Evaluation of definite integrals.

Literature: Collection of problems in mathematics. Part 1. Edited by A. V. Efimov, B. P. Demidovich.

The Fundamental Theorem of Calculus.

If $F(x)$ is one of the antiderivatives of a continuous function $f(x)$ on $[a, b]$, then the following Fundamental Theorem of Calculus holds: $$\int\limits_a^b f(x)\,dx=F(x)|_a^b=F(b)-F(a).$$

Examples:

Using the Fundamental Theorem of Calculus, compute the integrals:

6.324. $\int\limits_{-1}^2x^3,dx.$

Solution:

$$\int\limits_{-1}^2x^3\,dx=\left.\frac{x^4}{4}\right|_{-1}^2=\frac{(-1)^4}{4}-\frac{2^4}{4}=\frac{1}{4}-\frac{16}{4}=-\frac{15}{4}=-3,75.$$

Answer: $-\frac{15}{8}$.

6.331. $\int\limits_{-\pi/4}^0\frac{dx}{\cos^2 x}.$

Solution:

$$\int\limits_{-\pi/4}^0\frac{dx}{\cos^2 x}=\left.tg x\right|_{-\pi/4}^0=tg 0-tg(-\pi/4)=0-(-1)=1.$$

Answer: $1.$

6.335. $\int\limits_{1}^2\frac{dx}{2x-1}.$

Solution.

$$\int\limits_{1}^2\frac{dx}{2x-1}=\int\limits_1^2\frac{1}{2}d(\ln(2x-1))=\frac{1}{2}\left.\left(\ln(2x-1)\right)\right|_{1}^2=\ln3-\ln1=\ln 3.$$

Answer: $\ln 3.$

6.347. $\int\limits_0^{\pi/2}\cos^3\alpha\,d\alpha.$

Solution.

$$\int\limits_{0}^{\pi/2}\cos^3\alpha\,d\alpha=\int\limits_0^{\pi/2}\cos^2xd\sin x=\int\limits_0^{\pi/2}(1-\sin^2 x)d\sin x=\left.\left(\sin x-\frac{\sin^3 x}{3}\right)\right|_{0}^{\pi/2}=$$ $$=\sin\frac{\pi}{2}-\frac{\sin^3{\pi/2}}{3}-\left(\sin 0-\frac{\sin^30}{3}\right)=1-\frac{1}{3}=\frac{2}{3}.$$

Answer: $\frac{2}{3}.$

Properties of definite integrals:

1) If $f(x) \geq 0$ on the interval $[a, b]$, then $\int\limits_a^b f(x),dx \geq 0$.

2) If $f(x) \leq g(x)$ on $[a, b]$, then $\int\limits_a^b f(x),dx \leq \int\limits_a^b g(x),dx$.

3) $|\int\limits_a^b f(x),dx| \leq \int\limits_a^b |f(x)|,dx$.

4) If $f(x)$ is continuous on $[a, b]$, $m$ is the smallest and $M$ is the largest value of $f(x)$ on $[a, b]$, then $$m(b-a)\leq\int\limits_a^bf(x)\,dx\leq M(b-a)$$ (theorem on the estimation of definite integrals).

5) If $f(x)$ is continuous, and $g(x)$ is integrable on $[a, b]$ such that $g(x) \geq 0$, and $m$ and $M$ are the smallest and largest values of $f(x)$ on $[a, b]$, then $$m\int\limits_a^b g(x)\,dx\leq\int\limits_a^bf(x)g(x)dx\leq M\int\limits_a^bg(x)\,dx.$$ (обобщенная теорема об оценке определенного интеграла)

6) If $f(x)$ is continuous on $[a, b]$, then there exists a point $c \in (a, b)$ such that the equality holds:$$\int\limits_a^bf(x)dx=f(c)(b-a).$$ (mean value theorem)

The number $f(c)=\frac{1}{b-a}\int\limits_a^b f(x),dx$ is called the mean value of the function $f(x)$ on the interval $[a, b]$.

