Evaluation of definite integrals.

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: $$\underset{a}{\overset{b}{\int }}f(x)dx=F(x){|}_a^b=F(b)-F(a).$$

Examples:

Using the Fundamental Theorem of Calculus, compute the integrals:

1. $\underset{-1}{\overset{2}{\int }}{x}^3,dx.$

Solution:

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

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

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

Solution:

$$\underset{-\pi /4}{\overset{0}{\int }}\frac{dx}{{\cos }^{2}x}={tgx|}_{-\pi /4}^0=tg0-tg(-\pi /4)=0-(-1)=1.$$

Answer: $1.$

3. $\underset{1}{\overset{2}{\int }}\frac{dx}{2x-1}.$

Solution.

$$\underset{1}{\overset{2}{\int }}\frac{dx}{2x-1}=\underset{1}{\overset{2}{\int }}\frac{1}{2}d(\ln (2x-1))=\frac{1}{2}{(\ln (2x-1))|}_1^2=\ln 3-\ln 1=\ln 3.$$

Answer: $\ln 3.$

4 $\underset{0}{\overset{\pi /2}{\int }}{\cos }^3\alpha d\alpha .$

Solution.

$$\underset{0}{\overset{\pi /2}{\int }}{\cos }^3\alpha d\alpha =\underset{0}{\overset{\pi /2}{\int }}{\cos }^{2}xd\sin x=\underset{0}{\overset{\pi /2}{\int }}(1-{\sin }^{2}x)d\sin x={(\sin x-\frac{{\sin }^{3}x}{3})|}_0^{\pi /2}=$$ $$=\sin \frac{\pi }{2}-\frac{{\sin }^3\pi /2}{3}-(\sin 0-\frac{{\sin }^{3}0}{3})=1-\frac{1}{3}=\frac{2}{3}.$$

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

Properties of definite integrals:

1) If $f(x)\ge 0$ on the interval $[a,b]$, then $\underset{a}{\overset{b}{\int }}f(x),dx\ge 0$.

2) If $f(x)\le g(x)$ on $[a,b]$, then $\underset{a}{\overset{b}{\int }}f(x),dx\le \underset{a}{\overset{b}{\int }}g(x),dx$.

3) $|\underset{a}{\overset{b}{\int }}f(x),dx|\le \underset{a}{\overset{b}{\int }}|f(x)|,dx$.

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 $$m(b-a)\le \underset{a}{\overset{b}{\int }}f(x)dx\le M(b-a)$$ (theorem on the estimation of definite integrals).

5) If $f(x)$ is continuous, and $g(x)$ is integrable on $[a,b]$ such that $g(x)\ge 0$, and $m$ and $M$ are the smallest and largest values of $f(x)$ on $[a,b]$, then $$m\underset{a}{\overset{b}{\int }}g(x)dx\le \underset{a}{\overset{b}{\int }}f(x)g(x)dx\le M\underset{a}{\overset{b}{\int }}g(x)dx.$$ (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:$$\underset{a}{\overset{b}{\int }}f(x)dx=f(c)(b-a).$$ (mean value theorem)

The number $f(c)=\frac{1}{b-a}\underset{a}{\overset{b}{\int }}f(x),dx$ 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 $g(x)\ge 0$, then there exists a point $c\in (a,b)$ such that the equality holds: $$\underset{a}{\overset{b}{\int }}f(x)g(x)dx=f(c)\underset{a}{\overset{b}{\int }}g(x)dx$$ (generalized mean value theorem).

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

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

10) If the function $f(x)$ is continuous on the interval $[a,b]$, then the integral with a variable upper limit$$\mathrm{\Phi }(x)=\underset{a}{\overset{x}{\int }}f(t)dt$$ является первообразной для функции $f(x),$ т.е. $${\mathrm{\Phi }}^{\prime }(x)=(\underset{a}{\overset{x}{\int }}f(t)dt{)}^{\prime }=f(x),x\in [a,b].$$

11) If the functions $\varphi (x)$ and $\psi (x)$ are differentiable at the point $x\in (a,b)$ and $f(t)$ is continuous for $\varphi (a)\le t\le \psi (b)$, then $${(\underset{\varphi (x)}{\overset{\psi (x)}{\int }}f(t)dt)}_x^{\prime }=f(\psi (x)){\psi }^{\prime }(x)-f(\varphi (x)){\varphi }^{\prime }(x).$$

Examples.

1. Determine the sign of the integral without evaluating it: $\underset{-2}{\overset{1}{\int }}\sqrt[3]{x},dx.$

Solution.

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

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

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

2. Without computing the integrals, determine which of the integrals is greater, $\underset{1}{\overset{2}{\int }}\frac{dx}{x^2}$ or $\underset{1}{\overset{2}{\int }}\frac{dx}{x^3}$.

