following two examples, it will be demonstrated that the formal application of the Newton-Leibniz formula (without considering the integrand functions) can lead to an inc... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,847 |
Example 2.7. $\int_{-1}^{1} \frac{d}{d x}\left(\frac{1}{1+3^{1 / x}}\right) d x$.
| Solution. Formal application of the Newton-Leibniz formula is impossible here, since $F(x)=\frac{1}{\left(1+3^{1 / x}\right)}$ is an antiderivative of the integrand everywhere except $x=0$, but this point belongs to the interval of integration $[-1,1]$. To compute this integral, we will first use the additivity propert... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,848 |
Example 2.8. Calculate the integral $\int_{2}^{7}(x-3)^{2} d x$.
a) Using the substitution $z=x-3$.
b) Using the substitution $z=(x-3)^{2}$. | Solution.
a) If $z=x-3$, then $d z=d x, x_{1}=2$ corresponds to $z_{1}=$ $=-1, x_{2}=7$ corresponds to $z_{2}=4$. We obtain
$$
\int_{-1}^{4} z^{2} d z=\left.\frac{z^{3}}{3}\right|_{-1} ^{4}=\frac{4^{3}}{3}-\frac{(-1)^{3}}{3}=\frac{65}{3} \approx 21.67
$$
b) Formal use of the substitution $z=(x-3)^{2}, z_{1}=$ $=1, z... | \frac{65}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,849 |
Example 2.10. $I=\int_{0}^{\pi / 2} \sin ^{3} x \sin 2 x d x$.
| Solution. Let's make the substitution $z=\sin x$ and note that $\sin 2 x=2 \sin x \cos x$. We have
$$
\begin{aligned}
& I=\int_{0}^{\pi / 2} \sin ^{3} x \sin 2 x d x=2 \int_{0}^{\pi / 2} \sin ^{4} x \sin x d x= \\
& =2 \int_{0}^{\pi / 2} \sin ^{3} x d \sin x=2 \int_{0}^{1} z^{4} d z=\left.2 \frac{z^{5}}{5}\right|_{0} ... | \frac{2}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,850 |
Example 2.12. $I=\int_{0}^{4} x\left(1-x^{2}\right)^{1 / 5} d x$.
| Solution. The formal application of the trigonometric substitution $x=\sin t$ is impossible here, as with such a substitution, the variable $x$ cannot cover the interval $[0,4]$ for any $t$. It is preferable to use the substitution
$$
\begin{aligned}
& z=1-x^{2}, x d x=-0.5 d z, x_{1}=0 \rightarrow z_{1}=1, x_{2}=4 \r... | \frac{5}{12}((-15)^{\frac{6}{5}}-1^{\frac{6}{5}})\approx10.33 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,851 |
Example 2.13. $\int_{-1}^{1}|x| e^{|x|} d x$. | Solution. Here the integrand is an even function, which needs to be utilized, and then apply the integration by parts formula (2.3):
$$
\begin{gathered}
\int_{-1}^{1}|x| e^{|x|} d x=2 \int_{0}^{1} x e^{x} d x=\left|\begin{array}{l}
u=x, d u=d x \\
d v=e^{x} d x, v=\int e^{x} d x=e^{x}
\end{array}\right|= \\
=2\left(\l... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,852 |
Example 2.14. $\int_{0}^{(\pi / 2)^{2}} \sin \sqrt{x} d x$. | Solution. First, let's show an incorrect solution leading to a non-integrable function. Transform the integrand as follows:
$$
\begin{gathered}
\int_{0}^{(\pi / 2)^{2}} \frac{\sin \sqrt{x} \cdot \sqrt{x} d x}{\sqrt{x}}=\left|\begin{array}{l}
u=\sqrt{x}, d u=\frac{d x}{2 \sqrt{x}} \\
d v=\frac{\sin \sqrt{x} d x}{\sqrt{... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,853 |
Example 2.15. $I=\int_{-1}^{1}|x| e^{x^{2}} d x$. | Solution. Here the integrand is an even function, so
$$
\begin{gathered}
I=\int_{-1}^{1}|x| e^{x^{2}} d x=2 \int_{0}^{1} x e^{x^{2}} d x=\left|\begin{array}{l}
z=x^{2} \\
d z=2 x d x
\end{array}\right|= \\
=2 \int_{0}^{1} e^{z} d z=\left.e^{z}\right|_{0} ^{1}=e-1
\end{gathered}
$$ | e-1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,854 |
Example 2.16. Prove that if $u(x)$ and $v(x)$ have $n$ continuous derivatives on a finite interval, then the following formula holds:
$$
\begin{aligned}
\int_{a}^{b} u v^{(n)} d x & =\left(u v^{(n-1)}-u^{\prime} v^{(n-2)}+\ldots\right. \\
\left.\ldots+(-1)^{(n-1)} u^{(n-1)} v\right) & \left.\right|_{a} ^{b}-\int_{a}^{... | Solution. We will use the method of mathematical induction.