7) If $f(x)$ is continuous and integrable on $[a, b]$, and $g(x) \geq 0$, then there exists a point $c \in (a, b)$ such that the equality holds: $$\int\limits_a^bf(x)g(x)dx=f(c)\int\limits_a^bg(x)dx$$ (generalized mean value theorem).

8) If $f^2(x)$ and $g^2(x)$ are integrable on $[a, b]$, then$$|\int\limits_a^bf(x)g(x)dx|=\sqrt{\int\limits_a^bf^2(x)dx\int\limits_a^bg^2(x)dx}$$ (Cauchy-Schwarz inequality).

9) Integration of even and odd functions over symmetric limits. If the function $f(x)$ is even, then $\int\limits_{-a}^a f(x)dx = 2\int\limits_0^a f(x)dx$. If the function $f(x)$ is odd, then $\int\limits_{-a}^a f(x)dx = 0$.

10) If the function $f(x)$ is continuous on the interval $[a, b]$, then the integral with a variable upper limit$$\Phi(x)=\int\limits_a^x f(t)dt$$ является первообразной для функции $f(x),$ т.е. $$\Phi'(x)=(\int\limits_a^x f(t)dt)'=f(x),\quad x\in[a, b].$$

11) If the functions $\phi(x)$ and $\psi(x)$ are differentiable at the point $x \in (a, b)$ and $f(t)$ is continuous for $\phi(a) \leq t \leq \psi(b)$, then $$\left(\int\limits_{\phi(x)}^{\psi(x)} f(t)dt\right)_x'=f(\psi(x))\psi'(x)-f(\phi(x))\phi'(x).$$

Examples.

6.364. a) Determine the sign of the integral without evaluating it: $\int\limits_{-2}^1\sqrt[3]{x},dx.$

Solution.

Since the function $\sqrt[3]{x}$ is odd $(\sqrt[3]{-x}=-\sqrt[3]{x})$, using property 9, we have $\int\limits_{-2}^2\sqrt[3]{x},dx=0$. Next, we use the formula $$\int\limits_{-2}^1\sqrt[3]{x}\,dx=\int\limits_{-2}^2\sqrt[3]{x}\,dx-\int\limits_{1}^2\sqrt[3]{x}\,dx=-\int\limits_{1}^2\sqrt[3]{x}\,dx.$$

It's clear that $\sqrt[3]{x} > 0$ when $x \in [1, 2]$. Therefore, from the first property of definite integrals, it follows that $\int\limits_{1}^2\sqrt[3]{x},dx > 0$. Hence, $$\int\limits_{-2}^1\sqrt[3]{x}\,dx=-\int\limits_{1}^2\sqrt[3]{x}\,dx<0.$$

Answer: $\int\limits_{-2}^1\sqrt[3]{x},dx < 0.$

6.365. b) Without computing the integrals, determine which of the integrals is greater, $\int\limits_1^2\frac{dx}{x^2}$ or $\int\limits_1^2\frac{dx}{x^3}$.

Solution:

We will use the second property of definite integrals. On the interval $[1, 2]$, the inequality $\frac{1}{x^2} \geq \frac{1}{x^3}$ holds. Let's verify this: $$\frac{1}{x^2}\geq\frac{1}{x^3}\Rightarrow x^3\geq x^2\Rightarrow x\geq1.$$ Therefore, $$\int\limits_1^2\frac{dx}{x^2}\geq\int\limits_1^2\frac{dx}{x^3}.$$ It is easy to obtain a strict inequality by representing the given integrals as a sum of $$\int\limits_1^2\frac{dx}{x^2}=\int\limits_1^{3/2}\frac{dx}{x^2}+\int\limits_{3/2}^2\frac{dx}{x^2};$$ $$\int\limits_1^2\frac{dx}{x^3}=\int\limits_1^{3/2}\frac{dx}{x^3}+\int\limits_{3/2}^2\frac{dx}{x^3}.$$On the interval $[1, 3/2]$, the inequality $$\frac{1}{x^2}\geq\frac{1}{x^3}\Rightarrow\int\limits_1^{3/2}\frac{dx}{x^2}\geq\int\limits_{3/2}^2\frac{dx}{x^3};$$ On the interval $[3/2, 2]$, the inequality $$\frac{1}{x^2}>\frac{1}{x^3}\Rightarrow\int\limits_1^{3/2}\frac{dx}{x^2}>\int\limits_{3/2}^2\frac{dx}{x^3}.$$ Thus, $$\int\limits_1^2\frac{dx}{x^2}=\int\limits_1^{3/2}\frac{dx}{x^2}+\int\limits_{3/2}^2\frac{dx}{x^2}>\int\limits_1^{3/2}\frac{dx}{x^3}+\int\limits_{3/2}^2\frac{dx}{x^3}=\int\limits_1^2\frac{dx}{x^3}.$$

