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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{e^{\pi}-e^{x}}{\sin 5 x-\sin 3 x}$ | ## Solution
$\lim _{x \rightarrow \pi} \frac{e^{\pi}-e^{x}}{\sin 5 x-\sin 3 x}=\lim _{x \rightarrow \pi} \frac{-e^{\pi}\left(e^{x-\pi}-1\right)}{2 \sin \frac{5 x-3 x}{2} \cos \frac{5 x+3 x}{2}}=$
$=\lim _{x \rightarrow \pi} \frac{-e^{\pi}\left(e^{x-\pi}-1\right)}{2 \sin x \cos 4 x}=$
Substitution:
$x=y+\pi \Rightar... | \frac{e^{\pi}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,485 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{e^{\sin 2 x}-e^{\tan 2 x}}{\ln \left(\frac{2 x}{\pi}\right)}$ | ## Solution
$x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2}$
$x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0$
We obtain:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}} \frac{e^{\sin 2 x}-e^{\tan 2 x}}{\ln \left(\frac{2 x}{\pi}\right)}=\lim _{y \rightarrow 0} \frac{e^{\sin 2\left(y+\frac{\pi}{2}\... | -2\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,486 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{3^{2 x}-7^{x}}{\arcsin 3 x-5 x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{3^{2 x}-7^{x}}{\arcsin 3 x-5 x}=\lim _{x \rightarrow 0} \frac{\left(9^{x}-1\right)-\left(7^{x}-1\right)}{\arcsin 3 x-5 x}=$
$=\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 9}\right)^{x}-1\right)-\left(\left(e^{\ln 7}\right)^{x}-1\right)}{\arcsin 3 x-5 x}=$
$=\lim _{x \ri... | \ln\sqrt{\frac{7}{9}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,487 |
## Problem Statement
Calculate the limit of the function:
$\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin (x-h)}{h}$ | ## Solution
$\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin (x-h)}{h}=$
$=\lim _{h \rightarrow 0} \frac{2 \sin \frac{(x+h)-(x-h)}{2} \cos \frac{(x+h)+(x-h)}{2}}{h}=$
$=\lim _{h \rightarrow 0} \frac{2 \sin \frac{2 h}{2} \cos \frac{2 x}{2}}{h}=$
$=\lim _{h \rightarrow 0} \frac{2 \sin h \cos x}{h}=$
Using the substitu... | 2\cosx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,488 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(2-3^{\sin ^{2} x}\right)^{\frac{1}{\ln (\cos x)}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(2-3^{\sin ^{2} x}\right)^{\frac{1}{\ln (\cos x)}}= \\
& =\lim _{x \rightarrow 0}\left(e^{\ln \left(2-3^{\sin ^{2} x}\right)}\right)^{\frac{1}{\ln (\cos x)}}= \\
& =\lim _{x \rightarrow 0} e^{\ln \left(2-3^{\sin ^{2} x}\right) / \ln (\cos x)}= \\
& =\lim _{x... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,489 |
## problem statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{11 x+8}{12 x+1}\right)^{\cos ^{2} x}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{11 x+8}{12 x+1}\right)^{\cos ^{2} x}=\left(\frac{11 \cdot 0+8}{12 \cdot 0+1}\right)^{\cos ^{2} 0}=$
$=(8)^{\left(1^{2}\right)}=8^{1}=8$
## Problem Kuznetsov Limits 18-19 | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,490 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1}\left(2 e^{x-1}-1\right)^{\frac{3 x-1}{x-1}}$ | ## Solution
$\lim _{x \rightarrow 1}\left(2 e^{x-1}-1\right)^{\frac{3 x-1}{x-1}}=\lim _{x \rightarrow 1}\left(e^{\ln \left(2 e^{x-1}-1\right)}\right)^{\frac{3 x-1}{x-1}}=$
$=\lim _{x \rightarrow 1} e^{\frac{3 x-1}{x-1} \cdot \ln \left(2 e^{x-1}-1\right)}=\exp \left\{\lim _{x \rightarrow 1} \frac{3 x-1}{x-1} \cdot \ln... | e^4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,491 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1}(\arcsin x)^{\tan \pi x}$ | ## Solution
$\lim _{x \rightarrow 1}(\arcsin x)^{\operatorname{tg} \pi x}=(\arcsin 1)^{\operatorname{tg} \pi}=\left(\frac{\pi}{2}\right)^{0}=1$
## Problem Kuznetsov Limits 20-19 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,492 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \sqrt{2 \cos ^{2} x+\left(e^{x}-1\right) \sin \left(\frac{1}{x}\right)}$ | Solution
Since $\sin \left(\frac{1}{x}\right)_{\text {- is bounded, and }}\left(e^{x}-1\right) \rightarrow 0$, as $x \rightarrow 0$, then $\left(e^{x}-1\right) \sin \left(\frac{1}{x}\right) \rightarrow 0 \quad$, as $x \rightarrow 0$
Then:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \sqrt{2 \cos ^{2} x+\left(e^{x}-1... | \sqrt{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,493 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=(x-2)^{3}, y=4 x-8
$$ | ## Solution
From the graph, it can be seen that the area between the curves consists of two identical parts:
$$
S_{(0,4)}=2 S_{(2,4)}
$$
We will find the area of the part where \( x \in (2,4) \) as the difference of two integrals:
$$
\begin{aligned}
& S=\int_{2}^{4}(4 x-8) d x-\int_{2}^{4}(x-2)^{3} d x= \\
& =\left... | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,494 |
## Problem Statement
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=x \sqrt{9-x^{2}}, y=0, (0 \leq x \leq 3)
$$ | ## Solution
$$
\begin{aligned}
& S=\int_{0}^{3}\left(x \sqrt{9-x^{2}}\right) d x= \\
& =-\frac{1}{2} \int_{0}^{3}\left(9-x^{2}\right)^{\frac{1}{2}} d\left(9-x^{2}\right)= \\
& =-\left.\frac{... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,495 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=4-x^{2}, y=x^{2}-2 x
$$ | ## Solution
Find the abscissas of the points of intersection of the graphs of the functions:
$$
\begin{aligned}
& 4-x^{2}=x^{2}-2 x \\
& 2 x^{2}-2 x-4=0 \\
& x^{2}-x-2=0
\end{aligned}
$$
... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,496 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=\sin x \cdot \cos ^{2} x, y=0,\left(0 \leq x \leq \frac{\pi}{2}\right)
