problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
570. $y=x e^{x}$. | Solution. $1^{0}$. Find the derivative:
$$
y^{\prime}=\left(x e^{x}\right)^{\prime}=x^{\prime} e^{x}+\left(e^{x}\right)^{\prime} x=e^{x}+x e^{x}=e^{x}(1+x)
$$
$2^{0}$. Find the critical points: $e^{x}(1+x)=0, x=-1$.
$3^{\circ}$. Investigate the signs of the derivative to the left and right of the critical point:
$$... | -\frac{1}{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,070 |
582. $y=4x-x^{2}$. | Solution. $1^{0}$. Find the first derivative: $y^{\prime}=4-2 x$. $2^{0}$. Solving the equation $4-2 x=0$, we get the critical point $x=2$. $3^{0}$. Calculate the second derivative: $y^{\prime \prime}=-2$.
$4^{0}$. Since $y^{\prime \prime}$ is negative at any point, then $y^{\prime \prime}(2)=-2<0$. This means that th... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,071 |
584. $y=x^{5}$. | Solution. $1^{0}$. Find the first derivative: $y^{\prime}=5 x^{4}$.
$2^{0}$. Solving the equation $5 x^{4}=0$, we get the critical point $x=0$.
$3^{0}$. Calculate the second derivative and find its value at the critical point: $y^{\prime \prime}=20 x^{3}, y^{\prime \prime}(0)=0$.
$4^{\circ}$. Since the second deriva... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,073 |
585. $y=x^{6}$. | Solution. $1^{0} . y^{\prime}=6 x^{5}$.
$2^{0} .6 x^{5}=0$, i.e. $x=0$.
$3^{0} \cdot y^{\prime \prime}=30 x^{4}$.
$4^{0}$. Since $y^{\prime \prime}(0)=0$, the method of investigation using the second derivative is not applicable.
We will investigate the function for an extremum using the first derivative. If $x<0$,... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,074 |
587. $f(x)=\sin x+\cos x$ in the interval ( $0,2 \pi$ ). | Solution. $1^{0}$. We have $f^{\prime}(x)=\cos x-\sin x$.
$2^{0}$. Find the critical points: $\cos x-\sin x=0 ; 1-\operatorname{tg} x=0 ; \operatorname{tg} x=1$; $x_{1}=\pi / 4, x_{2}=5 \pi / 4$. | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,075 |
588. $y=x+\cos 2x$ in the interval $(0, \pi / 4)$. | Solution. $1^{\circ}$. We have $y^{\prime}=1-2 \sin 2 x$.
$2^{0}$. Determine the critical points belonging to the given interval: $1-2 \sin 2 x=0, \sin 2 x=1 / 2, 2 x=\pi / 6, x=\pi / 12$.
$3^{\circ}$. Find $y^{\prime \prime}=-4 \cos 2 x$.
$4^{0}$. Since $y^{\prime \prime}(\pi / 12)=-4 \cos (\pi / 6)=-2 \sqrt{3}<0$,... | \frac{\pi}{12}+\frac{\sqrt{3}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,076 |
589. $y=x^{2} e^{-x}$. | Solution. $1^{0} . y^{\prime}=2 x e^{-x}-x e^{-x}=x^{2} e^{-x}(2-x)$.
$2^{0} . y^{\prime}=0$ at $x_{1}=0$ and $x_{2}=2$.
$3^{0} . y^{\mu}=(2-2 x) e^{-x}-\left(2 x-x^{2}\right) e^{-x}-e^{-x}\left(2-4 x+x^{2}\right)$.
$4^{0} . y^{\prime \prime}(0)=2>0, y^{\prime \prime}(2)=e^{-2}(-2)<0 ; y_{\min }=y(0)=0, y_{\max }=y(... | y_{\}=0,y_{\max}=4e^{-2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,077 |
602. Find the greatest and the least values of the function $y=$ $=x^{5}-5 x^{4}+5 x^{3}+3$ on the interval $[-1,2]$. | Solution. $1^{0}$. Find the critical points belonging to the interval $(-1,2)$, and the values of the function at these points:
$$
\begin{gathered}
y^{\prime}=5 x^{4}-20 x^{3}+15 x^{2} ; 5 x^{4}-20 x^{3}+15 x^{2}=0 ; 5 x^{2}\left(x^{2}-4 x+3\right)=0 \\
x_{1}=0, x_{2}=1, x_{3}=3
\end{gathered}
$$
The critical point $... | y(1)=4,y(-1)=-8 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,078 |
603. Find the greatest and least values of the function $f(x)=\sqrt{100-x^{2}}$ on the interval $[-6,8]$. | Solution. $1^{0}$. Find the critical points on the interval $[-6,8]$:
$$
f^{\prime}(x)=\frac{-2 x}{2 \sqrt{100-x^{2}}}=-\frac{x}{\sqrt{100-x^{2}}}
$$
On the considered interval, we have only one critical point $x=0$; and $f(0)=10$.
$2^{j}$. Calculate the values of the function at the endpoints of the interval:
$$
f... | M=10,=6 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,079 |
610. Decompose the number 100 into two addends so that their product is the largest possible. | Solution. Let the first number be denoted by $x$. Then the second number is $100-x$. We then form the function $y=x(100-x)$. According to the condition, the argument $x$ must be such that the function takes its maximum value, i.e., the product on the right side of the equation is the largest. Therefore, we will find th... | 50,50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,080 |
611. Find the number which, when added to its square, gives the smallest sum. | Solution. Let the required number be $x$. Then we need to find such an $x$ for which the function $y=x+x^{2}$ has a minimum. We find this value of $x$:
$$
y^{\prime}=1+2 x ; 1+2 x=0 ; 2 x=-1 ; x=-1 / 2 ; y^{\prime \prime}=2, y^{\prime \prime}(-1 / 2)=2>0
$$
i.e., the minimum is achieved at $x=-1 / 2$. Therefore, the ... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,081 |
614. There is a square sheet of tin, the side of which $a=$ $=60 \mathrm{~cm}$. By cutting out equal squares from all its corners and folding up the remaining part, a box (without a lid) needs to be made. What should be the dimensions of the squares to be cut out so that the box has the maximum volume? | Solution. According to the condition, the side of the square $a=60$. Let the side of the squares cut out from the corners be denoted by $x$. The bottom of the box is a square with side $a-2 x$, and the height of the box is equal to the side $x$ of the cut-out square. Therefore, the volume of the box can be expressed by... | 10 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,082 |
615. From a round log of radius $R$, it is required to cut out a rectangular beam of maximum strength. It is known that the strength of the beam is directly proportional to the product of its width and the square of its height. What should be the dimensions of the beam so that its strength is maximized? | Solution. Let the width of the beam be denoted by $x$, its height by $y$, and its strength (in bending) by $J$, so we have $J=k x y^{2}$, where $k>0$ is a proportionality coefficient.