Answer: $\int\limits_1^2\frac{dx}{x^2}>\int\limits_1^2\frac{dx}{x^3}.$

6.366. в) To find the average value of the function on the given interval, $\cos x, \quad 0 \leq x \leq \frac{\pi}{2}$.

Solution:

We will use the 6th property of definite integrals. The average value of the function $f(x)$ on the interval $[a, b]$ is defined as $f(c)=\frac{1}{b-a}\int\limits_a^b f(x),dx.$

From here, we find $$\cos c=\frac{1}{\pi/2-0}\int\limits_0^{\pi/2}\cos x\,dx=\frac{2}{\pi}\left.\sin x\right|_0^{\pi/2}=\frac{2}{\pi}(1-0)=\frac{2}{\pi}.$$

Answer: $\frac{2}{\pi}.$

6.369. Estimate the integral $\int\limits_0^{2\pi}\frac{dx}{\sqrt{5+2\sin x}}.$

Solution:

Let's estimate the integrand:

$$-1\leq\sin x\leq 1\Rightarrow$$ $$3\leq 5+2\sin x\leq 7\Rightarrow$$ $$\sqrt 3\leq\sqrt{5+2\sin x}\leq 7\Rightarrow$$ $$\frac{1}{\sqrt 7}\leq\frac{1}{\sqrt{5+2\sin x}}\leq\frac{1}{\sqrt 3}.$$

From here and from the second property of definite integrals, it follows that

$$\int\limits_0^{2\pi}\frac{1}{\sqrt 7}dx\leq\int\limits_0^{2\pi}\frac{1}{\sqrt{5+2\sin x}}dx\leq\int\limits_0^{2\pi}\frac{1}{\sqrt 3}dx.$$

We find the limiting integrals: $$\int\limits_0^{2\pi}\frac{1}{\sqrt 7}dx=\frac{1}{\sqrt 7}(2\pi-0)=\frac{2\pi}{\sqrt 7};$$ $$\int\limits_0^{2\pi}\frac{1}{\sqrt 3}dx=\frac{1}{\sqrt 3}(2\pi-0)=\frac{2\pi}{\sqrt 3}.$$

Thus, $$\frac{2\pi}{\sqrt 7}\leq\int\limits_0^{2\pi}\frac{1}{\sqrt{5+2\sin x}}dx\leq\frac{2\pi}{\sqrt 3}.$$

Answer: $\frac{2\pi}{\sqrt 7}\leq\int\limits_0^{2\pi}\frac{1}{\sqrt{5+2\sin x}}dx\leq\frac{2\pi}{\sqrt 3}.$

6.370. б) Estimate the integral $\int\limits_0^1\sqrt{(1+x)(1+x^3)},dx$ using the Cauchy-Schwarz inequality.

Solution:

The Cauchy-Schwarz inequality gives us $$|\int\limits_0^1\sqrt{(1+x)(1+x^3)}dx|\leq\sqrt{\int\limits_0^1(1+x)dx\int\limits_0^1(1+x^3)dx}.$$

Let's compute each integral under the square root on the right-hand side:

$$\int\limits_0^1(1+x)dx=\left.\left(x+\frac{x^2}{2}\right)\right|_0^1=1+\frac{1}{2}-0=\frac{3}{2};$$ $$\int\limits_0^1(1+x^3)dx=\left.\left(x+\frac{x^4}{4}\right)\right|_0^1=1+\frac{1}{4}-0=\frac{5}{4};$$ From here we have $$|\int\limits_0^1\sqrt{(1+x)(1+x^3)}dx|\leq\sqrt{\frac{3}{2}\cdot\frac{5}{4}}=\frac{\sqrt{30}}{4}.$$