$$ | ## Solution
$$
\begin{aligned}
& S=S_{0 x} ; 0 \leq x \leq \pi / 2 ; 0 \leq y \leq \sin x \cos ^{2} x \\
& S=\int_{0}^{\pi / 2} y d x= \\
& =\int_{0}^{\pi / 2}\left(\sin x \cos ^{2} x-0\rig... | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,497 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=\sqrt{4-x^{2}}, y=0, x=0, x=1
$$ | ## Solution
The desired area \( S \) is:
\( S=\int_{0}^{1} \sqrt{4-x^{2}} d x \)
We make a trigonometric substitution:
\( \sqrt{4-x^{2}}=2 \sin t \), hence \( x=2 \cos t, d x=2 \sin t \... | \frac{\pi}{3}+\frac{\sqrt{3}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,498 |
## Problem Statement
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=x^{2} \sqrt{4-x^{2}}, y=0, (0 \leq x \leq 2)
$$ | ## Solution
Integral $14-6$
$S=\int_{0}^{2} x^{2} \sqrt{4-x^{2}} d x=\left|\begin{array}{c}x=2 \sin t \\ d x=2 \cos t d t \\ x=0 ; t=0 \\ x=2 ; t=\pi / 2\end{array}\right|=$
$$
\begin{a... | \pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,499 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=\cos x \cdot \sin ^{2} x, y=0,\left(0 \leq x \leq \frac{\pi}{2}\right)
$$ | ## Solution
At points $x=0$ and $x=\frac{\pi}{2}$, the graph of the function $y=\cos x \cdot \sin ^{2} x$ intersects the x-axis, since $y(0)=y\left(\frac{\pi}{2}\right)=0$.
The sketch of the desired area is as follows:
The desired area $S$ is:
$S=\int_{0}^{\ln x} \sqrt{e^{x}-1} d x$
We make a substitution of the variable:
$$
\sqrt{e^{x}-1}=t, \text { then } x=\ln \left(t^... | 2-\frac{\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,501 |
## Problem Statement
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=\frac{1}{x \sqrt{1+\ln x}}, y=0, x=1, x=e^{3}
$$ | ## Solution
Answer: 2
We construct the graphs:
We obtain a figure bounded above by the curve $\frac{1}{x \cdot \sqrt{1+\ln x}}$, on the left by the line $x=1$, on the right by the line ... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,502 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=\arccos x, y=0, x=0
$$ | ## Solution
$S=\int_{0}^{1} \arccos x d x$
We will use the integration by parts formula:
$$
\begin{aligned}
& \int_{a}^{b} u d v=\left.u \cdot v\right|_{a} ^{b}-\int_{a}^{b} v d u \\
& ... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,503 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=(x+1)^{2}, y^{2}=x+1
$$ | ## Solution
Integrals 14-11
$$
y=(x+1)^{2}, y^{2}=x+1
$$
The function $y^{2}=x+1$ can be represented as:
$$
y= \pm \sqrt{x+1}
$$
Since $y=(x+1)^{2} \geq 0$ for all $x$, we will use th... | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,504 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=2 x-x^{2}+3, y=x^{2}-4 x+3
$$ | ## Solution
Let's find the points of intersection of the graphs of the functions:
$$
2 x - x^{2} + 3 = x^{2} - 4 x + 3 \Rightarrow \left[\begin{array}{l}
x = 0 \\
x = 3
\end{array}\right]
$... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,505 |
## Problem Statement
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=x \sqrt{36-x^{2}}, y=0, (0 \leq x \leq 6)
$$ | ## Solution
Integrals 14-13
$S=\int_{0}^{6} x \sqrt{36-x^{2}} d x=$
$$
\begin{aligned}
& =\frac{1}{2} \int_{0}^{6} \sqrt{36-x^{2}} d\left(x^{2}\right)= \\
& =-\frac{1}{2} \int_{0}^{6}\l... | 72 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,506 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
x=\arccos y, x=0, y=0
$$ | ## Solution
$$
\begin{aligned}
& x=\arccos y ; y=0 ; x=0 \\
& y=\cos x ; y=0 ; x=0 \\
& \cos x=0 \Rightarrow x=\frac{\pi}{2}
\end{aligned}
$$
Source — "http://pluspi.org/wiki/index.php/?\?\?\?\?\?\?\?\?\?\?\? \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,507 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=\operatorname{arctg} x, y=0, x=\sqrt{3}
$$ | ## Solution
Find the abscissas of the points of intersection of the graphs of the functions
$y=\operatorname{arctg} x, y=0$ :
$$
\operatorname{arctg} x=0
$$
$$
x=0
$$
In the case of $y=\operatorname{arctg} x, x=\sqrt{3}$ it is obvious:
$$
x=\sqrt{3}
$$
Calculate the area:
$$ | ## Solution
The desired area $S$ is:
$S=\int_{0}^{2 \sqrt{2}} x^{2} \sqrt{8-x^{2}} d x$
We will use a trigonometric substitution:
$\sqrt{8-x^{2}}=2 \sqrt{2} \sin t$, hence
$x=2 \sqrt{2} \cos t, d x=2 \sqrt{2} \sin t d t$
When $x=0=2 \sqrt{2} \cos t \Rightarrow t=\frac{\pi}{2}$
and when $x=1=2 \sqrt{2} \cos t \Ri... | 4\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,509 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
x=\sqrt{e^{y}-1}, x=0, y=\ln 2
$$ | ## Solution
## Integrals 14-17
$$
x=\sqrt{e^{y}-1} \Rightarrow y=\ln \left(x^{2}+1\right)
$$
Let's find the limits of integration:
$$
\begin{aligned}
& x_{1}=0 \text { - given in the p... | 2-\frac{\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,510 |
## Problem Statement
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=x \sqrt{4-x^{2}}, y=0, (0 \leq x \leq 2)
$$ | ## Solution
Find the abscissas of the points of intersection of the graphs of the functions
$$
\begin{gathered}
y=x \sqrt{4-x^{2}}, y=0 \\
x \sqrt{4-x^{2}}=0 \\
x=0, x= \pm 2
\end{gathered}
$$
Calculate the area:
 ... | \frac{5}{3}-2\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,512 |
## Problem Statement
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=\frac{1}{1+\cos x}, y=0, x=\frac{\pi}{2}, x=-\frac{\pi}{2}
$$ | ## Solution
$$
\begin{aligned}
S= & \int_{-\pi / 2}^{\pi / 2} \frac{1}{1+\cos x} d x= \\
& =\int_{-\pi / 2}^{\pi / 2} \frac{1}{2 \cos ^{2} \frac{x}{2}} d x= \\