Thus, we need to find the maximum of the function $J=k x y^{2}$. To express the strength of the beam in terms of one of the unknowns, n... | \frac{2R}{\sqrt{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,083 |
616. An irrigation channel has the shape of an isosceles trapezoid, the lateral sides of which are equal to the smaller base. At what angle of inclination of the lateral sides is the cross-sectional area of the channel the largest? | Solution. Let the smaller base of the trapezoid be denoted by $a$, the angle of inclination of the lateral sides by $\alpha$, and the area of the section by $S$ (Fig. 124). According to the condition, $|A B|=|A D|=|B C|=a$.
; during this time, the ship will travel $24 v$ km. Therefore, the expenses per 1 km of the jo... | \sqrt[3]{\frac{}{2k}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,085 |
629. Investigate the convexity and concavity of the curve $y=$ $=x^{2}-x$. | Solution. $1^{0}$. Find the second derivative: $y^{\prime}=2 x-1 ; y^{\prime \prime}=2$.
$2^{0}$. Since the second derivative is positive for any $x$, the curve is concave up on the entire domain ( $-\infty, \infty$ ). | concaveupon(-\infty,\infty) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,086 |
630. Determine the intervals of convexity of the curve $y=x^{3}$. | Solution. $1^{0}$. Find the second derivative: $y^{\prime}=3 x^{2} ; y^{\prime \prime}=6 x$. It equals zero at the point $x=0$.
$2^{0}$. The point $x=0$ divides the domain of the function into intervals $(-\infty, 0)$ and $(0, \infty)$.
$3^{0}$. The condition for the convexity of the curve is $f^{\prime \prime}(x)<0$... | (-\infty,0) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,087 |
631. Find the intervals of convexity and concavity of the curve $y=x^{4}-2 x^{3}+36 x^{2}-x+7$ | 1. Solved. $1^{0}$. Let's find the second derivative:
$$
\begin{gathered}
y^{\prime}=4 x^{3}-6 x^{2}+72 x-1 ; \quad y^{\prime \prime}=12 x^{2}-12 x+72=12(x+2)(x-3) ; y^{\prime \prime}=0 \\
\text { at } x_{1}=-2 \text { and } x_{2}=3
\end{gathered}
$$
=x^{3}-3 x^{2}+5$. | Solution. $1^{0}$. Determine the first and second derivatives: $f^{\prime}(x)=$ $=3 x^{2}-6 x ; f^{\prime \prime}(x)=6 x-6=6(x-1)$.
$2^{0}$. From the equation $f^{\prime \prime}(x)=0$ we have $6(x-1)=0$, i.e., $x=1$ is a critical point of the second kind.
$3^{0}$. If $x<1$, then $f^{\prime \prime}(x)<0$ and in the in... | M(1;3) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,089 |
637. Investigate the convexity, concavity, and points of inflection of the curve $y=x^{3}-6 x^{2}+4$. | Solution. $1^{0} . y^{\prime}=3 x^{2}-12 x ; y^{\prime \prime}=6 x-12$.
$2^{0}$. From the equation $6 x-12=0$ we find $x=2$.
$3^{0}$. If $x<2$, then $y^{\prime \prime}<0$; hence, in the interval $(-\infty, 2)$ the curve is convex. If $x>2$, then $y^{\prime \prime}>0$; hence, in the interval $(2, \infty)$ the curve is... | (2,-12) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,090 |
638. Find the points of inflection of the curve $y=\frac{8}{x^{2}+4}$. | Solution. $1^{0} . y^{\prime}=-\frac{16 x}{\left(x^{2}+4\right)^{2}} ; y^{\prime \prime}=-\frac{16\left(x^{2}+4\right)^{2}-2\left(x^{2}+4\right) 2 x \cdot 16 x}{\left(x^{2}+4\right)^{4}}=$ $=\frac{16\left(x^{2}+4\right)-64 x^{2}}{\left(x^{2}+4\right)^{3}}=\frac{48 x^{2}-64}{\left(x^{2}+4\right)^{3}}$.
$2^{0} . \frac{4... | (\frac{2}{\sqrt{3}},\frac{3}{2})(-\frac{2}{\sqrt{3}},\frac{3}{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,091 |
639. Investigate the function $y=x^{3}-3 x^{2}-9 x+11$ for maximum, minimum, and point of inflection. | Solution. Let's find the first and second derivatives:
$$
f^{\prime}(x)=3 x^{2}-6 x-9=3\left(x^{2}-2 x-3\right) ; \quad f^{\prime \prime}(x)=6 x-6
$$
Set the first derivative to zero. From the equation $x^{2}-2 x-3=0$ we find $x_{1}=-1, x_{2}=3$. Substituting these values into the second derivative, we get $f^{\prime... | A(-1;16),B(3,-16),C(1;0) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,092 |
645. $y=x^{3}-12 x+4$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
645. $y=x^{3}-12 x+4$. | Solution. $1^{0}$. Domain of definition $(-\infty, \infty)$. The function is continuous throughout its domain of definition.
$2^{0}$. The function is neither even nor odd, since $f(-x) \neq f(x)$ and $f(-x) \neq -f(x)$.
$3^{0}$. If $x=0$, then $\dot{y}=4$, i.e., the graph of the function intersects the y-axis at the ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,094 |
646. $y=\frac{1}{4} x^{4}-\frac{3}{2} x^{2}$. | Solution. $1^{0}$. The domain of the function is the interval ( $-\infty, \infty$ ). There are no points of discontinuity.
$2^{0}$. Here $f(-x)=f(x)$, since $x$ only appears in even powers. Therefore, the function is even and its graph is symmetric with respect to the $O y$ axis.
$3^{0}$. To determine the points of i... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,095 | |
647. $y=e^{-x^{2}}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
647. $y=e^{-x^{2}}$. | Solution. $1^{0}$. The function is defined and continuous on the interval $(-\infty, \infty)$
Fig. 131
$2^{\prime \prime}$. The function is even, since $f(-x)=f(x)$. Its graph is symmetric... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,096 |
648. $y=\ln \left(x^{2}+1\right)$. | Solution. $1^{0}$. Domain of definition ( $-\infty, \infty$ ). There are no points of discontinuity since $x^{2}+1>0$ for any real $x$.
$2^{0}$. Since $y(-x)=\ln \left((-x)^{2}+1\right)=\ln \left(x^{2}+1\right)=y(x)$, the function is even; its graph is symmetric with respect to the y-axis.
$3^{0}$. If $x=0$, then $y=... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,097 |
649. $y=\frac{x^{3}}{3-x^{2}}$.
649. $y=\frac{x^{3}}{3-x^{2}}$. | Solution. $1^{0}$. The function is defined on the entire $0 x$ axis, except for the points $x=\sqrt{3}$ and $x=-\sqrt{3}$, where the function has a discontinuity.