Answer: $|I|\leq\frac{\sqrt{30}}{4}.$

6.374. Find the derivative of the following function: $\Phi(x)=\int\limits_0^x\frac{\sin t}{t},dt.$

Solution:

We use property 10:

$$\Phi'(x)=f(x)=\frac{\sin x}{x}.$$

Answer: $\frac{\sin x}{x}.$

6.376. Find the derivative of the following function: $\Phi(x)=\int\limits_x^0\frac{dt}{\sqrt{1+t^3}}.$

Solution:

$\Phi(x)=\int\limits_x^0\frac{dt}{\sqrt{1+t^3}}=-\int\limits_0^x\frac{dt}{\sqrt{1+t^3}}.$

We use property 10:

$$\Phi'(x)=f(x)=-\frac{1}{\sqrt{1+x^3}}.$$

Answer: $-\frac{1}{\sqrt{1+x^3}}.$

Change of variables in a definite integral.

If the function $f(x)$ is continuous on the interval $[a, b]$, and the function $x=\varphi(t)$ is continuously differentiable on the interval $[t_1, t_2]$, where $a=\varphi(t_1)$ and $b=\varphi(t_2)$, then$$\int\limits_a^bf(x)dx=\int\limits_{t_1}^{t_2}f(\varphi(t))\varphi'(t)\,dt.$$

Examples.

Compute the integrals using the indicated substitutions:

6.380. $\int\limits_1^6\frac{dx}{1+\sqrt{3x-2}},\quad 3x-2=t^2.$

Solution.

$$\int\limits_1^6\frac{dx}{1+\sqrt{3x-2}}=\left[\begin{array}{lcl}3x-2=t^2\Rightarrow x=\frac{1}{3}(t^2+2)\\dx=\frac{2}{3}dt\\x=1\Rightarrow t=1\\x=6\Rightarrow t=4\end{array}\right]=\int\limits_1^4\frac{2dt}{3(1+t)}=$$ $$=\left.\frac{2}{3}\ln|1+t|\right|_1^4=\frac{2}{3}\ln5-\frac{2}{3}\ln2=\frac{2}{3}\ln\frac{5}{2}.$$

Answer: $\frac{2}{3}\ln\frac{5}{2}.$

6.382. $\int\limits_0^{sh 1}\sqrt{x^2+1},\quad x=sh t.$

Solution.

$$\int\limits_0^{sh 1}\sqrt{x^2+1}dx=\left[\begin{array}{lcl}x=sh t\Rightarrow dx=ch t\\x=0\Rightarrow t=0\\x=sh 1\Rightarrow t=1\end{array}\right]=\int\limits_0^1\sqrt{sh^2 t+1}ch t\,dt=\int\limits_0^1ch^2t\,dt=$$ $$=\int\limits_0^1\frac{1}{2}(ch 2t+1)\,dt=\frac{1}{2}\left.\left(\frac{1}{2}sh 2t+t\right)\right|_0^1=\frac{1}{2}\left(\frac{1}{2}sh 2+1-\frac{1}{2}sh 0-0\right)=$$ $$=\frac{1}{4}\left(sh 2+2\right).$$

Answer: $\frac{1}{4}(sh 2+2).$

Compute the integrals using variable substitution:

6.391. $\int\limits_1^5\frac{dx}{x+\sqrt{2x-1}}.$

Solution.

$$\int\limits_1^{5}\frac{dx}{x+\sqrt{2x-1}}=\left[\begin{array}{lcl}t=\sqrt{2x-1}\Rightarrow x=\frac{1}{2}(t^2+1)\\dx=tdt\\x=1\Rightarrow t=1\\x=5\Rightarrow t=3\end{array}\right]=\int\limits_1^3\frac{tdt}{\frac{1}{2}(t^2+1)+t}=$$ $$=2\int\limits_1^3\frac{tdt}{t^2+2t+1}=2\int\limits_1^3\frac{tdt}{(t+1)^2}=2\int\limits_1^3\frac{t+1-1}{(t+1)^2dt}=2\int\limits_1^3\frac{dt}{1+t}-2\int\limits_1^3\frac{dt}{(t+1)^2}=$$ $$=\left.\left(2\ln|1+t|+2\frac{1}{t+1}\right)\right|_1^3=2\ln 4+\frac{1}{2}-2\ln 2-1=2\ln 2-\frac{1}{2}.$$