& =\int_{-\pi / 2}^{\pi / 2} \frac{1}{\cos ^{2} \frac{x}{2}} d\left(\frac{x}{2}\right)= \\
& =\left.\operatorname{tg}_{\frac{1}{2}}\right|_{-\pi / 2} ^{\pi / 2... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,513 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
x=(y-2)^{3}, x=4 y-8
$$ | ## Solution
$$
\begin{aligned}
& (y-2)^{3}=4(y-2) \\
& (y-2)\left[4-(y-2)^{2}\right]=0 \\
& \begin{array}{l}
4-y^{2}+4 y-4=0 \\
y(y-4)=0
\end{array} \\
& \frac{S}{2}=\int_{2}^{4}(4 y-8) d y-\int_{2}^{4}(y-3)^{3} d y= \\
& =\left.2 y^{2}\right|_{4} ^{2}-\left.8 y\right|_{4} ^{2}-\int_{2}^{4}\left(y^{3}-6 y^{2}+12 y-8\r... | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,514 |
## Problem Statement
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=\cos ^{5} x \cdot \sin 2 x, y=0,\left(0 \leq x \leq \frac{\pi}{2}\right)
$$ | ## Solution
Integrals $14-22$
$$
\begin{aligned}
& S=\int_{0}^{\frac{\pi}{2}} \cos ^{5} x \cdot \sin 2 x d x=\int_{0}^{\frac{\pi}{2}} \cos ^{5} x \cdot 2 \sin x \cos x d x= \\
& =-2 \int_... | \frac{2}{7} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,515 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=\frac{x}{\left(x^{2}+1\right)^{2}}, y=0, x=1
$$ | ## Solution
Find the abscissas of the points of intersection of the graphs of the functions
$$
\begin{gathered}
y=\frac{x}{\left(x^{2}+1\right)^{2}}, y=0 \\
\frac{x}{\left(x^{2}+1\right)^{2}}=0 \\
x=0
\end{gathered}
$$
In the case of $y=\frac{x}{\left(x^{2}+1\right)^{2}}, x=1$ it is obvious:
$$
x=1
$$
Calculate th... | \frac{1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,516 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
x=4-y^{2}, x=y^{2}-2 y
$$ | ## Solution
Let's find the limits of integration:
$$
\begin{aligned}
& \left(y^{2}-2 y\right)-\left(4-y^{2}\right)=0 \\
& 2 y^{2}-2 y-4=0 \\
& y^{2}-y-2=0
\end{aligned}
$$
Then the area ... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,517 |
## Problem Statement
Calculate the area of the figure bounded by the graphs of the functions:
$$
x=\frac{1}{y \sqrt{1+\ln y}}, x=0, y=1, y=e^{3}
$$ | ## Solution
The desired area $S$ is:
$S=\int_{1}^{e^{3}} \frac{1}{y \sqrt{\ln y+1}} d y$
We make a substitution of variables:
$t=\ln y$, hence
$d t=\frac{d y}{y}$
When $t=\ln 1 \Rightarrow t=0$ and when
$t=\ln e^{3} \Rightarrow t=3$
Then we get
The desired area \( S \) is:
\[
S=\int_{1}^{2} \frac{e^{1 / x}}{x^{2}} d x=-\left.e^{1 / x}\right|_{1} ^{2}=-e^{1 / 2}+e^{1}=e-\sqrt{e}
\]
Source — «http://... | \sqrt{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,519 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
x=\sqrt{4-y^{2}}, x=0, y=0, y=1
$$ | ## Solution
Find the ordinates of the points of intersection of the graphs of the functions
$x=\sqrt{4-y^{2}}, x=0$ :
$$
y= \pm 2
$$
In the case $x=\sqrt{4-y^{2}}, y=0$ it is obvious:
$$
y=0
$$
In the case $x=\sqrt{4-y^{2}}, y=1$ it is obvious:
$$
y=1
$$
Calculate the area:
^{2}, y^{2}=x-1
$$ | ## Solution
$$
y=(x-1)^{2}, y^{2}=x-1
$$
## Integrals 14-29
The function $y^{2}=x-1$ can be represented as two functions $y$ of $x$:
$$
\begin{aligned}
& y=\sqrt{x-1} \\
& y=-\sqrt{x-1}
\end{aligned}
$$
Based on the graphs of the functions, the required figure is bounded only by the graphs $y=(x-1)^{2}, y=\sqrt{x-... | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,522 |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=x^{2} \cdot \cos x, y=0,\left(0 \leq x \leq \frac{\pi}{2}\right)
$$ | ## Solution
Obviously, when $00
$$
And at points $x=0$ and $x=\frac{\pi}{2}$:
$$
x^{2} \cdot \cos x=0
$$
We calculate the area:
$$
S=\int_{0}^{\pi / 2} x^{2} \cdot \cos x d x=
$$
Using integration by parts:
$$
\begin{aligned}
& u=x^{2}, d v=\cos x d x \\
& d u=2 x d x, v=\sin x
\end{aligned}
$$
^{2}, x=y^{2}-4 y+3
$$ | ## Solution
Find the ordinates of the points of intersection of the graphs of the functions $x=4-(y-1)^{2}$
, $x=y^{2}-4 y+3:$
$$
4-(y-1)^{2}=y^{2}-4 y+3
$$
$$
\begin{aligned}
& 4-y^{2}... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,524 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=5(t-\sin t) \\
y=5(1-\cos t)
\end{array}\right. \\
& 0 \leq t \leq \pi
\end{aligned}
$$ | ## Solution
Let's find the derivatives with respect to $t$:
$$
\begin{aligned}
& x_{t}^{\prime}=5(1-\cos t) ; \quad y_{t}^{\prime}=5 \sin t \\
& L=5 \int_{0}^{\pi} \sqrt{(1-\cos t)^{2}+\sin ^{2} t} d t=5 \int_{0}^{\pi} \sqrt{1-2 \cos t+\cos ^{2} t+\sin ^{2} t} d t= \\
& =5 \int_{0}^{\pi} \sqrt{2-2 \cos t} d t=5 \int_... | 20 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,525 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=3(2 \cos t-\cos 2 t) \\
y=3(2 \sin t-\sin 2 t)
\end{array}\right. \\
& 0 \leq t \leq 2 \pi
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=3(2 \cos t-\cos 2... | 48 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,526 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=4(\cos t+t \sin t) \\
y=4(\sin t-t \cos t)
\end{array}\right. \\
& 0 \leq t \leq 2 \pi
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=4(\cos t+t \sin t... | 4\pi^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,527 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=\left(t^{2}-2\right) \sin t+2 t \cos t \\
y=\left(2-t^{2}\right) \cos t+2 t \sin t
\end{array}\right. \\
& 0 \leq t \leq \pi
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=\left(t^{2}-2\rig... | \frac{\pi^{3}}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,528 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=10 \cos ^{3} t \\
