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Example 3. Find the integral $\int \frac{x}{x^{3}+1} d x$. | Solution. In this case
$$
Q(x)=x^{3}+1=(x+1)\left(x^{2}-x+1\right)
$$
and the second factor does not factor further over the reals.
Based on formula (4.29), the decomposition of the given fraction is
$$
\frac{x}{x^{3}+1}=\frac{A}{x+1}+\frac{B x+C}{x^{2}-x+1}
$$
from which we have
$$
x=A\left(x^{2}-x+1\right)+(B x... | -\frac{1}{3}\ln|x+1|+\frac{1}{6}\ln(x^{2}-x+1)+\frac{1}{\sqrt{3}}\operatorname{arctg}\frac{2x-1}{\sqrt{3}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,721 |
Example 4. Find the integral $\int \frac{x^{3}+x^{2}+2}{x\left(x^{2}-1\right)^{2}} d x$. | Solution. We decompose the integrand into partial fractions. Since the denominator has roots $x_{1}=1$ and $x_{2}=-1$ of multiplicity 2 and a simple root $x_{3}=0$, the decomposition will take the form
$$
\frac{x^{3}+x^{2}+2}{x\left(x^{2}-1\right)^{2}}=\frac{A}{x}+\frac{B_{1}}{x-1}+\frac{B_{2}}{(x-1)^{2}}+\frac{C_{1}}... | \frac{x+3}{2(1-x^{2})}+\ln\frac{x^{2}}{\sqrt[4]{|x-1|^{3}|x+1|^{5}}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,722 |
Example 5. Find the integral $\int \frac{M x+N}{\left(x^{2}+p x+q\right)^{m}}(m=2,3, \ldots)$. | Solution. Let's extract the complete square of a binomial from the expression $x^{2}+p x+q$
$$
\begin{aligned}
x^{2}+p x+q & =x^{2}+2 \cdot \frac{p}{2} x+\left(\frac{p}{2}\right)^{2}+\left[q-\left(\frac{p}{2}\right)^{2}\right]= \\
& =\left(x+\frac{p}{2}\right)^{2}+\left(q-\frac{p^{2}}{4}\right)
\end{aligned}
$$
We as... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,723 |
Example 6. Find the integral $\int \frac{x^{4}+2 x^{2}+4}{\left(1+x^{2}\right)^{3}} d x$. | Solution. The denominator of the fraction $Q(x)=\left(1+x^{2}\right)^{3}$ has two imaginary roots $x= \pm i$ of multiplicity 3. The function is decomposed into the following simple fractions:
$$
\frac{x^{4}+2 x^{2}+4}{\left(1+x^{2}\right)^{3}}=\frac{M_{1} x+N_{1}}{1+x^{2}}+\frac{M_{2} x+N_{2}}{\left(1+x^{2}\right)^{2}... | \frac{11}{8}\operatorname{arctg}x+\frac{3}{4}\frac{x}{(1+x^{2})^{2}}+\frac{9}{8}\frac{x}{1+x^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,724 |
Example 1. Find the integral $\int \cos x \sin ^{2} x d x$. | Solution. Since $\cos x d x=d(\sin x)$, then based on formula (4. $5^{\prime}$ ) for $u=\sin x$ we obtain
$$
\int \cos x \sin ^{2} x d x=\int \sin ^{2} x d(\sin x)=\frac{\sin ^{3} x}{3}+C
$$ | \frac{\sin^{3}x}{3}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,725 |
Example 2. Find the integral $\int \cos ^{3} x \sin ^{2} x d x$. | Solution. In this case $m=3, n=2$. We get
$$
\begin{gathered}
\int \cos ^{3} x \sin ^{2} x d x=\int \cos ^{2} x \sin ^{2} x \cos x d x= \\
=\int\left(1-\sin ^{2} x\right) \sin ^{2} x d(\sin x)=\int\left(\sin ^{2} x-\sin ^{4} x\right) d(\sin x)= \\
=\int \sin ^{2} x d(\sin x)-\int \sin ^{4} x d(\sin x)=\frac{\sin ^{3} ... | \frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,726 |
Example 3. Find the integral $\int \sin ^{4} x d x$. | Solution. Here $n=4, m=0$. Transforming the integrand using the appropriate formulas, we find
$$
\begin{gathered}
\int \sin ^{4} x d x=\int\left(\sin ^{2} x\right)^{2} d x=\int\left(\frac{1-\cos 2 x}{2}\right)^{2} d x= \\
=\frac{1}{4} \int\left(1-2 \cos 2 x+\cos ^{2} 2 x\right) d x=\frac{1}{4} \int d x-\frac{1}{4} \in... | \frac{3}{8}x-\frac{1}{4}\sin2x+\frac{1}{32}\sin4x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,727 |
Example 4. Find the integral $\int \cos ^{5} x d x$. | Solution. In this case $n=0, m=5$. We separate the first-degree factor and express $\cos ^{2} x$ in terms of $\sin ^{2} x$:
$$
\cos ^{5} x=\cos ^{4} x \cos x=\left(\cos ^{2} x\right)^{2} \cos x=\left(1-\sin ^{2} x\right)^{2} \cos x
$$
Introducing a new variable $t$ by the formula $\sin x=t$, we get
$$
\begin{gathere... | \sinx-\frac{2}{3}\sin^{3}x+\frac{1}{5}\sin^{5}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,728 |
Example 5. Find the integral $\int \sin ^{4} x \cos ^{2} x d x$. | ## Solution.
$$
\begin{gathered}
\int \sin ^{4} x \cos ^{2} x d x=\int(\sin x \cos x)^{2} \sin ^{2} x d x=\int\left(\frac{\sin 2 x}{2}\right)^{2} \times \\
\times\left(\frac{1-\cos 2 x}{2}\right) d x=\frac{1}{8} \int \sin ^{2} 2 x(1-\cos 2 x) d x=\frac{1}{8} \int \sin ^{2} 2 x d x- \\
-\frac{1}{8} \int \sin ^{2} 2 x \... | \frac{1}{16}x-\frac{1}{64}\sin4x-\frac{1}{48}\sin^{3}2x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,729 |
Example 6. Find the integral $\int \sin 7 x \sin 3 x d x$. | Solution. Since
$$
\sin 3 x \sin 7 x=\frac{1}{2}[\cos (7-3) x-\cos (7+3) x]=
$$
$$
=\frac{1}{2}(\cos 4 x-\cos 10 x)
$$
then
$$
\begin{gathered}
\int \sin 7 x \sin 3 x d x=\frac{1}{2} \int[\cos 4 x-\cos 10 x] d x= \\
=\frac{1}{2} \int \cos 4 x d x-\frac{1}{2} \int \cos 10 x d x=\frac{1}{8} \sin 4 x-\frac{1}{20} \sin... | \frac{1}{8}\sin4x-\frac{1}{20}\sin10x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,730 |
Example 7. Find the integral $\int \cos 3 x \cos x d x$. | Solution. Transforming the integrand, we find
$$
\begin{gathered}
\int \cos 3 x \cos x d x=\frac{1}{2} \int(\cos 4 x+\cos 2 x) d x=\frac{1}{2} \int \cos 4 x d x+ \\
+\frac{1}{2} \int \cos 2 x d x=\frac{1}{8} \sin 4 x+\frac{1}{4} \sin 2 x+C
\end{gathered}
$$ | \frac{1}{8}\sin4x+\frac{1}{4}\sin2x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,731 |
Example 10. Find the integral $\int \frac{d x}{5 \cos ^{2} x+9 \sin ^{2} x}$. | Solution. The integrand does not change from replacing $\sin x$ with $(-\sin x)$, $\cos x$ with $(-\cos x)$, i.e., $R(-\sin x, -\cos x) \equiv R(\sin x, \cos x)$. We apply the substitution $\operatorname{tg} x=t$. Since in this case
$$
\sin x=\frac{t}{\sqrt{1+t^{2}}}, \cos x=\frac{1}{\sqrt{1+t^{2}}}, x=\operatorname{a... | \frac{1}{3\sqrt{5}}\operatorname{arctg}\frac{3\operatorname{tg}x}{\sqrt{5}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,732 |
Example 1. Find the integral $\int \frac{\sqrt{x+9}}{x} d x$. | Solution. This is an integral of the form (4.32), for which $\frac{a x+b}{c x+d}=x+9$ (i.e., $a=1, b=9, c=0, d=1$), $\frac{p_{1}}{q_{1}}=\frac{1}{2}$.