$2^{0}$. The function is odd, as $f(-x)=-f(x)$. Its graph is symmetric with respect to the origin. Therefore, we can investigate the function only for point... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,098 |
1. Find the antiderivative of the function $f(x)=4 x^{3}$. | Solution. Using the differentiation rule, one can guess that on the interval $(-\infty, \infty)$, the antiderivative is $F(x)=x^{4}$. Indeed, $F^{\prime}(x)=4 x^{3}$ for all $x \in(-\infty, \infty)$.
Remark. If it is said that $F(x)$ is an antiderivative of the function $f(x)$, but the interval is not specified, then ... | F(x)=x^4+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,099 |
3. Find the antiderivative of the function $y=x^{6}$ on the set $\mathbf{R}$. | Solution. The degree $x^{6}$ is obtained when differentiating $x^{7}$. Since $\left(x^{7}\right)^{\prime}=7 x^{6}$, to get a coefficient of 1 before $x^{6}$ when differentiating $x^{7}$, $x^{7}$ needs to be taken with a coefficient of $1 / 7$. Therefore, $F(x)=(1 / 7) x^{7}$. | \frac{1}{7}x^{7}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,100 |
5. Show that the function $F(x)=\frac{1}{2} \sin 2 x$ is an antiderivative of the function $f(x)=\cos 2 x$. | Solution. Since $F^{\prime}(x)=\left(\frac{1}{2} \sin 2 x\right)^{\prime}=\frac{1}{2} \cdot 2 \cos 2 x=\cos 2 x$, then $\frac{1}{2} \sin 2 x$ is an antiderivative of the function $\cos 2 x$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,101 |
15. Check the validity of the equality $\int \frac{d x}{x^{3}}=-\frac{1}{2 x^{2}}+C$. | Solution. We have $\left(-\frac{1}{2 x^{2}}+C\right)^{\prime}=\left(-\frac{1}{2} x^{-2}+C\right)^{\prime}=x^{-3}+0=\frac{1}{x^{3}}$. Since we obtained the integrand, the given equality is valid. | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,102 |
20. Represent the integral $\int\left(5 x^{2}-2 x^{3}\right) d x$ as an algebraic sum of integrals. | Solution. $\int\left(5 x^{2}-2 x^{3}\right) d x=5 \int x^{2} d x-2 \int x^{3} d x$. | 5\intx^{2}-2\intx^{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,103 |
22. Compute the integral $\int\left(5 \cos x+2 e^{x}\right) d x$. | Solution. $\quad \int\left(5 \cos x+2 e^{x}\right) d x=5 \int \cos x d x+2 \int e^{x} d x=5 \sin x+C_{1}+$ $+2 e^{x}+C_{2}=5 \sin x+2 e^{x}+C$.
Remark. When integrating the algebraic sum of functions, it is customary to write only one arbitrary constant, since the algebraic sum of arbitrary constants is a constant. | 5\sinx+2e^{x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,104 |
24. $\int x^{4} d x$. | Solution. Apply formula $\mathrm{I}$ for $n=4$ :
$$
\int x^{4} d x=\frac{x^{4+1}}{4+1}+C=\frac{1}{5} x^{5}+C
$$
Verification: $d\left(\frac{1}{5} x^{5}+C\right)=\frac{1}{5} \cdot 5 x^{4} d x=x^{4} d x$. We obtained the integrand; therefore, the integral is found correctly. | \frac{1}{5}x^{5}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,105 |
29. $\int x^{2 / 3} d x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
29. $\int x^{2 / 3} d x$. | Solution. $\int x^{2 / 3} d x=\frac{x^{2 / 3+1}}{2 / 3+1}+C=\frac{x^{5 / 3}}{5 / 3}+C=\frac{3}{5} x^{5 / 3}+C=\frac{3}{5} x \sqrt[3]{x^{2}}+$ $+C$.
Verification: $d\left(\frac{3}{5} x \sqrt[3]{x^{2}}+C\right)=d\left(\frac{3}{5} x^{5 / 3}+C\right)=\frac{3}{5} \cdot \frac{5}{3} x^{5 / 3-1} d x=$ $=x^{2 / 3} d x$, i.e., ... | \frac{3}{5}x^{5/3}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,106 |
32. $\int 8 x^{3} d x$.
The integral of $8x^3$ with respect to $x$ is calculated as follows:
\[
\int 8 x^{3} d x = 8 \int x^{3} d x = 8 \left( \frac{x^{4}}{4} \right) + C = 2x^{4} + C
\] | Solution. Applying property 2 and formula 1, we get
$$
\int 8 x^{3} d x=8 \int x^{3} d x=8 \cdot \frac{x^{4}}{4}+C=2 x^{4}+C
$$
Verification: $d\left(2 x^{4}+C\right)=2 \cdot 4 x^{3} d x=8 x^{3} d x$. The solution is correct. | 2x^4+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,107 |
35. $\int\left(5 x^{3}-2 x^{2}+3 x-8\right) d x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
35. $\int\left(5 x^{3}-2 x^{2}+3 x-8\right) d x$. | Solution. Applying properties 2 and 3, and then formula 1, we get
$$
\begin{aligned}
& \int\left(5 x^{3}-2 x^{2}+3 x-8\right) d x=5 \int x^{3} d x-2 \int x^{2} d x+3 \int x d x-8 \int d x= \\
= & 5 \cdot \frac{x^{4}}{4}-2 \cdot \frac{x^{3}}{3}+3 \cdot \frac{x^{2}}{2}-8 x+C=\frac{5}{4} x^{4}-\frac{2}{3} x^{3}+\frac{3}{... | \frac{5}{4}x^{4}-\frac{2}{3}x^{3}+\frac{3}{2}x^{2}-8x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,108 |
42. $\int(2 x-1)^{3} d x$.
The integral of $(2x-1)^3$ with respect to $x$. | Solution. $\int(2 x-1)^{3} d x=\int\left(8 x^{3}-12 x^{2}+6 x-1\right) d x=8 \int x^{3} d x-$ $-12 \int x^{2} d x+6 \int x d x-\int d x=2 x^{4}-4 x^{3}+3 x^{2}-x+C$. By differentiating the result, it is easy to verify that the solution is correct. | 2x^{4}-4x^{3}+3x^{2}-x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,110 |
45. $\int \frac{3 x^{3}-2 x^{2}+5 x}{2 x} d x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
45. $\int \frac{3 x^{3}-2 x^{2}+5 x}{2 x} d x$. | Solution. $\int \frac{3 x^{3}-2 x^{2}+5 x}{2 x} d x=\frac{3}{2} \int x^{2} d x-\int x d x+\frac{5}{2} \int d x=\frac{3}{2} \cdot \frac{x^{3}}{3}-$ $-\frac{x^{2}}{2}+\frac{5 x}{2}+C=\frac{1}{2} x^{3}-\frac{1}{2} x^{2}+\frac{5}{2} x+C$. | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,111 | |
48. $\int \frac{2 d x}{x}$.