1) The formula is valid for $n=1$, as it coincides with $(2.3)$.
2) Suppose the formula is valid for $k=n-1$, that is,
$$
\begin{aligned}
& \int_{a}^{b} u v^{(n-1)} d x=\left(u v^{(n-2)}-u^{\prime} v^{(n-3)}+\ldots\right. \\
& \left.\quad \ldots+(-1)^{(n-2)}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 30,855 |
Example 2.17. Estimate the upper and lower bounds of the integral $\int_{0}^{2} \sqrt{1+x^{3}} d x$. | Solution. Since the function $f(x)$ is monotonically increasing on the interval $[0,2]$, the minimum and maximum values are achieved at the endpoints of the interval. These values are respectively $m=1, M=3$. We obtain the following estimate for the integral in question:
$$
2 \leqslant \int_{0}^{2} \sqrt{1+x^{3}} d x ... | 2\leqslant\int_{0}^{2}\sqrt{1+x^{3}}\leqslant6 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,856 |
Example 2.18. Estimate the upper and lower bounds of the value of the integral $\int_{0}^{\pi / 6} \frac{d x}{1+3 \sin ^{2} x}$. | Solution. $\frac{1}{1+3 \sin ^{2}(\pi / 6)} \leqslant \frac{1}{1+3 \sin ^{2} x} \leqslant \frac{1}{1+3 \sin ^{2} 0}$.
From this, $\frac{2 \pi}{21} \leqslant \int_{0}^{\pi / 6} \frac{d x}{1+3 \sin ^{2} x} \leqslant \frac{\pi}{6}$. | \frac{2\pi}{21}\leqslant\int_{0}^{\pi/6}\frac{}{1+3\sin^{2}x}\leqslant\frac{\pi}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,857 |
Example 2.21. Find the limit $\lim _{x \rightarrow 0}\left(\left(\int_{0}^{x^{2}} \cos x d x\right) / x\right)$. | Solution. We will use L'Hôpital's rule, as there is an indeterminate form of type «0/0».
$$
\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \cos x d x}{x}=\lim _{x \rightarrow 0} \frac{\left(\int_{0}^{x^{2}} \cos x d x\right)^{\prime}}{x^{\prime}}=\lim _{x \rightarrow 0} \frac{\cos \left(x^{2}\right) \cdot 2 x}{1}=0
$$ | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,859 |
Example 2.22. Find the limit
$$
\lim _{x \rightarrow 0}\left(\int_{0}^{\operatorname{arctg} x} e^{\sin x} d x / \int_{0}^{x} \cos \left(x^{2}\right) d x\right)
$$ | Solution. According to L'Hôpital's rule and the rules for differentiating definite integrals, we transform the limit as follows:
$$
\lim _{x \rightarrow 0}\left(\int_{0}^{\operatorname{arctg} x} e^{\sin x} d x\right)^{\prime} /\left(\int_{0}^{x} \cos \left(x^{2}\right) d x\right)^{\prime}=\lim _{x \rightarrow 0} \frac... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,860 |
Example 2.23. $\int_{1}^{\infty} \frac{d x}{x^{3 / 2}}$. | Solution. By the definition of the improper integral with an infinite upper limit
$$
\int_{1}^{\infty} \frac{d x}{x^{3 / 2}}=\lim _{b \rightarrow \infty} \int_{1}^{b} \frac{d x}{x^{3 / 2}}=\left.\lim _{b \rightarrow \infty}\left(-\frac{1}{2 x^{1 / 2}}\right)\right|_{1} ^{b}=\frac{1}{2}
$$ | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,861 |
Example 2.24. $\int_{0}^{\infty} \frac{d x}{4+x^{2}}$. | Solution. We will use formula (2.5)