We apply the substitution
$$
x+9=t^{2}
$$
from which
$$
x=t^{2}-9, d x=2 t d t
$$
and
$$
\begin{gathered}
\int \frac{\sqrt{x+9}}{x} d x=\int \frac{t}{t^{2}-9} 2 t ... | 2\sqrt{x+9}+3\ln|\frac{\sqrt{x+9}-3}{\sqrt{x+9}+3}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,733 |
Example 2. Find the integral $\int \sqrt[3]{\frac{2-x}{2+x}} \cdot \frac{1}{(2-x)^{2}} d x$. | Solution. This is an integral of the form (4.32), for which $\frac{a x+b}{c x+d}=\frac{2-x}{2+x}$, $\frac{p_{1}}{q_{1}}=\frac{1}{3}$
We apply the substitution
$$
\frac{2-x}{2+x}=t^{3}
$$
Express $x, 2-x$ and $d x$ in terms of the new variable $t$:
$$
\begin{gathered}
2-x=t^{3}(2+x), 2-2 t^{3}=x+x t^{3}, 2-2 t^{3}=x... | \frac{3}{8}\sqrt[3]{(\frac{2+x}{2-x})^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,734 |
Example 3. Find the integral $\int \frac{\sqrt{x+1}+2}{(x+1)^{2}-\sqrt{x+1}} d x$. | Solution. Here $\frac{a x+b}{c x+d}=x+1, \frac{p_{1}}{q_{1}}=\frac{1}{2}$. Let
$$
x+1=t^{2}
$$
then
$$
\begin{aligned}
d x & =2 t d t \\
\int \frac{\sqrt{x+1}+2}{(x+1)^{2}-\sqrt{x+1}} d x & =2 \int \frac{t+2}{t^{4}-t} t d t=2 \int \frac{t+2}{t^{3}-1} d t
\end{aligned}
$$
Decomposing the integrand into partial fract... | 2(1+x^{\frac{1}{3}})^{\frac{3}{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,735 |
Example 6. Find the integral $\int \frac{d x}{\sqrt[4]{1+x^{4}}}$. | Solution. Rewriting the integral as
$$
\int \frac{d x}{\sqrt[4]{1+x^{4}}}=\int x^{0}\left(1+x^{4}\right)^{-\frac{1}{4}} d x
$$
we conclude that $m=0, n=4, p=-\frac{1}{4}$. Since $\frac{m+1}{n}+p=\frac{1}{4}-\frac{1}{4}=0$ (an integer), we have the third case of integrability. The substitution
$$
a x^{-n}+b=t^{s}
$$
... | \frac{1}{4}\ln|\frac{+1}{-1}|-\frac{1}{2}\operatorname{arctg}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,736 |
Example 1. Find the integral $\int \operatorname{sh}^{2} x d x$.
---
The text has been translated while preserving the original line breaks and format. | Solution. Since
$$
\operatorname{sh}^{2} x=\frac{1}{2}(\operatorname{ch} 2 x-1)
$$
then
$$
\begin{gathered}
\int \operatorname{sh}^{2} x d x=\frac{1}{2} \int(\operatorname{ch} 2 x-1) d x=\frac{1}{2} \int \operatorname{ch} 2 x d x-\frac{1}{2} \int d x= \\
=\frac{1}{4} \int \operatorname{ch} 2 x d(2 x)-\frac{1}{2} \in... | \frac{1}{4}\operatorname{sh}2x-\frac{1}{2}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,737 |
Example 2. Find the integral $\int \operatorname{sh}^{3} x d x$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Example 2. Find the integral $\int \operatorname{sh}^{3} x d x$. | Solution. Representing the integrand in the form
$$
\operatorname{sh}^{3} x=\operatorname{sh}^{2} x \operatorname{sh} x
$$
and taking into account that
$$
\operatorname{sh} x d x=d(\operatorname{ch} x), \operatorname{sh}^{2} x=\operatorname{ch}^{2} x-1
$$
we obtain
$$
\begin{aligned}
\int \operatorname{sh}^{3} x d... | \frac{\operatorname{ch}^{3}x}{3}-\operatorname{ch}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,738 |
Example 3. Find the integral $\int \operatorname{sh}^{2} x \operatorname{ch}^{2} x d x$. | Solution.
$$
\begin{gathered}
\int \sinh^{2} x \cosh^{2} x d x=\int(\sinh x \cosh x)^{2} d x=\int\left(\frac{1}{2} \sinh 2 x\right)^{2} d x= \\
=\frac{1}{4} \int \sinh^{2} 2 x d x=\frac{1}{4} \int \frac{\cosh 4 x-1}{2} d x=\frac{1}{8} \int \cosh 4 x d x-\frac{1}{8} \int d x= \\
=\frac{1}{32} \int \cosh 4 x d(4 x)-\fra... | \frac{1}{32}\sinh4x-\frac{1}{8}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,739 |
Example 4. Find the integral $\int \operatorname{coth}^{2} x d x$. | Solution.
$$
\begin{aligned}
\int \operatorname{cth}^{2} x d x & =\int \frac{\operatorname{ch}^{2} x}{\operatorname{sh}^{2} x} d x=\int \frac{1+\operatorname{sh}^{2} x}{\operatorname{sh}^{2} x} d x=\int \frac{d x}{\operatorname{sh}^{2} x}+\int d x= \\
& =-\operatorname{cth} x+x+C=x-\operatorname{cth} x+C
\end{aligned}... | x-\operatorname{cth}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,740 |
Example 5. Find the integral $\int \operatorname{ch}^{3} x \operatorname{sh} x d x$. | Solution.
$$
\int \operatorname{ch}^{3} x \operatorname{sh} x d x=\int \operatorname{ch}^{3} x d(\operatorname{ch} x)=\frac{\operatorname{ch}^{4} x}{4}+C
$$
## Problems
Find the integrals of hyperbolic functions:
1. $\int \operatorname{ch}^{2} x d x$.
2. $\int \operatorname{ch}^{3} x d x$.