48. $\int \frac{2 d x}{x}$.
(Note: The mathematical expression is the same in both languages, so it remains unchanged.) | Solution. Applying property 2 and formula 11, we find
$$
\int \frac{2 d x}{x}=2 \int \frac{d x}{x}=2 \ln |x|+C
$$
Verification: $\quad d(2 \ln x+C)=2 \cdot \frac{1}{x} d x=\frac{2 d x}{x}$. We obtained the integrand; therefore, the integral is found correctly. | 2\ln|x|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,112 |
51. $\int \frac{x^{3}+1}{x} d x$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
51. $\int \frac{x^{3}+1}{x} d x$. | Solution. Dividing the numerator term by term by $x$, we represent the integrand as the sum of two fractions:
$$
\int \frac{x^{3}+1}{x} d x=\int\left(x^{2}+\frac{1}{x}\right) d x
$$
We break the last integral into the sum of two integrals and apply formulas I and II. Then we get
$$
\int\left(x^{2}+\frac{1}{x}\right)... | \frac{x^{3}}{3}+\ln|x|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,113 |
54. $\int \frac{(3 x+1)^{2}}{x} d x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
54. $\int \frac{(3 x+1)^{2}}{x} d x$. | Solution. $\int \frac{(3 x+1)^{2}}{x} d x=\int \frac{9 x^{2}+6 x+1}{x} d x=9 \int x d x+6 \int d x+\int \frac{d x}{x}=$ $=9 \cdot \frac{x^{2}}{2}+6 x+\ln |x|+C=4.5 x^{2}+6 x+\ln |x|+C$. | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,114 | |
57. $\int \frac{2 x d x}{1+x^{2}}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
57. $\int \frac{2 x d x}{1+x^{2}}$. | Solution. The integrand $\frac{2 x}{1+x^{2}}$ is a fraction, the numerator of which is the differential of the denominator: $d\left(1+x^{2}\right)=$ $=2 x d x$. Then we get
$$
\int \frac{2 x d x}{1+x^{2}}=\int \frac{d\left(1+x^{2}\right)}{1+x^{2}}=\ln \left(1+x^{2}\right)+C
$$
Here, formula II is used again. The abso... | \ln(1+x^{2})+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,115 |
64. $\int\left(x^{5}+3 e^{x}\right) d x$
Translation:
64. $\int\left(x^{5}+3 e^{x}\right) d x$ | Solution. $\int\left(x^{5}+3 e^{x}\right) d x=\int x^{5} d x+3 \int e^{x} d x=\frac{1}{6} x^{6}+3 e^{x}+C$.
Here, properties 3 and 2 (decomposition into a sum of integrals and taking a constant factor outside the integral sign) were applied, and then for the first integral formula 1, for the second - formula III.
Ver... | \frac{1}{6}x^{6}+3e^{x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,116 |
67. $\int\left(x^{3}+2^{x}\right) d x$ | Solution. $\int\left(x^{3}+2^{x}\right) d x=\int x^{3} d x+\int 2^{x} d x=\frac{1}{4} x^{4}+\frac{2^{x}}{\ln 2}+C$. Since differentiating the result yields the integrand, the solution is correct. | \frac{1}{4}x^{4}+\frac{2^{x}}{\ln2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,117 |
72. $\int e^{3 x} d x$.
The integral of $e^{3x}$ with respect to $x$ is $\frac{1}{3}e^{3x} + C$, where $C$ is the constant of integration.
However, since the original text only contains the integral expression, I will keep it as is:
72. $\int e^{3 x} d x$. | Solution. According to the previous remark, the exponent should stand under the differential sign, but $d(3 x)=3 d x$. Therefore, it is necessary to add a factor of $1 / 3$. Then we get
$$
\int e^{3 x} d x=\frac{1}{3} \int e^{3 x} d(3 x)=\frac{1}{3} e^{3 x}+C
$$ | \frac{1}{3}e^{3x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,118 |
75. $\int 2 \sin x d x$
Translation:
75. $\int 2 \sin x d x$ | Solution. Directly by formula V we find
$$
\int 2 \sin x d x=2 \int \sin x d x=-2 \cos x+C
$$ | -2\cosx+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,119 |
77. $\int \frac{\sin 2 x}{\cos x} d x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
77. $\int \frac{\sin 2 x}{\cos x} d x$. | Solution. Applying the formula $\sin 2 x=2 \sin x \cos x$, we get
$$
\int \frac{\sin 2 x}{\cos x} d x=\int \frac{2 \sin x \cos x}{\cos x} d x=2 \int \sin x d x=-2 \cos x+C
$$ | -2\cosx+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,120 |
79. $\int\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2} d x$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
79. $\int\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2} d x$. | Solution. We have
$$
\begin{gathered}
\int\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2} d x=\int\left(\sin ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}+\cos ^{2} \frac{x}{2}\right) d x= \\
=\int(1+\sin x) d x=\int d x+\int \sin x d x=x-\cos x+C
\end{gathered}
$$
Here we applied the formulas $(a+b)^{2}=a^... | x-\cosx+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,121 |
81. $\int\left(7 x^{2}+3 \cos x-\sqrt[3]{x}\right) d x$ | Solution. $\int\left(7 x^{2}+3 \cos x-\sqrt[3]{x}\right) d x=7 \int x^{2} d x+3 \int \cos x d x-\int x^{1 / 3} d x=$ $=\frac{7 x^{3}}{3}+3 \sin x-\frac{3 x^{4 / 3}}{4}+C=\frac{7}{3} x^{3}+3 \sin x-\frac{3}{4} x \sqrt[3]{x}+C$. | \frac{7}{3}x^{3}+3\sinx-\frac{3}{4}x\sqrt[3]{x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,122 |
88. $\int \cos 5 x d x$
88. $\int \cos 5x \, dx$ | Solution. According to the remark made earlier, the argument of the integrand should be under the differential sign. Therefore,
$$
\int \cos 5 x d x=\frac{1}{5} \int \cos 5 x d(5 x)=\frac{1}{5} \sin 5 x+C
$$ | \frac{1}{5}\sin5x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,123 |
89. $\int \sin \frac{x}{2} d x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
89. $\int \sin \frac{x}{2} d x$. | Solution. $\int \sin \frac{x}{2} d x=2 \int \sin \frac{x}{2} d\left(\frac{x}{2}\right)=-2 \cos \frac{x}{2}+C$. | -2\cos\frac{x}{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,124 |
94. $\int \frac{3 d x}{\cos ^{2} x}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