y=10 \sin ^{3} t
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{2}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x_{t}^{\prime}\right)^{2}+\left(y_{t}^{\prime}\right)^{2}} d t
$$
From the equations of the curve, we find:
$$
\begin{aligned}
& x=10 \cos ^{3}(t) ; x_{t}^{\prime}=10 \cdot... | 15 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,529 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=e^{t}(\cos t+\sin t) \\
y=e^{t}(\cos t-\sin t)
\end{array}\right. \\
& 0 \leq t \leq \pi
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$:
$$
\begin{aligned}
& x^{\prime}(t)=e^{t}(\cos t+\sin t)+(\co... | 2(e^{\pi}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,530 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=3(t-\sin t) \\
y=3(1-\cos t)
\end{array}\right. \\
& \pi \leq t \leq 2 \pi
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x_{t}^{\prime}\right)^{2}+\left(y_{t}^{\prime}\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=3(t-\sin t) ; x... | 12 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,531 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=\frac{1}{2} \cos t-\frac{1}{4} \cos 2 t \\
y=\frac{1}{2} \sin t-\frac{1}{4} \sin 2 t
\end{array}\right. \\
& \frac{\pi}{2} \leq t \leq \frac{2 \pi}{3}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=\frac{1}{2} \cos ... | \sqrt{2}-1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,532 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=3(\cos t+t \sin t) \\
y=3(\sin t-t \cos t)
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{3}
\end{aligned}
$$ | ## Solution
$\left\{\begin{array}{l}x=3(\cos t+t \sin t) \\ y=3(\sin t-t \cos t)\end{array}\right.$
$x^{\prime}(t)=3(-\sin t+\sin t+t \cos t)=3 t \cos t$
$y^{\prime}(t)=3(\cos t-\cos t+t \sin t)=3 t \sin t$
$$
L=\int_{0}^{\frac{\pi}{3}} \sqrt{9 t^{2} \cos ^{2} t+9 t^{2} \sin ^{2} t} d t=\quad 3 \int_{0}^{\frac{\pi}... | \frac{\pi^{2}}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,533 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=\left(t^{2}-2\right) \sin t+2 t \cos t \\
y=\left(2-t^{2}\right) \cos t+2 t \sin t
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{3}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=\left(t^{2}-2\rig... | \frac{\pi^{3}}{81} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,534 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=6 \cos ^{3} t \\
y=6 \sin ^{3} t
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{3}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x_{t}^{\prime}\right)^{2}+\left(y_{t}^{\prime}\right)^{2}} d t
$$
From the equations of the curve, we find:
$$
\begin{aligned}
& x=6 \cos ^{3} t ; x_{t}^{\prime}=6 \cdot 3 ... | \frac{27}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,535 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=e^{t}(\cos t+\sin t) \\
y=e^{t}(\cos t-\sin t)
\end{array}\right. \\
& \frac{\pi}{2} \leq t \leq \pi
\end{aligned}
$$ | ## Solution
$$
L=\int_{\alpha}^{\beta} \sqrt{\left(x_{t}^{\prime}\right)^{2}+\left(y_{t}^{\prime}\right)^{2}} d t
$$
Find the derivatives with respect to $t$:
$$
\begin{aligned}
& x_{t}^{\prime}=e^{t}(\cos t+\sin t)+(\cos t-\sin t) e^{t}=2 e^{t} \cos t \\
& y_{t}^{\prime}=e^{t}(\cos t-\sin t)+(-\cos t-\sin t) e^{t}=... | 2(e^{\pi}-e^{\frac{\pi}{2}}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,536 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=2.5(t-\sin t) \\
y=2.5(1-\cos t)
\end{array}\right. \\
& \frac{\pi}{2} \leq t \leq \pi
\end{aligned}
$$ | ## Solution
The length of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x_{t}^{\prime}\right)^{2}+\left(y_{t}^{\prime}\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=2.5(t-\sin t) ; x_{t}^{\pr... | 5\sqrt{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,537 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=3.5(2 \cos t-\cos 2 t) \\
y=3.5(2 \sin t-\sin 2 t)
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{2}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=3.5(2 \cos t-\cos... | 14(2-\sqrt{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,538 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=6(\cos t+t \sin t) \\
y=6(\sin t-t \cos t)
\end{array}\right. \\
& 0 \leq t \leq \pi
\end{aligned}
$$ | ## Solution
We will use the formula for calculating the length of a curve segment:
$l=\int_{t_{0}}^{t_{1}} \sqrt{(d x)^{2}+(d y)^{2}}$
For the given problem, we find:
$d x=x^{\prime} d t=(6(\cos t+t \sin t))^{\prime} d t=6(-\sin t+\sin t+t \cos t) d t=6 t \cos t d t ;$
$d y=y^{\prime} d t=(6(\sin t-t \cos t))^{\pr... | 3\pi^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,539 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=\left(t^{2}-2\right) \sin t+2 t \cos t \\
y=\left(2-t^{2}\right) \cos t+2 t \sin t
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{2}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=\left(t^{2}-2\rig... | \frac{\pi^{3}}{24} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,540 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=8 \cos ^{3} t \\
y=8 \sin ^{3} t
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{6}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x_{t}^{\prime}\right)^{2}+\left(y_{t}^{\prime}\right)^{2}} d t
$$
From the equations of the curve, we find:
$$
\begin{aligned}
& x=8 \cos ^{3} t ; x_{t}^{\prime}=8 \cdot 3 ... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,541 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=e^{t}(\cos t+\sin t) \\
y=e^{t}(\cos t-\sin t)