Solution:

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

Answer: $\underset{1}{\overset{2}{\int }}\frac{dx}{x^2}>\underset{1}{\overset{2}{\int }}\frac{dx}{x^3}.$

3. To find the average value of the function on the given interval, $\cos x,0\le x\le \frac{\pi }{2}$.

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 $f(c)=\frac{1}{b-a}\underset{a}{\overset{b}{\int }}f(x),dx.$

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

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

4. Estimate the integral $\underset{0}{\overset{2\pi }{\int }}\frac{dx}{\sqrt{5+2\sin x}}.$

Solution:

Let's estimate the integrand:

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

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

$$\underset{0}{\overset{2\pi }{\int }}\frac{1}{\sqrt{7}}dx\le \underset{0}{\overset{2\pi }{\int }}\frac{1}{\sqrt{5+2\sin x}}dx\le \underset{0}{\overset{2\pi }{\int }}\frac{1}{\sqrt{3}}dx.$$

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

Thus, $$\frac{2\pi }{\sqrt{7}}\le \underset{0}{\overset{2\pi }{\int }}\frac{1}{\sqrt{5+2\sin x}}dx\le \frac{2\pi }{\sqrt{3}}.$$

Answer: $\frac{2\pi }{\sqrt{7}}\le \underset{0}{\overset{2\pi }{\int }}\frac{1}{\sqrt{5+2\sin x}}dx\le \frac{2\pi }{\sqrt{3}}.$

5. Estimate the integral $\underset{0}{\overset{1}{\int }}\sqrt{(1+x)(1+x^3)},dx$ using the Cauchy-Schwarz inequality.

Solution:

The Cauchy-Schwarz inequality gives us $$|\underset{0}{\overset{1}{\int }}\sqrt{(1+x)(1+x^3)}dx|\le \sqrt{\underset{0}{\overset{1}{\int }}(1+x)dx\underset{0}{\overset{1}{\int }}(1+x^3)dx}.$$

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

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

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

6. Find the derivative of the following function: $\mathrm{\Phi }(x)=\underset{0}{\overset{x}{\int }}\frac{\sin t}{t},dt.$

Solution:

We use property 10:

$${\mathrm{\Phi }}^{\prime }(x)=f(x)=\frac{\sin x}{x}.$$

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

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

Solution:

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

We use property 10:

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

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 $x=\phi (t)$ is continuously differentiable on the interval $[t_1,t_2]$, where $a=\phi (t_1)$ and $b=\phi (t_2)$, then$$\underset{a}{\overset{b}{\int }}f(x)dx=\underset{t_1}{\overset{t_2}{\int }}f(\phi (t)){\phi }^{\prime }(t)dt.$$

Examples.

Compute the integrals using the indicated substitutions:

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

Solution.

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

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

2. $\underset{0}{\overset{sh1}{\int }}\sqrt{x^2+1},x=sht.$

Solution.

$$\underset{0}{\overset{sh1}{\int }}\sqrt{x^2+1}dx=\left[\begin{array}{l}x=sht\Rightarrow dx=cht\\ x=0\Rightarrow t=0\\ x=sh1\Rightarrow t=1\end{array}\right]=\underset{0}{\overset{1}{\int }}\sqrt{s{h}^{2}t+1}chtdt=\underset{0}{\overset{1}{\int }}c{h}^{2}tdt=$$ $$=\underset{0}{\overset{1}{\int }}\frac{1}{2}(ch2t+1)dt=\frac{1}{2}{(\frac{1}{2}sh2t+t)|}_0^1=\frac{1}{2}(\frac{1}{2}sh2+1-\frac{1}{2}sh0-0)=$$ $$=\frac{1}{4}(sh2+2).$$

Answer: $\frac{1}{4}(sh2+2).$

Compute the integrals using variable substitution:

3. $\underset{1}{\overset{5}{\int }}\frac{dx}{x+\sqrt{2x-1}}.$

Solution.

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

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

4.$\underset{-1}{\overset{1}{\int }}\frac{xdx}{\sqrt{5-4x}}.$

Solution.

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

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

5. $\underset{\ln 2}{\overset{\ln 6}{\int }}\frac{e^x\sqrt{e^x-2}}{e^x+2}dx.$

Solution.

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

Answer: $4-\pi .$

6. Show that $\underset{e}{\overset{e^2}{\int }}\frac{dx}{\ln x}=\underset{1}{\overset{2}{\int }}\frac{e^x}{x},dx.$

Solution.

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

This completes the proof.

Integration by parts.

If the functions $u(x),,v(x)$ and their derivatives $u^{\prime }(x)$ and $v^{\prime }(x)$ are continuous on the interval $[a,b]$, then$$\underset{a}{\overset{b}{\int }}udv={uv|}_a^b-\underset{a}{\overset{b}{\int }}vdu.$$ (formula for integration by parts)

Examples.