94. $\int \frac{3 d x}{\cos ^{2} x}$. | Solution. Directly by formula VII we find
$$
\int \frac{3 d x}{\cos ^{2} x}=3 \int \frac{d x}{\cos ^{2} x}=3 \operatorname{tg} x+C
$$ | 3\operatorname{tg}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,125 |
101. $\int \frac{\cos 2 x}{\cos ^{2} x \sin ^{2} x} d x$
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
101. $\int \frac{\cos 2 x}{\cos ^{2} x \sin ^{2} x} d x$ | Solution. Applying the formula $\cos 2 x=\cos ^{2} x-\sin ^{2} x$, we get
$$
\begin{gathered}
\int \frac{\cos 2 x}{\cos ^{2} x \sin ^{2} x} d x=\int \frac{\cos ^{2} x-\sin ^{2} x}{\cos ^{2} x \sin ^{2} x} d x= \\
=\int \frac{d x}{\sin ^{2} x}-\int \frac{d x}{\cos ^{2} x}=-\operatorname{ctg} x-\operatorname{tg} x+C
\en... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,126 | |
103. $\int \frac{3-2 \cot^{2} x}{\cos ^{2} x} d x$. | Solution. $\int \frac{3-2 \operatorname{ctg}^{2} x}{\cos ^{2} x} d x=\int\left(\frac{3}{\cos ^{2} x}-\frac{2}{\sin ^{2} x}\right) d x=3 \operatorname{tg} x+2 \operatorname{ctg} x+$ $+C$. | 3\operatorname{tg}x+2\operatorname{ctg}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,127 |
107. $\int \frac{d x}{\cos ^{2}(2 x+1)}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
107. $\int \frac{d x}{\cos ^{2}(2 x+1)}$. | Solution. Due to the previously made remark, the argument should stand under the differential sign. Since $d(2 x+1)=2 d x$, then $d x=$ $=\frac{1}{2} d(2 x+1)$. Therefore,
$$
\int \frac{d x}{\cos ^{2}(2 x+1)}=\frac{1}{2} \int \frac{d(2 x+1)}{\cos ^{2}(2 x+1)}=\frac{1}{2} \operatorname{tg}(2 x+1)+C
$$ | \frac{1}{2}\operatorname{tg}(2x+1)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,128 |
108.
$$
\int \frac{d x}{\sin ^{2}\left(\frac{1}{5} x-4\right)}
$$ | Solution. Since $d\left(\frac{1}{5} x-4\right)=\frac{1}{5} d x$, then $d x=5 d\left(\frac{1}{5} x-4\right)$. Therefore
 | -5\cot(\frac{1}{5}x-4)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,129 |
115. $\int \sin x \cos x d x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
115. $\int \sin x \cos x d x$. | Solution. We will integrate, applying various techniques for transforming the integrand:
$$
\begin{gathered}
\int \sin x \cos x d x=\frac{1}{2} 2 \sin x \cos x d x=\frac{1}{2} \int \sin 2 x d x=\frac{1}{2} \cdot \frac{1}{2} \int \sin 2 x d(2 x)= \\
=-\frac{1}{4} \cos 2 x+C \\
\int \sin x \cos x d x=\int \sin x d(\sin ... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,130 | |
116. $\int \frac{2 d x}{3 \sqrt{1-x^{2}}}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
116. $\int \frac{2 d x}{3 \sqrt{1-x^{2}}}$.
116. $\int \frac{2 d x}{3 \sqrt{1-x^{2}}}$.
(Note: The mathematical expression is already in ... | Solution. Applying property 2 and formula $\mathrm{X}$, we find
$$
\int \frac{2 d x}{3 \sqrt{1-x^{2}}}=\frac{2}{3} \int \frac{d x}{\sqrt{1-x^{2}}}=\frac{2}{3} \arcsin x+C
$$
By differentiating the result, we obtain the integrand. Therefore, the solution is correct. | \frac{2}{3}\arcsinx+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,131 |
124. Find the function whose derivative is equal to $3 t^{2}-2 t+1$, given that at $t=2$ the function takes a value of 25. | Solution. $1^{0}$. From the condition, it follows that the desired function is an antiderivative of the function $3 t^{2}-2 t+1$; therefore, taking the indefinite integral of $3 t^{2}-2 t+1$, we find all antiderivatives of the given function:
$$
\int\left(3 t^{2}-2 t+1\right) d t=3 \int t^{2} d t-2 \int t d t+\int d t... | ^{3}-^{2}++19 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,132 |
129. Form the equation of the curve passing through the point ( $3 ; 4$ ), if the slope of the tangent to this curve at any point ( $x ; y$ ) is equal to $x^{2}-2 x$. | Solution. $1^{0}$. We know that
Fig. 137
$k=\operatorname{tg} \alpha=\frac{d y}{d x}$. In this case, we have $\frac{d y}{d x}=x^{2}-2 x ; d y=\left(x^{2}-2 x\right) d x$.
As a result of i... | \frac{1}{3}x^{3}-x^{2}+4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,133 |
135. The velocity of a point moving in a straight line is given by the equation $v=t^{2}-8 t+3$. Find the equation of motion of the point. | Solution. $1^{0}$. It is known that the velocity of a body in rectilinear motion is the derivative of the path $s$ with respect to time $t$, i.e., $v=\frac{d s}{d t}$, from which we have $d s=$ $=v d t$. Then we have $d s=\left(t^{2}-8 t+3\right) d t$.
$2^{0}$. To find the equation of motion, we integrate both sides o... | \frac{1}{3}^{3}-4^{2}+3+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,134 |
136. The velocity of a body is given by the equation $v=6 t^{2}+1$. Find the equation of motion, if in the time $t=3$ s the body has traveled a distance $s=60 \mathrm{M}$. | Solution. $1^{0}$. We have $d s=v d t=\left(6 t^{2}+1\right) d t$; then $s=\int\left(6 t^{2}+1\right) d t=$ $=2 t^{3}+t+C$.
$2^{0^{\circ}}$. Substituting the initial conditions $s=60 \mathrm{M}$, $t=3 \mathrm{c}$ into the found equation, we get $60=2 \cdot 3^{3}+3+C$, from which $C=3$.
$3^{\circ}$. The desired equati... | 2^{3}++3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,135 |
137. A body moves with a velocity $v=\left(3 t^{2}-1\right) \mathrm{m} / \mathrm{s}$. Find the law of motion $s(t)$, if at the initial moment the body was 5 cm away from the origin. | Solution. $1^{0}$. Since $d s=v d t=\left(3 t^{2}-1\right) d t$, then $s=\int\left(3 t^{2}-1\right) d t=$ $=t^{3}-t+C$.
$2^{0}$. From the condition, if $t=0$, then $s=5 \text{~cm}=0.05 \text{m}$. Substituting these values into the obtained equation, we have $s=t^{3}-t+C$, from which $0.05=C$.