3. $\int \operatorname{sh... | \frac{\operatorname{ch}^{4}x}{4}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,741 |
Example 1. Calculate the definite integral $\int_{1}^{4} x^{2} d x$. | Solution. By formula (5.3) we have
$$
\int_{1}^{4} x^{2} d x=\left.\frac{x^{3}}{3}\right|_{1} ^{4}=\frac{4^{3}}{3}-\frac{1^{3}}{3}=\frac{64}{3}-\frac{1}{3}=21
$$
Remark According to the Newton-Leibniz theorem, we can take any antiderivative of the integrand. In this case, instead of $\frac{x^{3}}{3}$, we could have t... | 21 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,742 |
Example 2. Calculate the integral $\int_{0}^{\frac{\pi}{2}} \cos x d x$. | Solution.
$$
\int_{0}^{\frac{\pi}{2}} \cos x d x=\left.\sin x\right|_{0} ^{\frac{\pi}{2}}=\sin \frac{\pi}{2}-\sin 0=1-0=1
$$ | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,743 |
Example 3. Calculate $\int_{4}^{9}\left(\frac{2 x}{5}+\frac{1}{2 \sqrt{x}}\right) d x$. | Solution. Based on properties 5 and 6 of the definite integral and formula (5.3), we obtain
$$
\begin{aligned}
& \int_{4}^{9}\left(\frac{2 x}{5}+\frac{1}{2 \sqrt{x}}\right) d x=\frac{2}{5} \int_{4}^{9} x d x+\int_{4}^{9} \frac{1}{2 \sqrt{x}} d x=\left.\frac{2}{5} \cdot \frac{x^{2}}{2}\right|_{4} ^{9}+ \\
& +\left.\sqr... | 14 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,744 |
Example 4. Calculate $\int_{0}^{5} x \sqrt{x+4} d x$. | Solution. Introduce a new variable by the formula $\sqrt{x+4}=t$. Determine $x$ and $d x$. Squaring both sides of the equation $\sqrt{x+4}=t$, we get $x+4=t^{2}$, from which $x=t^{2}-4$ and $d x=2 t d t$. Find the new limits of integration. Substituting the old limits into the formula $\sqrt{x+4}=t$, we get: $\sqrt{0+4... | \frac{506}{15} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,745 |
Example 5. Calculate $\int_{0}^{4} \frac{d x}{\sqrt{x}+1}$. | Solution. Let $\sqrt{x}=t$, then $x=t^{2}$ and $d x=2 t d t$. Substituting the old limits of integration into the formula $\sqrt{x}=t$, we get $\alpha=0$, $\beta=2$.
Therefore,
$$
\begin{gathered}
\int_{0}^{4} \frac{d x}{\sqrt{x}+1}=\int_{0}^{2} \frac{2 t d t}{t+1}=2 \int_{0}^{2} \frac{(t+1)-1}{t+1} d t=2 \int_{0}^{2... | 4-2\ln3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,746 |
Example 6. Calculate $\int_{1}^{e} \ln x d x$. | Solution. We apply the integration by parts formula (5.5).
Letting $u=\ln x, d v=d x$, we determine $d u=\frac{1}{x} d x, v=x$.
Therefore,
$$
\begin{gathered}
\int_{1}^{e} \ln x d x=\left.x \ln x\right|_{1} ^{e}-\int_{1}^{e} \frac{1}{x} x d x=\left.x \ln x\right|_{1} ^{e}-\int_{1}^{e} d x=\left.x \ln x\right|_{1} ^{... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,747 |
Example 7. Calculate $\int_{0}^{\pi} x \sin x d x$. | Solution. Integrating by parts, we get
$$
\begin{aligned}
& \int_{0}^{\pi} x \sin x d x=\int_{0}^{\pi} x d(-\cos x)=\left.x(-\cos x)\right|_{0} ^{\pi}-\int_{0}^{\pi}(-\cos x) d x= \\
= & -\left.x \cos x\right|_{0} ^{\pi}+\left.\sin x\right|_{0} ^{\pi}=-(\pi \cos \pi-0 \cdot \cos 0)+(\sin \pi-\sin 0)=\pi
\end{aligned}
... | \pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,748 |
Example 8. Determine the area bounded by the arc of the cosine curve from $x=-\frac{\pi}{2}$ to $x=\frac{\pi}{2}$ and the $O x$ axis. | Solution. Based on the geometric meaning of the definite integral, we conclude that the desired area is expressed by the integral
$$
S=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x d x
$$
Evaluating this integral, we get
$$
S=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x d x=\left.\sin x\right|_{-\frac{\pi}{2}} ^{\fra... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,749 |
Example 9. Evaluate the integral $I=\int_{0}^{\frac{\pi}{2}} \sqrt{1+\frac{1}{2} \cos ^{2} x} d x$. | Solution. Since $0 \leq \cos ^{2} x \leq 1$, then by formula (5.9) we find
$$
\frac{\pi}{2}<I<\frac{\pi}{2} \sqrt{\frac{3}{2}}
$$
i.e., $1.57<I<1.91$.
## Problems
Calculate the integrals:
1. $\int_{2}^{4}\left(x^{3}+x\right) d x$
2. $\int_{1}^{e} \frac{1}{x} d x$.
3. $\int_{0}^{\frac{\pi}{4}} \frac{d \alpha}{\cos ... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,750 |
Example 1. Determine the area bounded by the lines $x y=6$, $x=1, x=e, y=0$ (Fig. 5.5).
Fig. 5.5
 into formula (5.13), we find:
$$
S=\int_{1}^{e} \frac{6}{x} d x=6 \int_{1}^{e} \frac{d x}{x}=\left.6 \ln x\rig... | 6 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,751 |
Example 2. Calculate the area bounded by the curve $y=6 x-x^{2}-5$ and the $O x$ axis. | Solution. To determine the limits of integration, we find the points of intersection of the curve (parabola) with the $O x$ axis (Fig. 5.6). Solving the system $y=6 x-x^{2}-5, y=0$, we get $x_{1}=1, x_{2}=5$, hence, $a=1, b=5$.
Thus, the desired area is
$$
\begin{gathered}
S=\int_{1}^{5} y d x=\int_{1}^{5}\left(6 x-x... | \frac{32}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,752 |
Example 3. Determine the area bounded by the parabola $y=x^{2}-5 x+6$ and the coordinate axes (and adjacent to both axes). | Solution. Let's find the limits of integration. Since the figure is bounded on the left by the $O y$ axis, the lower limit of integration $a=0$. The curve $y=x^{2}-5 x+6$ intersects the $O x$ axis at two points: $x_{1}=2, x_{2}=3$ (Fig. 5.7), so the upper limit $b=2$. Thus,
$$
S=\int_{0}^{2}\left(x^{2}-5 x+6\right) d ... | \frac{14}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,753 |
Example 4. Calculate the area of the figure bounded by the curve $y^{2}=x^{3}$, the line $y=1$, and the $O y$ axis (Fig. 5.8). | Solution. We will determine the required area using formula (5.16). Substituting the values $c=0, d=1$ and the expression $x=y^{\frac{2}{3}}$, we get
$$
S=\int_{0}^{1} y^{\frac{2}{3}} d y=\left.\frac{y^{\frac{5}{3}}}{\frac{5}{3}}\right|_{0}^{1}=\left.\frac{3}{5} y^{\frac{5}{3}}\right|_{0}^{1}=\frac{3}{5}
$$
. The required area is calculated using formula (5.15). First, note that $y_{1}=\sqrt{x... | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,755 |
Example 6. Calculate the area bounded by the parabola $x=8 y-y^{2}-7$ and the $O y$ axis. | Solution. To determine the limits of integration, we find the points of intersection of the curve with the $O y$ axis.
Solving the system of equations $x=8 y-y^{2}-7, x=0$, we get $\quad y_{1}=1, \quad y_{2}=7$
(Fig. 5.10). By formula (5.16) we find
$$
\begin{aligned}
& S=\int_{1}^{7}\left(8 y-y^{2}-7\right) d y=8 \... | 36 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,756 |
Example 7. Determine the area bounded by one arch of the cycloid $x=a(t-\sin t), y=a(1-\cos t)$ and the $O x$ axis (see Fig. 5.11). | Solution. By formula (5.14) we get:
$$
S=\int_{0}^{2 \pi a} y d x=\int_{0}^{2 \pi} a(1-\cos t) a(1-\cos t) d t=a^{2} \int_{0}^{2 \pi}(1-2 \cos t+
$$
$$
\begin{gathered}
+\cos ^{2} t) d t=a^{2} \int_{0}^{2 \pi}\left(1-2 \cos t+\frac{1+\cos 2 t}{2}\right) d t=a^{2} \int_{0}^{2 \pi}\left(\frac{3}{2}-2 \cos t+ \\
+\frac{... | 3\pi^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,757 |
Example 8. Calculate the area bounded by the Bernoulli lemniscate $r^{2}=a^{2} \cos 2 \varphi$. | Solution. The curve is symmetric with respect to the coordinate axes (Fig. 5.12), so it is sufficient to determine one quarter of the desired area using formula (2.17):
$$
\frac{1}{4} S=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} a^{2} \cos 2 \varphi d \varphi=\frac{a^{2}}{4} \int_{0}^{\frac{\pi}{4}} \cos 2 \varphi d(2 \varp... | ^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,758 |
Example 1. Find the length of the arc of the astroid $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$. What is the length of the astroid when $a=1, a=\frac{2}{3}?$ | Solution. Since the astroid
Fig. 5.13 is symmetric with respect to the coordinate axes (Fig. 5.13), it is sufficient to compute the length of the arc $AB$ and multiply the result by 4.