\end{array}\right. \\
& 0 \leq t \leq 2 \pi
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is given by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\righ... | 2(e^{2\pi}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,542 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=2(2 \cos t-\cos 2 t) \\
y=2(2 \sin t-\sin 2 t)
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{3}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=2(2 \cos t-\cos 2... | 8(2-\sqrt{3}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,544 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=8(\cos t+t \sin t) \\
y=8(\sin t-t \cos t)
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{4}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=8(\cos t+t \sin t... | \frac{\pi^2}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,545 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=\left(t^{2}-2\right) \sin t+2 t \cos t \\
y=\left(2-t^{2}\right) \cos t+2 t \sin t
\end{array}\right. \\
& 0 \leq t \leq 2 \pi
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=\left(t^{2}-2\rig... | \frac{8\pi^{3}}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,546 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=4 \cos ^{3} t \\
y=4 \sin ^{3} t
\end{array}\right. \\
& \frac{\pi}{6} \leq t \leq \frac{\pi}{4}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x_{t}^{\prime}\right)^{2}+\left(y_{t}^{\prime}\right)^{2}} d t
$$
From the equations of the curve, we find:
$$
\begin{aligned}
& x=4 \cos ^{3} t ; x_{t}^{\prime}=4 \cdot 3 ... | \frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,547 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=e^{t}(\cos t+\sin t) \\
y=e^{t}(\cos t-\sin t)
\end{array}\right. \\
& 0 \leq t \leq \frac{3 \pi}{2}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$:
$$
\begin{aligned}
& x^{\prime}(t)=e^{t}(\cos t+\sin t)+e^{t... | 2(e^{3\pi/2}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,548 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=2(t-\sin t) \\
y=2(1-\cos t)
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{2}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x_{t}^{\prime}\right)^{2}+\left(y_{t}^{\prime}\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=2(t-\sin t) ; x... | 4(2-\sqrt{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,549 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=4(2 \cos t-\cos 2 t) \\
y=4(2 \sin t-\sin 2 t)
\end{array}\right. \\
& 0 \leq t \leq \pi
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=4(2 \cos t-\cos 2... | 32 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,550 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=2(\cos t+t \sin t) \\
y=2(\sin t-t \cos t)
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{2}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=2(\cos t+t \sin t... | \frac{\pi^2}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,551 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=\left(t^{2}-2\right) \sin t+2 t \cos t \\
y=\left(2-t^{2}\right) \cos t+2 t \sin t
\end{array}\right. \\
& 0 \leq t \leq 3 \pi
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=\left(t^{2}-2\rig... | 9\pi^{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,552 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=2 \cos ^{3} t \\
y=2 \sin ^{3} t
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{4}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x_{t}^{\prime}\right)^{2}+\left(y_{t}^{\prime}\right)^{2}} d t
$$
From the equations of the curve, we find:
$$
\begin{aligned}
& x=2 \cos ^{3} t ; x_{t}^{\prime}=2 \cdot 3 ... | \frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,553 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=e^{t}(\cos t+\sin t) \\
y=e^{t}(\cos t-\sin t)
\end{array}\right. \\
& \frac{\pi}{6} \leq t \leq \frac{\pi}{4}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$:
$$
\begin{aligned}
& x^{\prime}(t)=e^{t}(\cos t+\sin t)+e^{t... | 2(e^{\pi/4}-e^{\pi/6}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,554 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
e^{x \sin \frac{5}{x}}-1, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left... | f^{\}(0) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,555 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
e^{x \sin 5 x}-1, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,556 |
## Condition of the problem
To derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$$
y=\frac{1}{3 x+2}, x_{0}=2
$$ | ## Solution
Let's find $y^{\prime}:$
$y^{\prime}=\left(\frac{1}{3 x+2}\right)^{\prime}=-\frac{1}{(3 x+2)^{2}} \cdot 3=-\frac{3}{(3 x+2)^{2}}$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=-\frac{3}{(3 \cdot 2+2)^{2}}=-\frac{3}{8^{2}}=-\frac{3}{64}$
Since the function $y^{\prime}$ at the point $x_{0}$ has a fi... | -\frac{3}{64}\cdotx+\frac{7}{32} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,557 |
Condition of the problem
Find the differential $d y$.
$$
y=\ln \left|x^{2}-1\right|-\frac{1}{x^{2}-1}
$$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(\ln \left|x^{2}-1\right|-\frac{1}{x^{2}-1}\right)^{\prime} d x=\left(\frac{1}{x^{2}-1} \cdot 2 x+\frac{1}{\left(x^{2}-1\right)^{2}} \cdot 2 x\right) d x= \\
& =\frac{2 x}{x^{2}-1} \cdot\left(1+\frac{1}{x^{2}-1}\right) d x=\frac{2 x}{x^{2}-1} \cdot \frac{x... | \frac{2x^{3}}{(x^{2}-1)^{2}}\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,558 |
## Task Condition
Calculate approximately using the differential.