$3^{0}$. Then the desire... | ^{3}-+0.05 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,136 |
142. A body, initially at rest, falls with a constant acceleration $g$. Find the law of motion. | Solution. $1^{0}$. It is known that the acceleration of a body moving in a straight line is the derivative of the velocity $v$ with respect to time $t$. Therefore, $g=\frac{d v}{d t}$, i.e., $d v=g d t$. From this, after integration, we find
$$
v=\int g d t=g \int d t=g t+C
$$
$2^{0}$. Using the initial conditions $v... | \frac{^{2}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,137 |
143. A body moves with acceleration $a=\left(t^{2}+1\right)$ m $/ \mathrm{c}^{2}$. Find the law of motion of the body, if at the moment $t=1$ s the velocity $v=2 \mathrm{m} / \mathrm{c}$, and the path $s=4 \mathrm{M}$. | Solution. This problem is also solved in two steps. First, we need to find $v(t)$, knowing that $a=v^{\prime}(t)$, and then find $s(t)$, knowing that $v=s^{\prime}(t)$. | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,138 |
146. $\int(2 x+3)^{4} d x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
146. $\int(2 x+3)^{4} d x$. | Solved. $\int(2 x+3)^{4} d x=\left|\begin{array}{l}z=2 x+3 \\ d z=2 d x \\ d x=\frac{1}{2} d z\end{array}\right|=\frac{1}{2} \int z^{4} d z=0.1 z^{5}+C=$ $=0.1(2 x+3)^{5}+C$.
Verification: $\quad d\left(0.1(2 x+3)^{5}+C\right)=0.5(2 x+3)^{4}(2 x)^{\prime} d x=(2 x+3)^{4} d x$. | 0.1(2x+3)^{5}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,139 |
151. $\int \sqrt{x+1} d x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
151. $\int \sqrt{x+1} d x$. | Solution. $\int \sqrt{x+1} d x=\left|\begin{array}{c}z=x+1 \\ d z=d x\end{array}\right|=\int \sqrt{z} d z=\int z^{1 / 2} d z=\frac{2}{3} z^{3 / 2}+$ $+C=\frac{2}{3}(x+1) \sqrt{x+1}+C$.
Verification: $d\left(\frac{2}{3}(x+1) \sqrt{x+1}+C\right)=d\left(\frac{2}{3}(x+1)^{3 / 2}+C\right)=\frac{2}{3} \times$ $\times \frac{... | \frac{2}{3}(x+1)\sqrt{x+1}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,140 |
156. $\int \sqrt{1+x^{3}} x^{2} d x$
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
156. $\int \sqrt{1+x^{3}} x^{2} d x$ | Solution. $\int \sqrt{1+x^{3}} x^{2} d x=\left|\begin{array}{c}1+x^{3}=z, \\ x^{2} d x=\frac{1}{3} d z\end{array}\right|=\frac{1}{3} \int z^{1 / 2} d z=\frac{1}{3} \cdot \frac{2}{3} z^{3 / 2}+$ $+C=\frac{2}{9} z^{3 / 2}+C=\frac{2}{9}\left(1+x^{3}\right)^{3 / 2}+C=\frac{2}{9}\left(1+x^{3}\right) \sqrt{1+x^{3}}+C$.
Veri... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,141 | |
161. $\int \frac{x d x}{\sqrt{1-x^{2}}}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
161. $\int \frac{x d x}{\sqrt{1-x^{2}}}$. | Solution. $\int \frac{x d x}{\sqrt{1-x^{2}}}=\left|\begin{array}{l}t=1-x^{2} \\ d t=-2 x d x, \\ x d x=-\frac{1}{2} d t\end{array}\right|=-\frac{1}{2} \int \frac{d t}{\sqrt{t}}=-\frac{1}{2} \int t^{-1 / 2} d t=$
$=-\frac{1}{2} \cdot 2 t^{1 / 2}+C=-\sqrt{t}+C=-\sqrt{1-x^{2}}+C$. | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,142 | |
166. $\int \frac{\sqrt{1+\ln x}}{x} d x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
166. $\int \frac{\sqrt{1+\ln x}}{x} d x$. | Solution. $\int \frac{\sqrt{1+\ln x}}{x} d x=\left|\begin{array}{c}1+\ln x=t \\ \frac{1}{x} d x=d t\end{array}\right|=\int \sqrt{t} d t=\frac{2}{3} t^{3 / 2}+C=$ $=\frac{2}{3}(1+\ln x) \sqrt{1+\ln x}+C$. | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,143 | |
171. $\int \cos ^{3} x d x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
171. $\int \cos ^{3} x d x$. | Solution. First, we transform the integrand: $\cos ^{3} x=\cos ^{2} x \cdot \cos x=\left(1-\sin ^{2} x\right) \cos x$. Next, we find
$$
\begin{gathered}
\int \cos ^{3} x d x=\int\left(1-\sin ^{2} x\right) \cos x d x=\left|\begin{array}{l}
\sin x=t \\
\cos x d x=d t
\end{array}\right|=\int\left(1-t^{2}\right) d t= \\
=... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,144 | |
175. $\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
175. $\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x$. | Solution. $\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x=\left|\begin{array}{l}\sqrt{x}=t \\ \frac{d x}{2 \sqrt{x}}=d t \\ \frac{d x}{\sqrt{x}}=2 d t\end{array}\right|=2 \int \sin t d t=-2 \cos t+C=$ $=-2 \cos \sqrt{x}+C$.
This integral can also be found by bringing $\sqrt{x}$ under the differential sign:
$$
\int \frac{\si... | -2\cos\sqrt{x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,145 |
177. $\int \sin n x d x$.
177. $\int \sin n x d x$.
The translation is provided as requested, however, it seems the source text is a mathematical expression which is already in a universal format and does not change in translation. If you need an explanation or solution for the integral, please let me know. | Solution. $\int \sin n x d x=\left|\begin{array}{l}t=n x, \\ d t=n d x, \\ d x=\frac{1}{n} d t\end{array}\right|=\frac{1}{n} \int \sin t d t=-\frac{1}{n} \cos t+C=$ $=-\frac{1}{n} \cos n x+C$.
Therefore,
$$
\int \sin n x d x=-\frac{1}{n} \cos n x+C
$$ | -\frac{1}{n}\cosnx+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,146 |
178. $\int \cos n x d x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
178. $\int \cos n x d x$. | Solution. $\int \cos n x d x=\left|\begin{array}{l}t=n x, \\ d t=n d x, \\ d x=\frac{1}{n} d t\end{array}\right|=\frac{1}{n} \int \cos t d t=\frac{1}{n} \sin t+C=$ $=\frac{1}{n} \sin \pi x+C$.