Diff... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,759 |
Example 2. Calculate the length of the arc of the semicubical parabola $y^{2}=x^{3}$, intercepted by the line $x=5$. | Solution. The given arc consists of two parts symmetric with respect to the $O x$ axis (see Fig. 5.8).
Let's compute the length of one of them. By finding the derivative of the function $y^{2}=x^{3}$ and substituting it into formula (5.18), we get
$$
\begin{gathered}
\frac{1}{2} l=\int_{0}^{5} \sqrt{1+\left(\frac{3}{... | \frac{670}{27} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,760 |
Example 3. Find the length of the arc of one arch of the cycloid $x=a(t-\sin t)$, $y=a(1-\cos t)$. | Solution. A moving point describes one arch of a cycloid (Fig. 5.11) when $t$ changes from 0 to $2 \pi$. Therefore, in formula (5.19), $t_{1}=0, t_{2}=2 \pi$.
Let's find the expression for the integrand in formula (5.19). Differentiating the parametric equations of the cycloid, we get:
$$
x_{t}^{\prime}=a(1-\cos t) ;... | 8a | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,761 |
Example 4. Find the length of the arc of the involute (unwinding) of a circle: $x=a(t \sin t+\cos t), y=a(\sin t-t \cos t)$ from point $A(t=0)$ to an arbitrary point $M(t)$. | Solution. By formula (5.19) we get
$$
l=\int_{0}^{t} \sqrt{x'^2+y'^2} d t=a \int_{0}^{t} \sqrt{(t \cos t)^{2}+(t \sin t)^{2}} d t=a \int_{0}^{t} t d t=\frac{a t^{2}}{2}
$$
The cycloid is depicted in Fig. 5.14. | \frac{^{2}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,762 |
Example 6. Calculate the length of the curve $r=a \sin ^{3} \frac{\varphi}{3}$. | Solution. The entire curve is described by a point as $\varphi$ changes from 0 to $3 \pi$ (Fig. 5.16).
Fig. 5.16
$$
\text { Since } r^{\prime}=a \sin ^{2} \frac{\varphi}{3} \cos \frac{\varphi}{3} \text {, then based on formula (5.20) we find }
$$
$$
\begin{aligned}
& l=\int_{0}^{3 \pi} \sqrt{a^{2} \sin ^{6} \frac{\v... | \frac{3\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,763 |
Example 1. Find the volume of the body obtained by rotating around the $O x$ axis the curvilinear trapezoid bounded by the parabola $y^{2}=2 x$, the line $x=3$ and the $O x$ axis. | Solution. According to the condition of the problem, we determine the limits of integration $a=0, b=3$ (Fig. 5.20).
By formula (5.21), we find
$$
V_{x}=\pi \int_{0}^{3} y^{2} d x=\pi \int_{0}^{3} 2 x d x=\left.\pi x^{2}\right|_{0} ^{3}=9 \pi
$$
Fig. 5.22
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
$$
around the $O y$ axis. (This body is called an ellip... | Solution. Since in this case $c=-b, d=b$,
$$
x^{2}=a^{2}\left(1-\frac{y^{2}}{b^{2}}\right)
$$
then by formula (5.22) we determine
$$
V_{y}=\pi \int_{-b}^{b} a^{2}\left(1-\frac{y^{2}}{b^{2}}\right) d y=\pi a^{2} \int_{-b}^{b} d y-\pi \frac{a^{2}}{b^{2}} \int_{-b}^{b} y^{2} d y=
$$
$$
\begin{gathered}
=\left.\pi a^{2... | \frac{4}{3}\pi^{2}b | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,765 |
Example 1. The speed of a point $v=0.1 t^{3} m /$ sec. Find the path $s$ traveled by the point over the time interval $T=10$ sec, from the start of the motion. What is the average speed of the point over this interval? | Solution. By formula (5.23) we get
$$
s=\int_{0}^{10} 0.1 t^{3} d t=0.1 \cdot \left.\frac{t^{4}}{4}\right|_{0} ^{10}=250 \mathrm{M}
$$
Therefore,
$$
v_{\mathrm{cp}}=\frac{s}{T}=25 \mathrm{M} / \text{sec}
$$ | 25\mathrm{M}/ | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,766 |
Example 2. How much work is required to stretch a spring by $0.06 m$, if a force of 1 N stretches it by 0.01 m? | Solution. According to Hooke's law, the force $X$ stretching the spring by $x \mu$ is $X=k x$, where $k$ is the proportionality coefficient. Assuming $x=0.01$ m and $X=1$ N, we get $k=100$, therefore, $X=k x=100 x$. The sought work is determined by the formula (5.24):
$$
A=\int_{0}^{0.06} 100 x d x=\left.50 x^{2}\righ... | 0.18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,767 |
Example 3. Calculate the force of water pressure $P$ on a vertically submerged plate having the shape of a triangle $ABC$ with base $AC=b$ and height $BD=h$, assuming that the vertex $B$ of this triangle lies on the free surface of the liquid, and the base $AC$ is parallel to it (Fig. 5.23).
, we obtain
$$
P=\gamma \int_{0}^{h} x \frac{bx}{h} dx = \frac{\... | 9,80665\cdot10^{3}\frac{^{2}}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,768 |
Example 1. Show that the improper integral $\int_{0}^{+\infty} \frac{d x}{1+x^{2}}$ converges. | Solution. By definition
$$
\int_{0}^{+\infty} \frac{d x}{1+x^{2}}=\lim _{b \rightarrow+\infty} \int_{0}^{b} \frac{d x}{1+x^{2}}
$$
Since
$$
\begin{aligned}
& \lim _{b \rightarrow+\infty} \int_{0}^{b} \frac{d x}{1+x^{2}}=\left.\lim _{b \rightarrow+\infty} \operatorname{arctg} x\right|_{0} ^{b}= \\
= & \lim _{b \right... | \frac{\pi}{2} | Calculus | proof | Yes | Yes | olympiads | false | 32,769 |
Example 2. Show that the improper integrals $\int_{0}^{+\infty} x d x$, $\int_{0}^{+\infty} \sin x d x$ diverge. | Solution. Since
$$
\int_{0}^{+\infty} x d x=\lim _{b \rightarrow+\infty} \int_{0}^{b} x d x=\left.\lim _{b \rightarrow+\infty} \frac{x^{2}}{2}\right|_{0} ^{b}=\frac{1}{2} \lim _{b \rightarrow+\infty} b^{2}=+\infty
$$
the integral diverges.