$$
y=\frac{1}{\sqrt{x}}, x=4,16
$$ | ## Solution
If the increment $\Delta x = x - x_{0}$ of the argument $x$ is small in absolute value, then
$f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{0}\right) + f^{\prime}\left(x_{0}\right) \cdot \Delta x$
Choose:
$x_{0} = 4$
Then:
$\Delta x = 0.16$
Calculate:
$y(4) = \frac{1}{\sqrt{4}} = \frac{1}{2... | 0.49 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,559 |
## Task Condition
Find the derivative.
$$
y=\frac{x-1}{\left(x^{2}+5\right) \sqrt{x^{2}+5}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{x-1}{\left(x^{2}+5\right) \sqrt{x^{2}+5}}\right)^{\prime}=\left(\frac{x-1}{\sqrt{\left(x^{2}+5\right)^{3}}}\right)^{\prime}=\frac{1 \cdot \sqrt{\left(x^{2}+5\right)^{3}}-(x-1) \cdot \frac{3}{2} \cdot \sqrt{x^{2}+5} \cdot 2 x}{\left(x^{2}+5\right)^{3}}= \\
& =\fra... | \frac{-2x^{2}+3x+5}{\sqrt{(x^{2}+5)^{5}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,560 |
## Task Condition
Find the derivative.
$y=\frac{e^{x^{3}}}{1+x^{3}}$ | ## Solution
$y^{\prime}=\left(\frac{e^{x^{3}}}{1+x^{3}}\right)^{\prime}=\frac{e^{x^{3}} \cdot 3 x^{2} \cdot\left(1+x^{3}\right)-e^{x^{3}} \cdot 3 \cdot x^{2}}{\left(1+x^{3}\right)^{2}}=$
$=\frac{3 x^{2} \cdot e^{x^{3}} \cdot\left(1+x^{3}-1\right)}{\left(1+x^{3}\right)^{2}}=\frac{3 x^{5} \cdot e^{x^{3}}}{\left(1+x^{3}... | \frac{3x^{5}\cdote^{x^{3}}}{(1+x^{3})^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,561 |
## Task Condition
Find the derivative.
$$
y=\frac{1}{\sqrt{2}} \ln \left(\sqrt{2} \tan x+\sqrt{1+2 \tan^{2} x}\right)
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{\sqrt{2}} \ln \left(\sqrt{2} \tan x+\sqrt{1+2 \tan^{2} x}\right)\right)^{\prime}= \\
& =\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2} \tan x+\sqrt{1+2 \tan^{2} x}} \cdot\left(\sqrt{2} \tan x+\sqrt{1+2 \tan^{2} x}\right)^{\prime}= \\
& =\frac{1}{\sqrt{2}} \cdot \f... | \frac{1}{\cos^{2}x\sqrt{1+2\tan^{2}x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,562 |
## Condition of the problem
Find the derivative.
$$
y=\operatorname{ctg}(\cos 5)-\frac{1}{40} \cdot \frac{\cos ^{2} 20 x}{\sin 40 x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\operatorname{ctg}(\cos 5)-\frac{1}{40} \cdot \frac{\cos ^{2} 20 x}{\sin 40 x}\right)^{\prime}= \\
& =0-\frac{1}{40} \cdot\left(\frac{\cos ^{2} 20 x}{\sin 40 x}\right)^{\prime}=-\frac{1}{40} \cdot\left(\frac{\cos ^{2} 20 x}{2 \sin 20 x \cdot \cos 20 x}\right)^{\prime}
... | \frac{1}{4\sin^{2}20x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,563 |
## Task Condition
Find the derivative.
$y=\frac{2 x-5}{4} \cdot \sqrt{5 x-4-x^{2}}+\frac{9}{4} \cdot \arcsin \sqrt{\frac{x-1}{3}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{2 x-5}{4} \cdot \sqrt{5 x-4-x^{2}}+\frac{9}{4} \cdot \arcsin \sqrt{\frac{x-1}{3}}\right)^{\prime}= \\
& =\frac{2}{4} \cdot \sqrt{5 x-4-x^{2}}+\frac{2 x-5}{4} \cdot \frac{1}{2 \sqrt{5 x-4-x^{2}}} \cdot(5-2 x)+\frac{9}{4} \cdot \frac{1}{\sqrt{1-\left(\sqrt{\frac{x-... | \sqrt{5x-4-x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,564 |
## Problem Statement
Find the derivative.
$$
y=\frac{1}{4} \ln \left|\operatorname{th} \frac{x}{2}\right|-\frac{1}{4} \cdot \ln \frac{3+\operatorname{ch} x}{\operatorname{sh} x}
$$ | ## Solution
$y^{\prime}=\left(\frac{1}{4} \ln \left|\operatorname{th} \frac{x}{2}\right|-\frac{1}{4} \cdot \ln \frac{3+\operatorname{ch} x}{\operatorname{sh} x}\right)^{\prime}=$
$=\frac{1}{4} \cdot \frac{1}{\operatorname{th} \frac{x}{2}} \cdot \frac{1}{\operatorname{ch}^{2} \frac{x}{2}} \cdot \frac{1}{2}-\frac{1}{4}... | \frac{1}{2\operatorname{sh}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,565 |
## Task Condition
Find the derivative.