Answer: $2\ln 2-\frac{1}{2}.$

6.393.$\int\limits_{-1}^1\frac{xdx}{\sqrt{5-4x}}.$

Solution.

$$\int\limits_{-1}^{1}\frac{xdx}{\sqrt{5-4x}}=\left[\begin{array}{lcl}t=\sqrt{5-4x}\Rightarrow x=\frac{1}{4}(5-t^2)\\dx=-\frac{t}{2}dt\\x=-1\Rightarrow t=3\\x=1\Rightarrow t=1\end{array}\right]=-\int\limits_3^1\frac{(5-t^2)tdt}{8t}=$$ $$=\frac{1}{8}\int\limits_1^3{(5-t^2)dt}=\frac{1}{8}\left.\left(5t-\frac{t^3}{3}\right)\right|_1^3=\frac{1}{8}(15-9-5+\frac{1}{3})=\frac{1}{6}.$$

Answer: $\frac{1}{6}.$

6.394. $\int\limits_{\ln 2}^{\ln 6}\frac{e^x\sqrt{e^x-2}}{e^x+2}\,dx.$

Solution.

$$\int\limits_{\ln 2}^{\ln 6}\frac{e^x\sqrt{e^x-2}}{e^x+2}dx=\left[\begin{array}{lcl}t=\sqrt{e^x-2}\Rightarrow x=\ln(t^2+2)\\dx=\frac{2t}{t^2+2}dt\\x=\ln 2\Rightarrow t=0\\x=\ln 6\Rightarrow t=2\end{array}\right]=\int\limits_0^2\frac{(t^2+2)2t^2dt}{(t^2+2)(t^2+4)}=$$ $$=2\int\limits_0^2\frac{t^2}{t^2+4}dt=2\int\limits_0^2\frac{t^2+4-4}{t^2+4}dt=2\left.\left(t-\frac{4}{2}arctg\frac{t}{2}\right)\right|_0^2=2(2-2\frac{\pi}{4})=4-\pi.$$

Answer: $4-\pi.$

6.396. Show that $\int\limits_e^{e^2}\frac{dx}{\ln x}=\int\limits_1^2\frac{e^x}{x},dx.$

Solution.

$$\int\limits_{e}^{e^2}\frac{dx}{\ln x}dx=\left[\begin{array}{lcl}t=\ln x\Rightarrow x=e^t\\dx=e^t\,dt\\x=e\Rightarrow t=1\\x=e^2\Rightarrow t=2\end{array}\right]=\int\limits_1^2\frac{(e^tdt}{t}.$$ Let's change variables to $t=x.$ We get $$\int\limits_1^2\frac{(e^tdt}{t}\int\limits_1^2\frac{e^x\,dx}{x}.$$

This completes the proof.

Integration by parts.

If the functions $u(x),, v(x)$ and their derivatives $u'(x)$ and $v'(x)$ are continuous on the interval $[a, b]$, then$$\int\limits_a^budv=\left.uv\right|_a^b-\int\limits_a^bvdu.$$ (formula for integration by parts)

Examples.