Consider the second integral:
$$
\int_{0}^{+\infty} \sin x d x=\lim _{b \rig... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,770 |
Example 3. Investigate the convergence of the improper integral
\[
\int_{1}^{+\infty} \frac{d x}{x^{\alpha}}(\alpha \neq-1)
\] | Solution. By definition
\[
\begin{aligned}
\int_{1}^{+\infty} \frac{d x}{x^{\alpha}} & =\lim _{b \rightarrow+\infty} \int_{1}^{b} \frac{d x}{x^{\alpha}}=\left.\lim _{b \rightarrow+\infty} \frac{x^{-\alpha+1}}{-\alpha+1}\right|_{1} ^{b}= \\
& =\lim _{b \rightarrow+\infty}\left(\frac{b^{-\alpha+1}}{-\alpha+1}-\frac{1}{-... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,771 |
Example 4. Using the comparison test, show that the integral $\int_{1}^{\infty} \frac{d x}{\sqrt{1+x^{4}}}$ converges. | Solution. Transform the integrand:
$$
\frac{1}{\sqrt{1+x^{4}}}=\frac{1}{\sqrt{x^{4}\left(\frac{1}{x^{4}}+1\right)}}=\frac{1}{x^{2} \sqrt{1+\frac{1}{x^{4}}}}
$$
Since as $x \rightarrow \infty$
$$
\frac{1}{x^{2} \sqrt{1+\frac{1}{x^{4}}}}<\frac{1}{x^{2}}
$$
and $\int_{1}^{\infty} \frac{1}{x^{2}} d x$ converges (see ex... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,772 |
Example 6. Investigate for which values of $\alpha>0$ the improper integral $\int_{a}^{b} \frac{d x}{(x-a)^{\alpha}}(b>a)$ converges. | Solution. If $\alpha \neq 1$, then the integral
$$
\int_{a+\varepsilon}^{b} \frac{d x}{(x-a)^{\alpha}}=\frac{1}{1-\alpha}\left[(b-a)^{1-\alpha}-\varepsilon^{1-\alpha}\right]
$$
as $\varepsilon \rightarrow 0$ has a limit of $\infty$ or a finite number $\frac{1}{1-\alpha}(b-a)^{1-\alpha}$ depending on whether $\alpha>1... | \alpha<1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,773 |
Example 1. Find the domain of the function $z=A x+B y$ | Solution. The function of two variables $z=A x+B y$ is defined for all $x$ and $y$. Therefore, its domain of existence will be the entire plane Oxy. (The geometric representation of this function is a plane in space.) | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,775 |
Example 2. Find the domain of the function $x^{2}+y^{2}+z^{2}=9$. | Solution. Solving this equation for $z$, we obtain two functions
$$
z= \pm \sqrt{9-x^{2}-y^{2}}
$$
These functions are defined when the expression under the square root is non-negative, i.e.,
$$
9-x^{2}-y^{2} \geq 0 \text { or } x^{2}+y^{2} \leq 9
$$
The last inequality is satisfied by the coordinates of all points... | x^{2}+y^{2}\leq9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 32,776 |
Example 3. Find the domain of the function $z=\sqrt{x y}$.
---
The text has been translated while preserving the original line breaks and format. | Solution. The function is defined when the expression under the root is non-negative, i.e., $x y \geq 0$. This is possible in two cases: 1) $x \geq 0, y \geq 0 ; 2) x \leq 0, y \leq 0$. The first condition is satisfied by the coordinates of all points lying in the first quadrant and on the coordinate axes, the second b... | \ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,777 |
Example 4. Find the domain of the function
$$
z=\frac{1}{\sqrt{9-x^{2}-y^{2}}}
$$ | Solution. The function is defined when the expression under the square root is positive (unlike in the previous example, equality to zero is excluded here), i.e.
$$
9-x^{2}-y^{2}>0 \text { or } x^{2}+y^{2}<9 \text {. }
$$
The latter inequality is satisfied by points lying inside a circle of radius $R=3$ (boundary poi... | x^{2}+y^{2}<9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,778 |
Example 5. Determine the increments of the function $z=x y$, when $x$ and $y$ change from the point $M_{0}(1,2)$ to the points: $M_{1}(1,1 ; 2), \quad M_{2}(1 ; 1,9)$, $M_{3}(1,1 ; 2,2)$. | ## Solution.
1. When $x$ and $y$ change from point $M_{0}(1,2)$ to point $M_{1}(1.1, 2)$, only the argument $x$ undergoes an increment, with $\Delta x=1.1-1=0.1$. The partial increment of the function with respect to $x$ is determined by formula (6.1):
$$
\begin{gathered}
\Delta_{x} z=(x+\Delta x) y-x y=x y+\Delta x ... | 0.2,-0.1,0.42 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,779 |
Example 6. Find the level curves of the function $z=x y$. | Solution. A level line of the function $z=f(x, y)$ is the geometric locus of points in the plane $O x y$ for which the given function has the same value: the equation of the level line is
$$
f(x, y)=c
$$
In this case, we have $x y=c$. The level lines are hyperbolas when $c \neq 0$. | x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,780 |
Example 8. Given the function $f(x, y)=\frac{2 x+y}{x+2 y}$. Prove that:
1) $f(1,2)=\frac{1}{f(2,1)}$
2) $f(c, c)=-f(-c, c)$ | Solution. Since
$$
f(1,2)=\frac{2 \cdot 1+2}{1+2 \cdot 2}=\frac{4}{5}, f(2,1)=\frac{2 \cdot 2+1}{2+2 \cdot 1}=\frac{5}{4}
$$
then
$$
f(1,2) \cdot f(2,1)=1, f(1,2)=\frac{1}{f(2,1)}
$$
Since
$$
\begin{aligned}
f(c, c)=\frac{2 c+c}{c+2 c}=\frac{3 c}{3 c} & =1, f(-c, c)=\frac{2(-c)+c}{-c+2 c}= \\
= & \frac{-c}{c}=-1
\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,781 |
Example 1. Find $\lim _{\substack{x \rightarrow 2 \\ y \rightarrow 0}} \frac{\sin x y}{y}$. | Solution. At the point $M(2,0)$, the function $z=\frac{\sin x y}{y}$ is undefined.
By multiplying and dividing the given function by $x \neq 0$, we get
$$
\frac{\sin x y}{y}=\frac{x \sin x y}{x y}=x \frac{\sin x y}{x y}
$$
Taking the limit in the last equality, we obtain
$$
\lim _{\substack{x \rightarrow 2 \\ y \ri... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,782 |
Example 2. Given the function
$$
f(x, y)=x \sin \frac{1}{y}+y \sin \frac{1}{x} \quad\left(x^{2}+y^{2} \neq 0\right), \quad f(0, y)=0, \quad f(x, 0)=0
$$
find $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} f(x, y)$. | Solution. Let $\varepsilon>0$, then for $|x|<\frac{\varepsilon}{2},|y|<\frac{\varepsilon}{2}$ we get $\rho=$
$$
=\sqrt{\left(\frac{\varepsilon}{2}\right)^{2}+\left(\frac{\varepsilon}{2}\right)^{2}}=\frac{\sqrt{2}}{2} \varepsilon . \text { We form the difference } f(x, y)-0 \text { and estimate it: }
$$
$$
|f(x, y)-0|... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,783 |
Example 3. Find $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x^{2} y}{x^{2}+y^{2}}$. | Solution. The given function can be represented as:
$$
f(x, y)=\frac{x^{2} y}{x^{2}+y^{2}}=\frac{x y}{x^{2}+y^{2}} x .
$$
Since
$$
\left|\frac{x y}{x^{2}+y^{2}}\right| \leq \frac{1}{2}
$$
(this can be derived from the inequality $(x-y)^{2} \geq 0$ : $x^{2}-2 x y+y^{2} \geq 0$, $\left.x^{2}+y^{2} \geq 2 x y, \frac{1... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,784 |
Example 4. Find $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$. | Solution. Dividing the numerator and the denominator by $x^{2}$, we get
$$
\frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\frac{1-\left(\frac{y}{x}\right)^{2}}{1+\left(\frac{y}{x}\right)^{2}}
$$
Since the ratio $\frac{y}{x}$ does not have a limit as $x \rightarrow 0$ and $y \rightarrow 0$ arbitrarily, the given function also does n... | \frac{1-k^{2}}{1+k^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,785 |
Example 6. Show that the function $z=x y$ is continuous at any point in the plane $O x y$. | Solution. First of all, the function $z=x y$ is defined for all $x$ and $y$, i.e., at all points in the plane $O x y$. The total increment of the function is given by the formula
$$
\Delta z=x \Delta y+y \Delta x+\Delta x \Delta y
$$
Taking the limit as $\Delta x \rightarrow 0, \Delta y \rightarrow 0$, we get
$$
\li... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,786 |
Example 7. Find the points of discontinuity of the function
$$
f(x, y)=\frac{3}{4-x^{2}-y^{2}}
$$ | Solution. The function is undefined where the denominator is zero, i.e., at points where $4-x^{2}-y^{2}=0$. This function is discontinuous at every point on the circle $x^{2}+y^{2}=4$. Therefore, the line of discontinuity is the circle $x^{2}+y^{2}=4$.