$y=x^{3^{x}} \cdot 2^{x}$ | ## Solution
$y=x^{3^{x}} \cdot 2^{x}$
$\ln y=\ln \left(x^{3^{x}} \cdot 2^{x}\right)=\ln \left(x^{3^{x}}\right)+x \ln 2=3^{x} \cdot \ln (x)+x \ln 2$
$\frac{y^{\prime}}{y}=\left(3^{x} \cdot \ln (x)+x \ln 2\right)^{\prime}=3^{x} \cdot \ln 3 \cdot \ln (x)+3^{x} \cdot \frac{1}{x}+\ln 2$
$y^{\prime}=y \cdot\left(3^{x} \c... | x^{3^{x}}\cdot2^{x}\cdot(3^{x}\cdot\ln3\cdot\ln(x)+\frac{3^{x}}{x}+\ln2) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,566 |
## Task Condition
Find the derivative.
$$
y=\ln \frac{1+\sqrt{-3-4 x-x^{2}}}{-x-2}-\frac{2}{x+2} \sqrt{-3-4 x-x^{2}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \frac{1+\sqrt{-3-4 x-x^{2}}}{-x-2}-\frac{2}{x+2} \sqrt{-3-4 x-x^{2}}\right)^{\prime}= \\
& =\frac{-x-2}{1+\sqrt{-3-4 x-x^{2}}} \cdot\left(\frac{1+\sqrt{-3-4 x-x^{2}}}{-x-2}\right)^{\prime}-2\left(\frac{\sqrt{-3-4 x-x^{2}}}{x+2}\right)^{\prime}= \\
& =\frac{-x-2}{1+... | \frac{2x^{2}+8x+9+\sqrt{-3-4x-x^{2}}}{(x+2)\cdot\sqrt{-3-4x-x^{2}}\cdot(1+\sqrt{-3-4x-x^{2}})} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,567 |
## Problem Statement
Find the derivative.
$$
y=x \ln (\sqrt{1-x}+\sqrt{1+x})+\frac{1}{2}(\arcsin x-x)
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(x \ln (\sqrt{1-x}+\sqrt{1+x})+\frac{1}{2}(\arcsin x-x)\right)^{\prime}= \\
& =\ln (\sqrt{1-x}+\sqrt{1+x})+x \cdot \frac{1}{\sqrt{1-x}+\sqrt{1+x}} \cdot\left(\frac{1}{2 \sqrt{1-x}} \cdot(-1)+\frac{1}{2 \sqrt{1+x}}\right)+ \\
& +\frac{1}{2} \cdot\left(\frac{1}{\sqrt{1-x^... | \ln(\sqrt{1-x}+\sqrt{1+x}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,568 |
## Task Condition
Find the derivative.
$$
y=\frac{3^{x}(4 \sin 4 x+\ln 3 \cdot \cos 4 x)}{16+\ln ^{2} 3}
$$ | ## Solution
$y^{\prime}=\left(\frac{3^{x}(4 \sin 4 x+\ln 3 \cdot \cos 4 x)}{16+\ln ^{2} 3}\right)^{\prime}=$
$=\frac{1}{16+\ln ^{2} 3} \cdot\left(3^{x} \cdot \ln 3 \cdot(4 \sin 4 x+\ln 3 \cdot \cos 4 x)+3^{x}(16 \cos 4 x-4 \ln 3 \cdot \sin 4 x)\right)=$ $=\frac{3^{x}}{16+\ln ^{2} 3} \cdot\left(4 \ln 3 \cdot \sin 4 x+... | 3^{x}\cdot\cos4x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,569 |
## Condition of the problem
Find the derivative $y_{x}^{\prime}$.
$\left\{\begin{array}{l}x=(\arcsin t)^{2} \\ y=\frac{t}{\sqrt{1-t^{2}}}\end{array}\right.$ | ## Solution
$x_{t}^{\prime}=\left((\arcsin t)^{2}\right)^{\prime}=2 \arcsin t \cdot \frac{1}{\sqrt{1-t^{2}}}=\frac{2 \arcsin t}{\sqrt{1-t^{2}}}$
$y_{t}^{\prime}=\left(\frac{t}{\sqrt{1-t^{2}}}\right)^{\prime}=\frac{1 \cdot \sqrt{1-t^{2}}-t \cdot \frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t)}{1-t^{2}}=$
$=\frac{1-t^{2}+t^{2}... | \frac{1}{2(1-^{2})\arcsin} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,570 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{
\begin{array}{l}
x=\ln \left(1+t^{2}\right) \\
y=t-\operatorname{arctg} t
\end{array}
\right.
\]
$t_{0}=1$ | ## Solution
Since $t_{0}=1$, then
$x_{0}=\ln \left(1+1^{2}\right)=\ln 2$
$y_{0}=1-\operatorname{arctg} 1=1-\frac{\pi}{4}=\frac{4-\pi}{4}$
Let's find the derivatives:
$x_{t}^{\prime}=\left(\ln \left(1+t^{2}\right)\right)^{\prime}=\frac{1}{1+t^{2}} \cdot 2 t=\frac{2 t}{1+t^{2}}$
$y_{t}^{\prime}=(t-\operatorname{arc... | \begin{aligned}& | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,571 |
## Problem Statement
Find the $n$-th order derivative.
$y=\frac{5 x+1}{13(2 x+3)}$ | ## Solution
$$
\begin{aligned}
& y=\frac{5 x+1}{13(2 x+3)} \\
& y^{\prime}=\left(\frac{5 x+1}{13(2 x+3)}\right)^{\prime}=\frac{5 \cdot(2 x+3)-(5 x+1) \cdot 2}{13(2 x+3)^{2}}= \\
& =\frac{10 x+15-10 x-2}{13(2 x+3)^{2}}=\frac{13}{13(2 x+3)^{2}}=\frac{1}{(2 x+3)^{2}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime... | y^{(n)}=\frac{(-1)^{n-1}\cdotn!\cdot2^{n-1}}{(2x+3)^{n+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,572 |
## Task Condition
Find the derivative of the specified order.
$$
y=(3 x-7) \cdot 3^{-x}, y^{I V}=?
$$ | ## Solution
$y^{\prime}=\left((3 x-7) \cdot 3^{-x}\right)^{\prime}=\left((3 x-7) \cdot\left(e^{\ln 3}\right)^{-x}\right)^{\prime}=$
$=\left((3 x-7) \cdot e^{-x \cdot \ln 3}\right)^{\prime}=3 e^{-x \cdot \ln 3}-\ln 3 \cdot(3 x-7) \cdot e^{-x \cdot \ln 3}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(3 e^... | (7\ln3-12-3\ln3\cdotx)\cdot\ln^{3}3\cdot3^{-x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,573 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=t-\sin t \\
y=2-\cos t
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(t-\sin t)^{\prime}=1-\cos t$
$y_{t}^{\prime}=(2-\cos t)^{\prime}=\sin t$
## We obtain:
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{\sin t}{1-\cos t}$
$\left(y_{x}^{\prime}\right)_{t}^{\prime}=\left(\frac{\sin t}{1-\cos t}\right)^{\prime}=\frac{\cos t \cdot(1-\cos t)-\si... | -\frac{1}{(1-\cos)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,574 |
## Problem Statement
Show that the function $y$ satisfies equation (1).