Therefore,
$$
\int \cos n x d x=\frac{1}{n} \sin n x+C
$$
It is useful to remember formulas (1) and (2) and use them as tabl... | \int\cosnx=\frac{1}{n}\sinnx+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,147 |
182. $\int \tan x d x$ | Solution. First, we transform the integrand: $\operatorname{tg} x=\frac{\sin x}{\cos x}$. Next, we find
$$
\int \lg x d x=\int \frac{\sin x d x}{\cos x}=\left|\begin{array}{c}
\cos x=z \\
-\sin x d x=d z \\
\sin x d x=-d z
\end{array}\right|=
$$
$$
=-\int \frac{d z}{z}=-\ln |z|+C=-\ln |\cos x|+C
$$ | -\ln|\cosx|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,148 |
191. $\int \frac{d x}{(\arcsin x)^{5} \sqrt{1-x^{2}}}$. | Solution. $\int \frac{d x}{(\arcsin x)^{5} \sqrt{1-x^{2}}}=\left|\begin{array}{l}\arcsin x=t \\ \frac{d x}{\sqrt{1-x^{2}}}=d t\end{array}\right|=\int \frac{d t}{t^{5}}=\int t^{-5} d t=$ $=-\frac{1}{4} t^{-4}+C=-\frac{1}{4(\arcsin x)^{4}}+C$. | -\frac{1}{4(\arcsinx)^{4}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,149 |
192. $\int \frac{\operatorname{arctan} x}{1+x^{2}} d x$. | Solution. $\int \frac{\operatorname{arctg} x}{1+x^{2}} d x=\left|\begin{array}{l}\operatorname{arctg} x=t \\ \frac{1}{1+x^{2}} d x=d t\end{array}\right|=\int t d t=\frac{1}{2} t^{2}+C=$ $=\frac{1}{2}(\operatorname{arctg} x)^{2}+C$. | \frac{1}{2}(\operatorname{arctg}x)^{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,150 |
199. $\int \frac{d x}{a^{2}+x^{2}}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
199. $\int \frac{d x}{a^{2}+x^{2}}$. | Solution. Let's bring the given integral to the standard form $\mathrm{X}$:
$$
\int \frac{d x}{a^{2}+x^{2}}=\int \frac{d x}{a^{2}\left(1+\frac{x^{2}}{a^{2}}\right)}=\frac{1}{a^{2}} \int \frac{d x}{1+\left(\frac{x}{a}\right)^{2}}
$$
We will use the substitution $x=a t$; $d x=a d t$. Then we get
$$
\int \frac{d x}{a^{... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,151 | |
200. $\int \frac{d x}{\sqrt{a^{2}-x^{2}}}$.
200. $\int \frac{d x}{\sqrt{a^{2}-x^{2}}}$.
(Note: The original text and the translation are identical as the text was already in English.) | Solution. Let's bring this integral to the standard form IX:
$$
\int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\frac{1}{a} \int \frac{d x}{\sqrt{1-\left(\frac{x}{a}\right)^{2}}}
$$
Using the substitution $x=a t ; d x=a d t$. Therefore,
$$
\int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\frac{1}{a} \int \frac{a d t}{\sqrt{1-t^{2}}}=\int \f... | \arcsin\frac{x}{}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,152 |
207. $\int x \cos x d x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
207. $\int x \cos x d x$. | The integral contains the product of two functions $x$ and $\cos x$. The substitution method does not provide a way to find this integral. Let $x=u, \cos x d x=d v ;$ then $d x=d u ; v=\sin x$. We apply the integration by parts formula:
$$
\int x \cos x d x=x \sin x-\int \sin x d x=x \sin x+\cos x+C
$$
By setting $x=... | x\sinx+\cosx+C | Calculus | proof | Yes | Yes | olympiads | false | 32,153 |
208. $\int x e^{x} d x$ | S o l u t i o n. $\int x e^{x} d x=\left|\begin{array}{l}u=x, d v=e^{x} d x \\ d u=d x, v=e^{x}\end{array}\right|=x e^{x}-\int e^{x} d x=x e^{x}-e^{x}+C$. | xe^{x}-e^{x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,154 |
211. $\int x^{2} \sin x d x$ | Solution. We have
$$
\int x^{2} \sin x d x=\left|\begin{array}{l}
u=x^{2}, d v=\sin x d x \\
d u=2 x d x, v=-\cos x
\end{array}\right|=-x^{2} \cos x+2 \int x \cos x d x
$$
To find the integral obtained on the right side of the equation, we integrate by parts again:
$$
\int x \cos x d x=x \sin x+\cos x+C
$$
(see the... | \intx^{2}\sinx=-x^{2}\cosx+2x\sinx+2\cosx+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,155 |
212. $\int \arctan x \, dx$. | Solution. $\int \operatorname{arctg} x d x=\left|\begin{array}{l}u=\operatorname{arctg} x, \\ d u=\frac{d x}{1+x^{2}}, v=x\end{array}\right|=x \operatorname{arctg} x-\int \frac{x d x}{1+x^{2}}=$ $=x \operatorname{arctg} x-\frac{1}{2} \int \frac{d\left(1+x^{2}\right)}{1+x^{2}}=x \operatorname{arctg} x-\frac{1}{2} \ln \l... | x\operatorname{arctg}x-\frac{1}{2}\ln(1+x^{2})+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,156 |
221. Using the equality $S^{\prime}(x)=f(x)$, calculate the area of the curvilinear trapezoid bounded by the lines $y=x^{2}, x=1$, $x=2, y=0$. | Solution. Let $x \in [1,2]$ (Fig. 141). Since $S'(x) = f(x)$, then $S'(x) = x^2$. Therefore, $S(x)$ is an antiderivative of the function $f(x) = x^2$.
Let's find the set of all antiderivatives: $S(x) = \frac{x^3}{3} + C$. The value of $C$ can be found from the condition $S(1) = 0$; we have $0 = \frac{1}{3} + C$, from ... | \frac{7}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,157 |
222. Calculate the area of the curvilinear trapezoid bounded by the lines $f(x)=1 / x, x=1, x=2, y=0$. | Solution. Let's construct a trapezoid (Fig. 142). Determine the antiderivative of the function \( f(x) = 1 / x \):
$$
\int \frac{1}{x} d x = \ln x + C
$$
One of the antiderivatives when \( C = 0 \) is \( F(x) = \ln x \). Then the desired area can be found using formula (1):
$$
S = F(2) - F(1) = \ln 2 - \ln 1 = \ln 2... | \ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,158 |
223. Calculate the area of the figure bounded by the lines $y=$ $=2 x-x^{2}$ and $y=0$ (Fig. 143). | Solution. Let's find the points of intersection of the curve $2x - x^2$ with the x-axis: $2x - x^2 = 0; x(2 - x) = 0; x_1 = 0, x_2 = 2$. Therefore, $a = 0$, $b = 2$.