## Problems
Find the limits of the functions:
1. $\lim _{\substa... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,787 |
Example 2. Find the partial derivatives of the function
$$
u=\frac{x}{x^{2}+y^{2}+z^{2}}
$$ | Solution. Using the rules for finding partial derivatives, we get:
$$
\frac{\partial u}{\partial x}=\left(\frac{x}{x^{2}+y^{2}+z^{2}}\right)^{\prime} x=\frac{x^{\prime}\left(x^{2}+y^{2}+z^{2}\right)-\left(x^{2}+y^{2}+z^{2}\right)_{x}^{\prime} x}{\left(x^{2}+y^{2}+z^{2}\right)^{2}}=
$$
$$
\begin{aligned}
& =\frac{x^{2... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,788 | |
Example 5. How will the volume of a rectangular parallelepiped with dimensions $a=8 \, m, b=6 \, m, c=3 \, m$ change when its length and width
increase by $10 \, \mathrm{cm}$ and $5 \, \mathrm{cm}$ respectively, and its height decreases by $15 \, \mathrm{cm}$. | Solution. The volume of a parallelepiped is expressed by the formula $V=x y z$, where $x, y, z,$ are its dimensions. The increment of volume can be approximately calculated by the formula
$$
\Delta V \approx d V, \text { where } d V=y z d x+x z d y+x y d z
$$
Since according to the condition $x=8, y=6, z=3, d x=0.1, ... | -4.2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 32,789 |
Example 6. Calculate approximately $1.02^{3.01}$. | Solution. Consider the function $z=x^{y}$. The desired number can be considered as the incremental value of this function at $x=1, y=3$, $\Delta x=0.02, \Delta y=0.01$.
The initial value of the function $z=1^{3}=1$. Based on formulas (7.4) and (7.5), we get
$$
\Delta z \approx d z=y x^{y-1} \Delta x+x^{y} \ln x \Delt... | 1.06 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,790 |
Example 1. Find the second-order partial derivatives of the function $z=\left(x^{2}+y^{2}\right)^{2}$. | Solution. First, we find the first-order partial derivatives:
$$
\frac{\partial z}{\partial x}=2\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}\right)_{x}^{\prime}=2\left(x^{2}+y^{2}\right) 2 x=4 x^{3}+4 x y^{2}
$$
$$
\frac{\partial z}{\partial y}=2\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}\right)_{y}^{\prime}=2\left(x^{2}+... | \frac{\partial^{2}z}{\partialx\partialy}=\frac{\partial^{2}z}{\partialy\partialx}=8xy | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,791 |
Example 2. Given the function $u=x^{2} y^{3}$. Find its third-order partial derivatives.
---
The original text has been translated into English, maintaining the original line breaks and formatting. | Solution. Differentiating, we find:
$$
\begin{gathered}
u_{x}^{\prime}=2 x y^{3}, u_{y}^{\prime}=3 x^{2} y^{2}, u_{x x}^{\prime \prime}=2 y^{3}, u_{x y}^{\prime \prime}=6 x y^{2} \\
u_{y x}^{\prime \prime}=6 x y^{2}, u_{y y}^{\prime \prime}=6 x^{2} y, u_{x x x}^{\prime \prime \prime}=0, u_{x y x}^{\prime \prime \prime... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,792 |
Example 3. Prove that the function
$$
u=\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}
$$
satisfies the equation $\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}+\frac{\partial^{2} u}{\partial z^{2}}=0$. | Solution. First, we find the partial derivatives of the given function $u=\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}:$
$$
\begin{gathered}
\frac{\partial u}{\partial x}=-x\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}} \\
\frac{\partial^{2} u}{\partial x^{2}}=3 x^{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,793 |
Example 1. Find the first and second derivatives of the implicit function
$$
\ln \sqrt{x^{2}+y^{2}}-\operatorname{arctg} \frac{y}{x}=0
$$ | Solution. In this case, $F(x, y)=\ln \sqrt{x^{2}+y^{2}}-\operatorname{arctg} \frac{y}{x}$. The derivative is defined by formula (7.16). We find the partial derivatives of this function with respect to $x$ and $y$:
$$
\begin{gathered}
F_{x}^{\prime}=\left[\frac{1}{2} \ln \left(x^{2}+y^{2}\right)-\operatorname{arctg} \f... | \frac{2(x^{2}+y^{2})}{(x-y)^{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,794 |
Example 2. Find the total differential of the function $z=z(x, y)$, given by the equation $e^{x y z}-\operatorname{arctg} \frac{x y}{z}=0$. | Solution. In this case, $F(x, y, z)=e^{x y z}-\operatorname{arctg} \frac{x y}{z}$. First, we find the partial derivatives of the function $F(x, y, z)$:
$$
\begin{gathered}
F_{x}^{\prime}=\frac{\partial F}{\partial x}=y z e^{x y z}-\frac{1}{1+\frac{x^{2} y^{2}}{z^{2}}} \cdot \frac{y}{z}=y z\left(e^{x y z}-\frac{1}{x^{2... | \frac{1-(x^{2}y^{2}+z^{2})e^{xyz}}{1+(x^{2}y^{2}+z^{2})e^{xyz}}\cdot\frac{z}{xy}(y+x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,795 |
Example 3. Find the total differential of the function $z=z(x, y)$, given by the equation $z^{2}-2 x y=c$. | Solution. Let's take the differentials of both sides of the given equation:
$$
\begin{gathered}
d\left(z^{2}-2 x y\right)=d c, d\left(z^{2}\right)-d(2 x y)=d c \\
2 z d z-2(x d y+y d x)=0, z d z-(x d y+y d x)=0 \\
z d z=x d y+y d x, d z=\frac{y}{z} d x+\frac{x}{z} d y
\end{gathered}
$$
## Problems
Find the derivativ... | \frac{y}{z}+\frac{x}{z} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,796 |
Example 1. Find $\frac{d u}{d x}$, if $u=e^{z-2 y}$, where $z=\sin x, y=x^{2}$. | Solution. The function $u=e^{z-2 y}$ depends on two intermediate arguments $v_{1}=y, v_{2}=z$. According to formula (7.21), which in this case takes the form
$$
\frac{d u}{d x}=\frac{\partial u}{\partial y} \cdot \frac{d y}{d x}+\frac{\partial u}{\partial z} \cdot \frac{d z}{d x}
$$
we obtain
$$
\begin{gathered}
\fr... | e^{\sinx-2x^{2}}(\cosx-4x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,797 |
Example 2. Find the total differential of the function
$$
u=x^{2} \operatorname{arctg} \frac{y}{x}-y^{2} \operatorname{arctg} \frac{x}{y}
$$ | Solution. The function $u$ of two variables $x$ and $y$ can be represented by the following formula:
$$
u \equiv F\left(v_{1}, v_{2}, v_{3}, v_{4}\right)=v_{1} v_{2}-v_{3} v_{4}
$$
where
$$
v_{1}=x^{2}, v_{2}=\operatorname{arctg} \frac{y}{x}, v_{3}=y^{2}, v_{4}=\operatorname{arctg} \frac{x}{y}
$$
The total differen... | (2x\operatorname{arctg}\frac{y}{x}-y)+(x-2y\operatorname{arctg}\frac{x}{y}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,798 |
Example 1. Find the direction vector of the normal to the ellipsoid $x^{2}+2 y^{2}+3 z^{2}=6$ at the point $M_{0}(1,-1,1)$. | Solution. First of all, the point $M_{0}$ lies on the ellipsoid, which can be verified by substituting its coordinates into the given equation. This equation can be rewritten as
$$
x^{2}+2 y^{2}+3 z^{2}-6=0
$$
By comparing this equation with equation (8.1), we conclude that
$$
F(x, y, z)=x^{2}+2 y^{2}+3 z^{2}-6
$$
... | {2,-4,6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,799 |
Example 2. Form the equations of the normal and the tangent plane to the sphere $x^{2}+y^{2}+z^{2}-2 x+4 y-6 z+5=0$ at the point $M_{0}(3,-1,5)$. | Solution. Let's find the partial derivatives of the function $F(x, y, z) = x^{2} + y^{2} + z^{2} - 2x + 4y - 6z + 5$:
$$
\frac{\partial F}{\partial x} = 2x - 2, \frac{\partial F}{\partial y} = 2y + 4, \frac{\partial F}{\partial z} = 2z - 6
$$
Evaluating these derivatives at the point $M_{0}(3, -1, 5)$, we get the dir... | 2x+y+2z-15=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 32,800 |
Example 3. At what point is the tangent plane to the elliptic paraboloid $z=2 x^{2}+4 y^{2} \quad$ parallel to the plane $8 x-32 y-2 z+3=0$? Write the equations of the normal and the tangent plane at this point. | Solution. In this case, the surface is given by an equation solved for $z$, i.e., an equation of the form (8.5), where $f(x, y)=2 x^{2}+4 y^{2}$.