Compute the integrals using the method of integration by parts:

1. $\underset{0}{\overset{1}{\int }}x{e}^{x}dx.$

Solution.

$$\underset{0}{\overset{1}{\int }}x{e}^{x}dx=\left[\begin{array}{l}u=x\Rightarrow du=dx\\ dv=e^{x}dx\Rightarrow v=e^x\end{array}\right]=x{e}^x{|}_0^1-\underset{0}{\overset{1}{\int }}{e}^{x}dx=$$ $$=e-e^x{|}_0^1=e-e+1=1.$$

Answer: $1.$

2. $\underset{1}{\overset{e}{\int }}{\ln }^{2}xdx.$

Solution.

$$\underset{1}{\overset{e}{\int }}{\ln }^{2}xdx=\left[\begin{array}{l}u={\ln }^{2}x\Rightarrow du=\frac{2\ln x}{x}dx\\ dv=dx\Rightarrow v=x\end{array}\right]=x{\ln }^{2}x{|}_1^e-\underset{1}{\overset{e}{\int }}2\frac{x\ln x}{x}dx=$$ $$=\left[\begin{array}{l}u=\ln x\Rightarrow du=\frac{dx}{x}\\ dv=dx\Rightarrow v=x\end{array}\right]=x{\ln }^{2}x{|}_1^e-2x\ln x{|}_1^e+2\underset{1}{\overset{e}{\int }}\frac{x}{x}dx=(x{\ln }^{2}x-2x\ln x+2x){|}_1^e=$$ $$=e-2e+2e-(0-0+2)=e-2.$$

Answer: $e-2.$

3.$\underset{0}{\overset{\pi /2}{\int }}{e}^x\cos xdx.$

Solution.

$$\underset{0}{\overset{\pi /2}{\int }}{e}^x\cos xdx=\left[\begin{array}{l}u=\cos x\Rightarrow du=-\sin xdx\\ dv=e^{x}dx\Rightarrow v=e^x\end{array}\right]=e^x\cos x{|}_0^{\pi /2}+\underset{0}{\overset{\pi /2}{\int }}{e}^x\sin xdx=$$ $$=\left[\begin{array}{l}u=\sin x\Rightarrow du=\cos xdx\\ dv=e^{x}dx\Rightarrow v=e^x\end{array}\right]=e^x\cos x{|}_0^{\pi /2}+e^x\sin x{|}_0^{pi/2}-\underset{0}{\overset{\pi /2}{\int }}{e}^x\cos xdx.$$ Обозначим $I=\underset{0}{\overset{\pi /2}{\int }}{e}^x\cos xdx.$ 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:

1. $\underset{0}{\overset{1}{\int }}(\sqrt{x}+\sqrt[3]{x^2}),dx.$

2. $\underset{1}{\overset{2}{\int }}{e}^x,dx.$

3 $\underset{0}{\overset{\pi /4}{\int }}{\sin }^2\phi ,d\phi .$

4. $\underset{0}{\overset{2}{\int }}\frac{2x-1}{2x+1},dx.$

5.) Determine the sign of the integral without calculating it: $\underset{-1}{\overset{1}{\int }}{x}^3{e}^x,dx.$

6. Without calculating the integrals, determine which of the integrals is greater: $\underset{1}{\overset{2}{\int }}\frac{dx}{\sqrt{1+x^2}}$ or $\underset{1}{\overset{2}{\int }}\frac{dx}{x}.$

7. Find the average value of the function on the given interval: $x^3,,,0\le x\le 1.$

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

9. Estimate the integral $\underset{0}{\overset{1}{\int }}\sqrt{(1+x)(1+x^3)}dx,$ using the generalized theorem for estimating integrals.

Find the derivatives of the following functions:

10. $\mathrm{\Phi }(x)=\underset{1/x}{\overset{\sqrt{x}}{\int }}\sin t^2,dt.$

11. $\mathrm{\Phi }(x)=\underset{x^2}{\overset{x^3}{\int }}\frac{dt}{\ln t},(x>0).$

Compute the integrals using the specified substitutions:

12. $\underset{ln3}{\overset{\ln 8}{\int }}\frac{dx}{\sqrt{e^x+1}},e^x+1=t^2.$

13. $\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{3+2\cos x},\tan \frac{x}{2}=t.$

Compute the integrals using variable substitution:

14. $\underset{-2}{\overset{0}{\int }}\frac{dx}{\sqrt{x+3}+\sqrt{(x+3{)}^3}}.$

15. $\underset{0}{\overset{3}{\int }}{x}^2\sqrt{9-x^2},dx.$

16. Show that $\underset{1/\sqrt{2}}{\overset{1}{\int }}\frac{dx}{\arcsin x}=\underset{\pi /4}{\overset{\pi /2}{\int }}\frac{\cos x}{x},dx.$

Compute the integrals using integration by parts:

17. $\underset{0}{\overset{1}{\int }}\frac{\arcsin x}{\sqrt{1+x}},dx.$

18. $\underset{0}{\overset{\pi /4}{\int }}{e}^{3x}\sin 4x,dx.$

19. $\underset{1}{\overset{e}{\int }}x\ln x,dx.$

Tags: Change of variables, Integration by parts, Properties of definite integrals, The Fundamental Theorem of Calculus, calculus, integral, mathematical analysis