$$
\int_{0}^{\infty} \frac{d x}{4+x^{2}}=\lim _{b \rightarrow \infty} \int_{0}^{b} \frac{d x}{4+x^{2}}=\left.\lim _{b \rightarrow \infty}(0.5 \operatorname{arctg}(0.5 x))\right|_{0} ^{b}=\frac{\pi}{4}
$$ | \frac{\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,862 |
Example 2.25. $\int_{-\infty}^{0} e^{x} d x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
Example 2.25. $\int_{-\infty}^{0} e^{x} d x$. | Solution. According to formula (2.6), we get
$$
\int_{-\infty}^{0} e^{x} d x=\lim _{a \rightarrow-\infty} \int_{a}^{0} e^{x} d x=\left.\lim _{a \rightarrow -\infty} e^{x}\right|_{a} ^{0}=1
$$ | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,863 |
Example 2.26. $\int_{-\infty}^{+\infty} \frac{d x}{x^{2}+6 x+18}$. | Solution. Apply formula (2.7)
$$
\int_{-\infty}^{+\infty} \frac{d x}{x^{2}+6 x+18}=\lim _{\substack{a \rightarrow-\infty \\ b \rightarrow \infty}} \int_{a}^{b} \frac{d x}{x^{2}+6 x+18}=
$$
$$
=\lim _{\substack{a \rightarrow-\infty \\ b \rightarrow \infty}} \int_{a}^{b} \frac{d x}{(x+3)^{2}+9}=\left.\lim _{\substack{a... | \frac{\pi}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,864 |
Example 2.27. Investigate the convergence of the integral $I=$ $=\int_{0}^{\infty} \frac{x d x}{1+x^{2} \cos ^{2} x}$. | Solution. Since $\frac{x}{1+x^{2} \cos ^{2} x} \geqslant \frac{x}{1+x^{2}}$, and the "smaller" integral
$$
\int_{0}^{\infty} \frac{x d x}{1+x^{2}}=\lim _{b \rightarrow \infty} \int_{0}^{b} \frac{x d x}{1+x^{2}}=0.5 \lim _{b \rightarrow \infty}\left(\ln \left(1+b^{2}\right)-\ln 1\right)=\infty
$$
diverges, then by the... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,865 |
Example 2.28. Compute $I_{n}=\int_{0}^{\infty} x^{n} e^{-x} d x$. | Solution. Using the formula for integration by parts, we find
$$
\int_{0}^{\infty} x^{n} e^{-x} d x=-\left.\lim _{b \rightarrow \infty} x^{n} e^{-x}\right|_{0} ^{b}+n \int_{0}^{\infty} x^{n-1} e^{-x} d x
$$
from here
$$
I_{n}=\lim _{b \rightarrow \infty} \int_{0}^{b} x^{n} e^{-x} d x=\lim _{b \rightarrow \infty}\lef... | I_{n}=n! | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,866 |
Example 2.29. Prove that the integral
$$
I=\int_{0}^{\infty} \frac{d x}{\left(1+x^{2}\right)\left(1+x^{a}\right)}
$$
does not depend on $a$. | Solution. Let's divide the integration region $x \in [0, \infty)$ into two subregions $x \in [0,1]$ and $x \in [1, \infty)$. Then
$$
I=\int_{0}^{1} \frac{d x}{\left(1+x^{2}\right)\left(1+x^{a}\right)}+\int_{1}^{\infty} \frac{d x}{\left(1+x^{2}\right)\left(1+x^{a}\right)}=I_{1}+I_{2}
$$
In the first integral, we perfo... | \frac{\pi}{4} | Calculus | proof | Yes | Yes | olympiads | false | 30,867 |
Example 2.30. Prove that for $m>0$ and for $a>0$ the integrals $\int_{a}^{\infty} \frac{\sin x d x}{x^{m}}$ and $\int_{a}^{\infty} \frac{\cos x d x}{x^{m}}$ converge. | Solution. We will use Dirichlet's test.