The equation of the tangent plane at any point $M$ according to formula (8.6) is
$$
z-z_{0}=4 x_{0}\left(x-x_{0}\right)+8 y_{0}\left(y-y_{0}\right)
$$
or
$$
4 x_{0}\left(... | x_{0}=1,y_{0}=-2,z_{0}=18 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,801 |
Example 1. Investigate the function for extremum
$$
f(x, y)=x^{3}+y^{3}+9 x y
$$ | Solution. We find the first and second partial derivatives:
$$
\begin{gathered}
f_{x}^{\prime}(x, y)=3 x^{2}+9 y ; f_{y}^{\prime}(x, y)=3 y^{2}+9 x \\
f_{x x}^{\prime \prime}(x, y)=6 x ; f_{x y}^{\prime \prime}(x, y)=9 ; f_{y y}^{\prime \prime}(x, y)=6 y
\end{gathered}
$$
Setting the first derivatives to zero, we obt... | 27 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,802 |
Example 2. Find the extremum of the function
$$
f(x, y)=x^{3}+3 x y^{2}-18 x^{2}-18 x y-18 y^{2}+57 x+138 y+290
$$ | Solution. We find the first and second partial derivatives:
$$
\begin{gathered}
f_{x}^{\prime}(x, y)=3 x^{2}+3 y^{2}-36 x-18 y+57 \\
f_{y}^{\prime}(x, y)=6 x y-18 x-36 y+138 \\
f_{x x}^{\prime \prime}(x, y)=6 x-36 ; f_{x y}^{\prime \prime}(x, y)=6 y-18 ; f_{y y}^{\prime \prime}(x, y)=6 x-36
\end{gathered}
$$
By setti... | \maxf(x,y)=19,\f(x,y)=10 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,803 |
Example 1. Find the maximum and minimum values of the function $z=f(x, y)=2 x^{2}-2 y^{2}$ in the circle $x^{2}+y^{2} \leq 9$. | Solution. The given function has partial derivatives:
$$
f_{x}^{\prime}(x, y)=4 x ; f_{y}^{\prime}(x, y)=-4 y
$$
Setting these derivatives to zero, we get a system of equations from which we find $x_{0}=0, y_{0}=0$. The value of the function at the critical point $M_{0}(0,0)$ is zero:
$$
z_{0}=f(0,0)=2 \cdot 0-2 \cd... | z_{\text{max}}=18,\;z_{\text{}}=-18 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,804 |
Example 2. Find the maximum and minimum values of the function $z=f(x, y)=x^{3}+y^{3}+6 x y \quad$ in the rectangle with vertices $A(-3,-3), B(-3,2), C(1,2), D(1,-3)$. | Solution. We take the partial derivatives of the given function:
$$
f_{x}^{\prime}(x, y)=3 x^{2}+6 y ; f_{y}^{\prime}(x, y)=3 y^{2}+6 x
$$
From the system of equations
$$
\left.\left.\begin{array}{c}
3 x^{2}+6 y=0 \\
3 y^{2}+6 x=0
\end{array}\right\} \text { or } \begin{array}{r}
x^{2}+2 y=0 \\
y^{2}+2 x=0
\end{arra... | -55 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,805 |
Example 1. Determine the order of the differential equation:
1) $y^{\prime \prime}-3 y^{\prime}+2 y-4=0$
2) $x(1+x) y^{\prime}-(1+2 x) y-(1+2 x)=0$;
3) $y^{\mathrm{IV}}-16 y^{\prime \prime}=0$
4) $y^{\prime \prime \prime}-6 y^{\prime \prime}+11 y^{\prime}-6 y=0$. | Solution. The first equation is a second-order differential equation, since the order of the highest derivative in it is 2, the second is a first-order equation, as it contains only the first derivative. (Note that in the first equation, the coefficients of $y, y^{\prime}, y^{\prime \prime}$ and the free term are numbe... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,806 |
Example 2. Show that the function $y=e^{2 x}$ is a solution to the differential equation $y^{\prime \prime \prime}-8 y=0$. | Solution. Let's find the third derivative of the given function:
$$
y^{\prime}=2 e^{2 x}, y^{\prime \prime}=4 e^{2 x}, y^{\prime \prime \prime}=8 e^{2 x}
$$
Substituting the expressions for $y$ and $y^{\prime \prime \prime}$ into the differential equation, we get the identity $8 e^{2 x}-8 e^{2 x}=0$. This means that ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,807 |
Example 4. Verify that the function $y=C_{1} e^{x}+C_{2} e^{x}+\frac{1}{3} x+\frac{1}{9}$ is a solution to the equation $y^{\prime \prime}-4 y^{\prime}+3 y=x-1$. | Solution. Let's find the derivatives
$$
y^{\prime}=C_{1} e^{x}+C_{2} e^{x}+\frac{1}{3}, y^{\prime \prime}=C_{1} e^{x}+C_{2} e^{x}
$$
and, substituting the expressions for $y, y^{\prime}$ and $y^{\prime \prime}$ into the given equation, we obtain the identity
$$
\begin{aligned}
& \left(C_{1} e^{x}+C_{2} e^{x}\right)-... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,808 |
Example 5. Show that the function $y=C_{1} e^{3 x}+C_{2} e^{x}+\frac{1}{3} x+\frac{1}{9}$ is the general solution of the equation $y^{\prime \prime}-4 y^{\prime}+3 y=x-1$. | Solution. The given function contains two independent arbitrary constants (their number cannot be reduced, as in the previous example). If we show that the function satisfies the equation, then this will mean that it is the general solution of the given differential equation.