We find the antiderivative of the function $f(x) = 2x - x^2$; we have $f(x) = \int (2x - x^2) dx = x^2 - \frac{1}{3} x^3 + C$. When $C = 0$, we get $F(x)... | \frac{4}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,159 |
228. Given the function $f(x)=2 x+4$. Find the increment of any of its antiderivatives when $x$ changes from -2 to 0. | Solution. Let's find the antiderivative of the given function:
$$
F=\int(2 x+4) d x=x^{2}+4 x+C
$$
Consider, for example, the antiderivatives $F_{1}=x^{2}+4 x, F_{2}=x^{2}+4 x+2$, $F_{3}=x^{2}+4 x-1$, and compute the increment of each of them on the interval $[-2,0]: \quad F_{1}(0)-F_{1}(-2)=0-(-4)=4 ; \quad F_{2}(0)... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,160 |
238. Calculate the area of the curvilinear trapezoid bounded by the $O x$ axis, the lines $x=-1, x=2$, and the parabola $y=9-x^{2}$ (Fig. 146). | Solution. Since the function $y=9-x^{2}$ takes positive values on the interval $[-1,2]$, to find the desired area $S$
Fig. 146
-\left(-\frac{1}{3}\right)=$ $=-\frac{26}{3}=-8 \frac{2}{3}$. | -\frac{26}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,163 |
244. Find $\int_{1}^{2}\left(5 x^{4}+2 \dot{x}-8\right) d x$. | Solution. $\int_{1}^{2}\left(5 x^{4}+2 x-8\right) d x=\int_{1}^{2} 5 x^{4} d x+\int_{1}^{2} 2 x d x-\int_{1}^{2} 8 d x=\left.x^{5}\right|_{1} ^{2}+$ $+x^{2}-\left.8 x\right|_{1} ^{2}=\left(2^{5}-1^{5}\right)+\left(2^{2}-1^{2}\right)-8(2-1)=26$. | 26 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,164 |
245. Find $\int_{\pi / 2}^{\pi} \frac{2 \sin x d x}{(1-\cos x)^{2}}$. | Solution. Let's use the substitution $u=1-\cos x$, from which $d u=$ $=\sin x d x$. Then we will find the new limits of integration; substituting into the equation $u=1-\cos x$ the values $x_{1}=\pi / 2$ and $x_{2}=\pi$, we will respectively obtain $u_{1}=1-\cos (\pi / 2)=1$ and $u_{2}=1-\cos \pi=2$. The solution is wr... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,165 |
246. $\int 8 x^{3} d x$ | Solution. $\int_{1}^{3} 8 x^{3} d x=\left.8 \cdot \frac{x^{4}}{4}\right|_{1} ^{3}=\left.2 x^{4}\right|_{1} ^{3}=2\left(3^{4}-1^{4}\right)=160$. | 160 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,166 |
247. $\int_{0}^{2} 3 x^{4} d x . \quad$ 248. $\int_{-1}^{\sqrt{3}} 4 t^{3} d t . \quad$ 249. $\int_{0}^{1} \sqrt[3]{x} d x$. | Solution. $\int_{0}^{1} \sqrt[3]{x} d x=\int_{0}^{1} x^{1 / 3} d x=\left.\frac{3}{4} x^{4 / 3}\right|_{0} ^{1}=\frac{3}{4}\left(1^{1 / 3}-0^{4 / 3}\right)=\frac{3}{4}$. | \frac{3}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,167 |
252. $\int_{-1}^{1}\left(2 x+3 x^{2}+4 x^{3}+5 x^{4}\right) d x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
252. $\int_{-1}^{1}\left(2 x+3 x^{2}+4 x^{3}+5 x^{4}\right) d x$. | Solution. $\int_{-1}^{1}\left(2 x+3 x^{2}+4 x^{3}+5 x^{4}\right) d x=\left.\left(x^{2}+x^{3}+x^{4}+x^{5}\right)\right|_{-1} ^{1}=$ $=\left(1^{2}+1^{3}+1^{4}+1^{5}\right)-\left((-1)^{2}+(-1)^{3}+(-1)^{4}+(-1)^{5}\right)=4$. | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,168 | |
257. $\int_{0}^{1} e^{2 x} d x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
257. $\int_{0}^{1} e^{2 x} d x$. | Solution. $\int_{0}^{1} e^{2 x} d x=\left.\frac{1}{2} e^{2 x}\right|_{0} ^{1}=\frac{1}{2}\left(e^{2}-1\right)=0.5(7.36-1) \approx 3.18$. | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,169 | |
264. $\int_{0}^{\pi / 3} \sin x d x$. 265. $\int_{\pi / 6}^{\pi / 2} \cos x d x . \quad 266 . \int_{0}^{\pi / 4} \cos 2 x d x$. | Solution. $\int_{0}^{\pi / 4} \cos 2 x d x=\left.\frac{1}{2} \sin 2 x\right|_{0} ^{\pi / 4}=\frac{1}{2}\left(\sin 2 \cdot \frac{\pi}{4}-\sin 2 \cdot 0\right)=$ $=\frac{1}{2} \sin \frac{\pi}{2}=\frac{1}{2} \cdot 1=\frac{1}{2}$. | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,171 |
271. $\int_{\pi / 4}^{\pi / 3} \frac{d x}{\sin ^{2} x}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
271. $\int_{\pi / 4}^{\pi / 3} \frac{d x}{\sin ^{2} x}$. | Solution. $\int_{\pi / 4}^{\pi / 3} \frac{d x}{\sin ^{2} x}=-\left.\operatorname{ctg} x\right|_{\pi / 4} ^{\pi / 3}=-\left(\operatorname{ctg} \frac{\pi}{3}-\operatorname{ctg} \frac{\pi}{4}\right)=$ $=-\left(\frac{\sqrt{3}}{3}-1\right)=\frac{3-\sqrt{3}}{3}$. | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,172 | |
274. $\int_{0}^{1 / 2} \frac{d x}{\sqrt{1-x^{2}}}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
274. $\int_{0}^{1 / 2} \frac{d x}{\sqrt{1-x^{2}}}$.
The above text is already in a mathematical format and does not require transl... | Solution. $\int_{0}^{1 / 2} \frac{d x}{\sqrt{1-x^{2}}}=\left.\arcsin x\right|_{0} ^{1 / 2}=\frac{\pi}{6}$. | \frac{\pi}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,173 |
275. $\int_{-1}^{\sqrt{3} / 2} \frac{d x}{\sqrt{1-x^{2}}}$
276. $\int_{0}^{0.5} \frac{d x}{\sqrt{1-4 x^{2}}}$
277. $\int_{0}^{\sqrt{3}} \frac{d x}{1+x^{2}}$. | Solution. $\quad \int_{0}^{\sqrt{3}} \frac{d x}{1+x^{2}}=\operatorname{arctg} x \int_{0}^{\sqrt{3}}=\operatorname{arctg} \sqrt{3}-\operatorname{arctg} 0=\pi / 3-0=$ $=\pi / 3$. | \pi/3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,174 |
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