Compute the integrals using the method of integration by parts:

6.399. $\int\limits_0^1xe^xdx.$

Solution.

$$\int\limits_0^1xe^xdx=\left[\begin{array}{lcl}u=x\Rightarrow du=dx\\dv=e^xdx\Rightarrow v=e^x\end{array}\right]=xe^x|_0^1-\int\limits_0^1 e^xdx=$$ $$=e-e^x|_0^1=e-e+1=1.$$

Answer: $1.$

6.402. $\int\limits_1^e\ln^2 x\,dx.$

Solution.

$$\int\limits_1^e\ln^2 x\,dx=\left[\begin{array}{lcl}u=\ln^2x\Rightarrow du=\frac{2\ln x}{x}dx\\dv=dx\Rightarrow v=x\end{array}\right]=x\ln^2x|_1^e-\int\limits_1^e 2\frac{x\ln x}{x}dx=$$ $$=\left[\begin{array}{lcl}u=\ln x\Rightarrow du=\frac{dx}{x}\\dv=dx\Rightarrow v=x\end{array}\right]=x\ln^2x|_1^e-2x\ln x|_1^e+2\int\limits_1^e \frac{x}{x}dx=(x\ln^2 x-2x\ln x+2x)|_1^e=$$ $$=e-2e+2e-(0-0+2)=e-2.$$

Answer: $e-2.$

6.408.$\int\limits_0^{\pi/2}e^x\cos x\,dx.$

Solution.

$$\int\limits_0^{\pi/2}e^x\cos x\,dx=\left[\begin{array}{lcl}u=\cos x\Rightarrow du=-\sin x dx\\dv=e^xdx\Rightarrow v=e^x\end{array}\right]=e^x\cos x|_0^{\pi/2}+\int\limits_0^{\pi/2} e^x\sin x\,dx=$$ $$=\left[\begin{array}{lcl}u=\sin x\Rightarrow du=\cos x\,dx\\dv=e^xdx\Rightarrow v=e^x\end{array}\right]=e^x\cos x|_0^{\pi/2}+e^x\sin x|_0^{pi/2}-\int\limits_0^{\pi/2}e^x\cos x dx.$$ Обозначим $I=\int\limits_0^{\pi/2}e^x\cos x dx.$ Получили уравнение $$I=e^x\cos x|_0^{\pi/2}+e^x\sin x|_0^{pi/2}-I\Rightarrow I=\frac{1}{2}(e^x\cos x+e^x\sin x)|_0^{\pi/2}=\frac{1}{2}(e^{\pi/2}-1).$$

Answer: $\frac{1}{2}(e^{\pi/2}-1).$

Homework.

Using the fundamental theorem of calculus, compute the following integrals:

6.327. $\int\limits_0^1(\sqrt x+\sqrt[3]{x^2}),dx.$

6.332. $\int\limits_1^2e^x,dx.$

6.337. $\int\limits_0^{\pi/4}\sin^2\varphi,d\varphi.$

6.351. $\int\limits_0^2\frac{2x-1}{2x+1},dx.$

6.364. (b) Determine the sign of the integral without calculating it: $\int\limits_{-1}^1x^3e^x,dx.$

6.365. (a) Without calculating the integrals, determine which of the integrals is greater: $\int\limits_1^2\frac{dx}{\sqrt{1+x^2}}$ or $\int\limits_1^2\frac{dx}{x}.$

6.366. (a) Find the average value of the function on the given interval: $x^3,,, 0\leq x \leq 1.$

6.368. Estimate the integral $\int\limits_{-1}^1\sqrt{8+x^3}dx.$

6.370. (a) Estimate the integral $\int\limits_0^1\sqrt{(1+x)(1+x^3)}dx,$ using the generalized theorem for estimating integrals.

Find the derivatives of the following functions:

6.375. $\Phi(x)=\int\limits_{1/x}^{\sqrt x}\sin t^2,dt.$

6.377. $\Phi(x)=\int\limits_{x^2}^{x^3}\frac{dt}{\ln t},\quad (x>0).$

Compute the integrals using the specified substitutions:

6.381. $\int\limits_{ln 3}^{\ln 8}\frac{dx}{\sqrt{e^x+1}},\quad e^x+1=t^2.$

6.383. $\int\limits_0^{\pi/2}\frac{dx}{3+2\cos x},\quad \tan\frac{x}{2}=t.$

Compute the integrals using variable substitution:

6.387. $\int\limits_{-2}^{0}\frac{dx}{\sqrt{x+3}+\sqrt{(x+3)^3}}.$

6.395. $\int\limits_0^3x^2\sqrt{9-x^2},dx.$