Since
$$
y^{\prime}=3 C_{1} e^{3 x}+C_{2}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,809 |
Example 6. Given the general solution $y=C_{1} \sin 2 x+C_{2} \cos 2 x$ of the differential equation $y^{\prime \prime}+4 y=0$. What particular solutions are obtained when $C_{1}=2, C_{2}=3$? For what values of the parameters $C_{1}$ and $C_{2}$ do the particular solutions $y=\sin 2 x, y=\cos 2 x$ result? | Solution. Substituting the values $C_{1}=2, C_{2}=3$ into the formula for the general solution, we obtain the particular solution $y=2 \sin 2 x+3 \cos 2 x$. The particular solution $y=\sin 2 x$ is obtained from the general solution when $C_{1}=1, C_{2}=0$, and the particular solution $y=\cos 2 x$ when $C_{1}=0, C_{2}=1... | 2\sin2x+3\cos2x,\,\sin2x\,(C_{1}=1,C_{2}=0),\,\cos2x\,(C_{1}=0,C_{2}=1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,810 |
Example 8. Find the differential equation for which the function $y=C_{1} x+C_{2}$, depending on two arbitrary constants, is the general solution. | Solution. Differentiating the given function twice, we eliminate the parameters $C_{1}$ and $C_{2}$:
$$
y^{\prime}=C_{1}, y^{\prime \prime}=0
$$
The equation $y^{\prime \prime}=0$ satisfies the condition of the problem. | y^{\\}=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,811 |
Example 9. Form a differential equation whose general solution is
$$
y=C_{1} x+\frac{C_{2}}{x}
$$ | Solution. Differentiating the given function, we get:
$$
y^{\prime}=C_{1}-\frac{C_{2}}{x^{2}}, y^{\prime \prime}=C_{1}-\frac{2 C_{2}}{x^{3}}
$$
From the equations
$$
y=C_{1} x+\frac{C_{2}}{x}, y^{\prime}=C_{1}-\frac{C_{2}}{x^{2}}
$$
we eliminate $C_{1}$. Multiplying the second equation by $x$ and subtracting the fi... | xy^{\}-y+y^{\\}x^{2}=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,812 |
Example 10. Find the differential equation of the family of curves $y=C x^{3}$. | Solution. Differentiating the given function, we obtain $y^{\prime}=3 C x^{2}$. Substituting the expression $C=\frac{y}{x^{3}}$, obtained from the given equation, we find the required differential equation
$$
y^{\prime}=3 \frac{y}{x^{3}} \cdot x^{2}, y^{\prime}=\frac{3 y}{x}
$$
## Problems
1. Determine the order of ... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,813 |
Example 2. Solve the equation
$$
\frac{d y}{d x}=-\frac{y}{x}(x \neq 0)
$$
Find the particular solution that satisfies the condition: $y=3$ when $x=2$. | Solution. Separating the variables, we get
$$
\frac{d y}{y}+\frac{d x}{x}=0
$$
Integrating, we find
$$
\ln |y|+\ln |x|=C_{1}
$$
The constant $C_{1}$ can be written as
$$
C_{1}=\ln |C|(C \neq 0)
$$
(since any positive or negative number $C_{1}$ can be represented as the natural logarithm of another, positive numbe... | x6 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,814 |
Example 3. Integrate the differential equation
$$
\left(1+x^{2}\right) d y-2 x y d x=0
$$
Find a particular solution that satisfies the condition: $y=1$ when $x=0$. Solution. The given equation is a separable equation (the coefficient of $d y$ is a function of $x$ only, and the coefficient of $d x$ is a product of fu... | Solution. By factoring out the corresponding multipliers, the given equation can be written as:
$$
x\left(y^{2}+1\right) d x+y\left(1-x^{2}\right) d y=0
$$
from which it is clear that this is a separable equation. Dividing both sides of the last equation by the product $\left(y^{2}+1\right)\left(1-x^{2}\right) \neq 0... | 1+y^{2}=C(1-x^{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,815 |
Example 5. Find the general solution of the differential equation
$$
d r-r d \varphi=0
$$ | Solution. In this equation, the unknown function is denoted by the letter $r$, and its argument by the letter $\varphi$. Separating the variables, we get
$$
\frac{d r}{r}=d \varphi
$$
from which
$$
\ln r=\varphi+C_{1} .
$$
From the last equation, we find
$$
r=e^{\varphi+C_{1}}=C e^{\varphi}
$$
where $C=e^{C_{1}}$... | Ce^{\varphi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,816 |
Example 1. Integrate the homogeneous differential equation
$$
x d y-\left(y+\sqrt{x^{2}+y^{2}}\right) d x=0
$$ | Solution. The coefficients of $d y$ and $d x$ are respectively:
$$
Q(x, y)=x ; P(x, y)=-\left(y+\sqrt{x^{2}+y^{2}}\right) .
$$
The functions $P(x, y)$ and $Q(x, y)$ are homogeneous functions of the first degree. Indeed,
$$
\begin{gathered}
Q(k x, k y)=k x=k Q(x, y) \\
P(k x, k y)=-\left(k y+\sqrt{k^{2} x^{2}+k^{2} y... | x^{2}-2CC^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,817 |
Example 2. Find the general integral of the homogeneous equation
$$
\left(x^{2}-y^{2}\right) d y-2 y x d x=0
$$ | Solution. In this case, we have:
$$
Q(x, y)=x^{2}-y^{2} ; P(x, y)=-2 x y .
$$
These functions are homogeneous functions of the second degree. Indeed:
$$
\begin{gathered}
Q(k x, k y)=(k x)^{2}-(k y)^{2}=k^{2}\left(x^{2}-y^{2}\right) \equiv k^{2} Q(x, y) \\
P(k x, k y)=-2(k x)(k y)=k^{2}(-2 x y) \equiv k^{2} P(x, y)
\... | x^{2}+y^{2}=Cy | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,818 |
Example 3. Integrate the homogeneous equation
$$
\left(y^{4}-2 x^{3} y\right) d x+\left(x^{4}-2 x y^{3}\right) d y=0
$$ | Solution. The functions
$$
P(x, y)=y^{4}-2 x^{3} y, Q(x, y)=x^{4}-2 x y^{3}
$$
are homogeneous functions of the fourth degree. Let $y=u x$, then $d y=u d x+x d u$ and the equation (after canceling by $x^{4} \neq 0$) will take the form
$$
\left(u^{4}+u\right) d x-\left(1-2 u^{3}\right) x d u=0
$$
or
$$
\frac{d x}{x... | x^{3}+y^{3}=Cxy | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,819 |
Example 4. Integrate the differential equation
$$
(4 y-3 x-5) y^{\prime}+7 x-3 y+2=0
$$ | Solution. This is an equation of the form (9.15)
$$
y^{\prime}=\frac{-7 x+3 y-2}{-3 x+4 y-5}
$$
here $\Delta=\left|\begin{array}{ll}-7 & 3 \\ -3 & 4\end{array}\right|=-28+9=-19 \neq 0$.
We introduce new variables by the formulas (9.17):
$$
\left.\begin{array}{l}
x=u+h ; \\
y=v+k,
\end{array}\right\}
$$
where $h, k... | 2y^{2}-3xy+\frac{7}{2}x^{2}+2x-5C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,820 |
Example 1. Find the general solution of the linear differential equation
$$
y^{\prime}+y-k x=0
$$ | Solution. The given equation is of the form (9.20), where $p(x)=1, q(x)=k x$. Let $y=u v$, then $y^{\prime}=u^{\prime} v+u v^{\prime}$. Substituting these expressions into the original equation, we get
$$
u^{\prime} v+u v^{\prime}+u v-k x=0 \text { or } u\left(v^{\prime}+v\right)+u^{\prime} v-k x=0
$$
We choose $v$ a... | k(x-1)+Ce^{-x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,821 |
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