$y=a \cdot \operatorname{tg} \sqrt{\frac{a}{x}-1}$
$a^{2}+y^{2}+2 x \sqrt{a x-x^{2}} \cdot y^{\prime}=0$. | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(a \cdot \tan \sqrt{\frac{a}{x}-1}\right)^{\prime}=a \cdot \frac{1}{\cos ^{2} \sqrt{\frac{a}{x}-1}} \cdot \frac{1}{2 \sqrt{\frac{a}{x}-1}} \cdot\left(-\frac{a}{x^{2}}\right)= \\
& =-\frac{a^{2}}{2 x^{2} \sqrt{\frac{a}{x}-1} \cdot \cos ^{2} \sqrt{\frac{a}{x}-1}}=-\frac{a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 47,575 |
## Task Condition
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O x \cdot$
$$
y=-x^{2}+5 x-6, y=0
$$ | ## Solution
Since
the $O x$
integrals $21-1$
is
the
axis of rotation, the volume
is found
by
the formula:
$V=\pi \cdot \int_{a}^{b} y^{2} d x$
Let's find the limits of integrati... | \frac{\pi}{30} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,576 |
## Problem Statement
Calculate the volumes of solids formed by rotating the figures bounded by the graphs of the functions. The axis of rotation is $O x$.
$$
2 x - x^{2} - y = 0, \quad 2 x^{2} - 4 x + y = 0
$$ | ## Solution
Since the $O x$ axis is the axis of rotation, the volume is found using the formula:
$$
V=\pi \cdot \int_{a}^{b} y^{2} d x
$$
Let's find the limits of integration:
$$
y=2 x-... | \frac{16\pi}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,577 |
## Problem Statement
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O x$.
$$
y=3 \sin x, y=\sin x, 0 \leq x \leq \pi
$$ | ## Solution
Since the $O x$ axis is the axis of rotation, the volume is found using the formula:
$$
V=\pi \cdot \int_{a}^{b} y^{2} d x
$$
Integration limits:
$x_{1}=0 ; x_{2}=\pi$
We will find the volume of the solid as the difference in volumes of two solids of revolution:
Integrals 21-3
$$
V=\pi \cdot \int_{a}^{b} y^{2} d x
$$
Let's find the limits of ... | 6\pi^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,579 |
## Problem Statement
Calculate the volumes of solids formed by rotating the figures bounded by the graphs of the functions. The axis of rotation is $O x$.
$$
y=\sin ^{2} x, x=\frac{\pi}{2}, y=0
$$ | ## Solution
Since
the $O x$
axis
is
the
axis
of
rotation, the volume
is found
using Integrals $21-5$
by
the formula:
$$
V=\pi \cdot \int_{a}^{b} y^{2} d x
$$
Let's find the li... | \frac{3\pi^{2}}{16} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,580 |
## Task Condition
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O x$.
$$
x=\sqrt[3]{y-2}, x=1, y=1
$$ | ## Solution
## Since
the $O x$
axis
is
the
axis
of
rotation, the volume is found
by
the formula:
$$
V=\pi \cdot \int_{a}^{b} y^{2} d x
$$
Let's find the limits of integ... | \frac{44}{7}\cdot\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,581 |
## Problem Statement
Calculate the volumes of solids formed by rotating the figures bounded by the graphs of the functions. The axis of rotation is $O x$.
$$
y=x e^{x}, y=0, x=1
$$ | ## Solution
Since the $O x$ axis is the axis of rotation, the volume is found using the formula:
$$
V=\pi \cdot \int_{a}^{b} y^{2} d x
$$
Let's find the limits of integration: From the problem statement, we already have: $x_{2}=1$. Now let's find the lower limit:
$$
y=x e^{x}=0 \Rightarrow x_{1}=0
$$
Let's find th... | \frac{\pi(e^2-1)}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,582 |
## Task Condition
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O x$.
$$
y=2 x-x^{2}, y=-x+2, x=0
$$ | ## Solution
## Since
the $O x$
axis
is
the
axis
of
rotation for Integrals $21-8$, the volume is
found
by
the
formula:
$$
V=\pi \cdot \int_{a}^{b} y^{2} d x
$$
Deter... | \frac{9}{5}\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,583 |
## Problem Statement
Calculate the volumes of solids formed by rotating the figures bounded by the graphs of the functions. The axis of rotation is $O x$.
$$
y=2 x-x^{2}, y=-x+2
$$ | ## Solution
Since the $O x$ axis
is the axis of rotation, the volume is
found using the formula:
$$
V=\pi \cdot \int_{a}^{b} y^{2} d x
$$
Let's find the limits of integration:
$$
y=2... | \frac{1}{5}\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,584 |
## Task Condition
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O x$.
$$
y=e^{1-x}, y=0, x=0, x=1
$$ | ## Solution
Since the $O x$ axis is the axis of rotation, the volume is found using the formula:
$$
V=\pi \cdot \int_{a}^{b} y^{2} d x
$$
From the problem statement, we already have the limits of integration: $x_{1}=0 ; x_{2}=1$
Let's find the volume of the solid of revolution:
$$
\begin{aligned}
& V=\pi \cdot \in... | 2\pi\cdot(e^{2}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,585 |
## Problem Statement
Calculate the volumes of solids formed by rotating the figures bounded by the graphs of the functions. The axis of rotation is $O x$.
$$
y=x^{2}, y^{2}-x=0
$$ | ## Solution
Since the axis
$O x$ is the axis of rotation, the volume is found using the formula:
Let's find the limits of integration:
$$
y=x^{2}=\sqrt{x} \Rightarrow x_{1}=0 ; x_{2}=1... | \frac{3\pi}{10} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,586 |
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