Let $f_{1}(x)=\sin x, f_{2}(x)=\cos x, v(x)=x^{-m}$. We have $\left|\int_{a}^{b} \sin x d x\right|=|\cos a-\cos b| \leqslant 2,\left|\int_{a}^{b} \cos x d x\right|=|\sin b-\sin a| \leqslant 2$.
Thus, the functions $\sin x$ and $\cos x$ have bounded primitives for any $b>a$, and... | proof | Calculus | proof | Yes | Yes | olympiads | false | 30,868 |
Example 2.31. Investigate the convergence of the integral
$$
I=\int_{1}^{\infty} \frac{\sin ^{2} x d x}{x}
$$ | Solution. We transform the integrand using the formula $\sin ^{2} x=\frac{1-\cos 2 x}{2}$. We have $I=$ $=\frac{1}{2}\left(\int_{1}^{\infty} \frac{d x}{x}-\int_{1}^{\infty} \frac{\cos 2 x d x}{x}\right)$. The integral $\int_{1}^{\infty} \frac{\cos 2 x d x}{x}$ converges by Dirichlet's test. Indeed, $f(x)=\frac{1}{x} \r... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,869 |
Example 2.32. Prove that the integral $I=\int_{1}^{\infty} \frac{x \cos x d x}{\left(1+x^{2}\right) \sqrt{4+x^{2}}}$ converges. | Solution. We will use Abel's criterion. We have
$$
u(x)=\frac{\cos x}{1+x^{2}}
$$
The integral $\int_{1}^{\infty} u(x) d x=\int_{1}^{\infty} \frac{\cos x d x}{1+x^{2}}$ converges by the comparison test in the form of an inequality: $\left|\frac{\cos x}{1+x^{2}}\right| \leqslant \frac{1}{x^{2}}$; the "larger" integral... | proof | Calculus | proof | Yes | Yes | olympiads | false | 30,870 |
Example 2.34. $I=$ V.p. $\int_{-\infty}^{\infty} \frac{(1+x) d x}{1+x^{2}}$. | Solution.
$$
\begin{gathered}
I=\lim _{a \rightarrow \infty}\left(\int_{-a}^{a} \frac{d x}{1+x^{2}}-\int_{-a}^{a} \frac{x d x}{1+x^{2}}\right)= \\
=\left.\lim _{a \rightarrow \infty}\left(\operatorname{arctg} x+0.5 \ln \left(1+x^{2}\right)\right)\right|_{-a} ^{a}= \\
=\lim _{a \rightarrow \infty}\left(\operatorname{ar... | \pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,871 |
Example 2.36. $I=\mathrm{V}$.p. $\int_{-\infty}^{\infty} \operatorname{arctg} x d x$.
Translating the above text into English, while preserving the original text's line breaks and format, yields:
Example 2.36. $I=\mathrm{V}$.p. $\int_{-\infty}^{\infty} \arctan x d x$. | Solution.
$$
\begin{gathered}
I=\lim _{a \rightarrow \infty} \int_{-a}^{a} \operatorname{arctg} x d x=\left|\begin{array}{l}
\left.u=\operatorname{arctg} x . d u=\frac{d x}{1+x^{2}} \right\rvert\,= \\
d v=x, v=x
\end{array}\right|= \\
=\lim _{a \rightarrow \infty}\left(\left.x \operatorname{arctg} x\right|_{-a} ^{a}-\... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,873 |
Example 2.37. Find the average value of the function $u(x)=$ $=\sin ^{2}(2 x)$ on the half-interval $[0, \infty)$. | Solution.
$$
\begin{gathered}
M(u)=\lim _{x \rightarrow \infty}\left(\int_{0}^{x} \frac{\sin ^{2}(2 x) d x}{x}\right)=\lim _{x \rightarrow \infty} 0.5\left(\int_{0}^{x} \frac{1-\cos (4 x) d x}{x}\right)= \\
=\lim _{x \rightarrow \infty}\left(\frac{0.5 x-0.125 \sin 4 x}{x}\right)=0.5
\end{gathered}
$$ | 0.5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,874 |
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