NCERT Solutions for Exercise 7.9 Class 12 Maths Chapter 7 - Integrals

Question 1: Evaluate the definite integral

${\int }_{-1}^1(x+1)dx$

Answer:

Given integral: $I={\int }_{-1}^1(x+1)dx$

Consider the integral $\int (x+1)dx$

$\int (x+1)dx=\frac{x^2}{2}+x$

So, we have the function of $x$ , $f(x)=\frac{x^2}{2}+x$

Now, by 4the Second fundamental theorem of calculus, we have

$I=f(1)-f(-1)$

$=(\frac{1}{2}+1)-(\frac{1}{2}-1)$

$=\frac{1}{2}+1-\frac{1}{2}+1$

$=2$

Question 2: Evaluate the definite integral

${\int }_2^3\frac{1}{x}dx$

Answer:

Given integral: $I={\int }_2^3\frac{1}{x}dx$

Consider the integral ${\int }_2^3\frac{1}{x}dx$

$\int \frac{1}{x}dx=\log |x|$

So, we have the function of $x$ , $f(x)=\log |x|$

Now, by Second fundamental theorem of calculus, we have

$I=f(3)-f(2)$

$=\log |3|-\log |2|=\log \frac{3}{2}$

Question 3: Evaluate the definite integral

${\int }_1^2(4x^3-5x^2+6x+9)dx$

Answer:

Given integral: $I={\int }_1^2(4x^3-5x^2+6x+9)dx$

Consider the integral $I=\int (4x^3-5x^2+6x+9)dx$

$\int (4x^3-5x^2+6x+9)dx=4\frac{x^4}{4}-5\frac{x^3}{3}+6\frac{x^2}{2}+9x$

$=x^4-\frac{5x^3}{3}+3x^2+9x$

So, we have the function of $x$ , $f(x)=x^4-\frac{5x^3}{3}+3x^2+9x$

Now, by Second fundamental theorem of calculus, we have

$I=f(2)-f(1)$

$=\{2^4-\frac{5(2{)}^3}{3}+3(2{)}^2+9(2)\}-\{1^4-\frac{5(1{)}^3}{3}+3(1{)}^2+9(1)\}$

$=\{16-\frac{40}{3}+12+18\}-\{1-\frac{5}{3}+3+9\}$

$=\{46-\frac{40}{3}\}-\{13-\frac{5}{3}\}$

$=\{33-\frac{35}{3}\}=\left\{\frac{99-35}{3}\right\}$

$=\frac{64}{3}$

Question 4: Evaluate the definite integral

${\int }_0^{\frac{\pi }{4}}\sin 2xdx$

Answer:

Given integral: ${\int }_0^{\frac{\pi }{4}}\sin 2xdx$

Consider the integral $\int \sin 2xdx$

$\int \sin 2xdx=\frac{-\cos 2x}{2}$

So, we have the function of $x$ , $f(x)=\frac{-\cos 2x}{2}$

Now, by Second fundamental theorem of calculus, we have

$I=f(\frac{\pi }{4})-f(0)$

$=\frac{-\cos 2(\frac{\pi }{4})}{2}+\frac{\cos 0}{2}$

$=\frac{1}{2}-0$

$=\frac{1}{2}$

Question 5: Evaluate the definite integral

${\int }_0^{\frac{\pi }{2}}\cos 2xdx$

Answer:

Given integral: ${\int }_0^{\frac{\pi }{2}}\cos 2xdx$

Consider the integral $\int \cos 2xdx$

$\int \cos 2xdx=\frac{\sin 2x}{2}$

So, we have the function of $x$ , $f(x)=\frac{\sin 2x}{2}$

Now, by Second fundamental theorem of calculus, we have

$I=f(\frac{\pi }{2})-f(0)$

$=\frac{1}{2}\{\sin 2(\frac{\pi }{2})-\sin 0\}$

$=\frac{1}{2}\{0-0\}=0$

Question 6: Evaluate the definite integral

${\int }_4^5{e}^{x}dx$

Answer:

Given integral: ${\int }_4^5{e}^{x}dx$

Consider the integral $\int e^{x}dx$

$\int e^{x}dx=e^x$

So, we have the function of $x$ , $f(x)=e^x$

Now, by Second fundamental theorem of calculus, we have

$I=f(5)-f(4)$

$=e^5-e^4$

$=e^4(e-1)$

Question 7: Evaluate the definite integral

${\int }_0^{\frac{\pi }{4}}\tan xdx$

Answer:

Given integral: ${\int }_0^{\frac{\pi }{4}}\tan xdx$

Consider the integral $\int \tan xdx$

$\int \tan xdx=-\log |\cos x|$

So, we have the function of $x$ , $f(x)=-\log |\cos x|$

Now, by Second fundamental theorem of calculus, we have

$I=f(\frac{\pi }{4})-f(0)$

$=-\log |\cos \frac{\pi }{4}|+\log |\cos 0|$

$=-\log |\cos \frac{1}{\sqrt{2}}|+\log |1|$

$=-\log \left|\frac{1}{\sqrt{2}}\right|+0=-\log (2{)}^{-\frac{1}{2}}$

$=\frac{1}{2}\log (2)$

Question 8: Evaluate the definite integral

${\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}\text{cosec}xdx$

Answer:

Given integral: ${\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}\text{cosec}xdx$

Consider the integral $\int \text{cosec}xdx$

$\int \text{cosec}xdx=\log |\text{cosec}x-\text{cot}x|$

So, we have the function of $x$ , $f(x)=\log |\text{cosec}x-\text{cot}x|$

Now, by Second fundamental theorem of calculus, we have

$I=f(\frac{\pi }{4})-f(\frac{\pi }{6})$

$=\log |\text{cosec}\frac{\pi }{4}-\text{cot}\frac{\pi }{4}|-\log |\text{cosec}\frac{\pi }{6}-\text{cot}\frac{\pi }{6}|$

$=\log |\sqrt{2}-1|-\log |2-\sqrt{3}|$

$=\log \left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)$

Question 9: Evaluate the definite integral

${\int }_0^1\frac{dx}{\sqrt{1-x^2}}$

Answer:

Given integral: ${\int }_0^1\frac{dx}{\sqrt{1-x^2}}$

Consider the integral $\int \frac{dx}{\sqrt{1-x^2}}$

$\int \frac{dx}{\sqrt{1-x^2}}={\sin }^{-1}x$

So, we have the function of $x$ , $f(x)={\sin }^{-1}x$

Now, by Second fundamental theorem of calculus, we have

$I=f(1)-f(0)$

$={\sin }^{-1}(1)-{\sin }^{-1}(0)$

$=\frac{\pi }{2}-0$

$=\frac{\pi }{2}$

Question 10: Evaluate the definite integral

${\int }_0^1\frac{dx}{1+x^2}$

Answer:

Given integral: ${\int }_0^1\frac{dx}{1+x^2}$

Consider the integral $\int \frac{dx}{1+x^2}$

$\int \frac{dx}{1+x^2}={\tan }^{-1}x$

So, we have the function of $x$ , $f(x)={\tan }^{-1}x$

Now, by Second fundamental theorem of calculus, we have

$I=f(1)-f(0)$

$={\tan }^{-1}(1)-{\tan }^{-1}(0)$

$=\frac{\pi }{4}-0$

$=\frac{\pi }{4}$

Question 11: Evaluate the definite integral

${\int }_2^3\frac{dx}{x^2-1}$

Answer:

Given integral: ${\int }_2^3\frac{dx}{x^2-1}$

Consider the integral $\int \frac{dx}{x^2-1}$

$\int \frac{dx}{x^2-1}=\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|$

So, we have the function of $x$ , $f(x)=\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|$

Now, by Second fundamental theorem of calculus, we have

$I=f(3)-f(2)$

$=\frac{1}{2}\{\log \left|\frac{3-1}{3+1}\right|-\log \left|\frac{2-1}{2+1}\right|\}$

$=\frac{1}{2}\{\log \left|\frac{2}{4}\right|-\log \left|\frac{1}{3}\right|\}$

$=\frac{1}{2}\{\log \frac{1}{2}-\log \frac{1}{3}\}$

$=\frac{1}{2}\log \frac{3}{2}$

Question 12: Evaluate the definite integral

${\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx$

Answer:

Given integral: ${\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx$

Consider the integral $\int {\cos }^{2}xdx$

$\int {\cos }^{2}xdx=\int \frac{1+\cos 2x}{2}dx=\frac{x}{2}+\frac{\sin 2x}{4}$

$=\frac{1}{2}(x+\frac{\sin 2x}{2})$

So, we have the function of $x$ , $f(x)=\frac{1}{2}(x+\frac{\sin 2x}{2})$

Now, by Second fundamental theorem of calculus, we have

$I=f(\frac{\pi }{2})-f(0)$

$=\frac{1}{2}\{(\frac{\pi }{2}-\frac{\sin \pi }{2})-(0+\frac{\sin 0}{2})\}$

$=\frac{1}{2}\{\frac{\pi }{2}+0-0-0\}$

$=\frac{\pi }{4}$

Question 13: Evaluate the definite integral

${\int }_2^3\frac{xdx}{x^2+1}$

Answer:

Given integral: ${\int }_2^3\frac{xdx}{x^2+1}$

Consider the integral $\int \frac{xdx}{x^2+1}$

$\int \frac{xdx}{x^2+1}$

$=\frac{1}{2}\int \frac{2x}{x^2+1}dx$

$=\frac{1}{2}\log (1+x^2)$

So, we have the function of $x$ , $f(x)=\frac{1}{2}\log (1+x^2)$

Now, by Second fundamental theorem of calculus, we have

$I=f(3)-f(2)$

$=\frac{1}{2}\{\log (1+(3{)}^2)-\log (1+(2{)}^2)\}$

$=\frac{1}{2}\{\log (10)-\log (5)\}$

$=\frac{1}{2}\log \left(\frac{10}{5}\right)$

$=\frac{1}{2}\log 2$

Question 14: Evaluate the definite integral

${\int }_0^1\frac{2x+3}{5x^2+1}dx$

Answer:

Given integral: ${\int }_0^1\frac{2x+3}{5x^2+1}dx$

Consider the integral $\int \frac{2x+3}{5x^2+1}dx$

Multiplying by 5 both in numerator and denominator:

$\int \frac{2x+3}{5x^2+1}dx=\frac{1}{5}\int \frac{5(2x+3)}{5x^2+1}dx$

$=\frac{1}{5}\int \frac{10x+15}{5x^2+1}dx$

$=\frac{1}{5}\int \frac{10x}{5x^2+1}dx+3\int \frac{1}{5x^2+1}dx$

$=\frac{1}{5}\int \frac{10x}{5x^2+1}+3\int \frac{1}{5(x^2+\frac{1}{5})}dx$

$=\frac{1}{5}\log (5x^2+1)+\frac{3}{5}\times \frac{1}{\frac{1}{\sqrt{5}}}{\tan }^{-1}\frac{x}{\frac{1}{\sqrt{5}}}$

$=\frac{1}{5}\log (5x^2+1)+\frac{3}{\sqrt{5}}{\tan }^{-1}(\sqrt{5}x)$

So, we have the function of $x$ , $f(x)=\frac{1}{5}\log (5x^2+1)+\frac{3}{\sqrt{5}}{\tan }^{-1}(\sqrt{5}x)$

Now, by Second fundamental theorem of calculus, we have

$I=f(1)-f(0)$

$=\{\frac{1}{5}\log (1+5)+\frac{3}{\sqrt{5}}{\tan }^{-1}(\sqrt{5})\}-\{\frac{1}{5}\log (1)+\frac{3}{\sqrt{5}}{\tan }^{-1}(0)\}$

$=\frac{1}{5}\log 6+\frac{3}{\sqrt{5}}{\tan }^{-1}\sqrt{5}$

Question 15: Evaluate the definite integral

${\int }_0^{1}x{e}^{x^2}dx$

Answer:

Given integral: ${\int }_0^{1}x{e}^{x^2}dx$

Consider the integral $\int x{e}^{x^2}dx$

Putting $x^2=t$ which gives, $2xdx=dt$

As, $x\to 0,t\to 0$ and as $x\to 1,t\to 1$ .

So, we have now:

$\therefore I=\frac{1}{2}{\int }_0^1{e}^{t}dt$

$=\frac{1}{2}\int e^{t}dt=\frac{1}{2}{e}^t$

So, we have the function of $x$ , $f(x)=\frac{1}{2}{e}^t$

Now, by Second fundamental theorem of calculus, we have

$I=f(1)-f(0)$

$=\frac{1}{2}{e}^1-\frac{1}{2}{e}^0$

$=\frac{1}{2}(e-1)$

Question 16: Evaluate the definite integral

${\int }_1^2\frac{5x^2}{x^2+4x+3}$

Answer:

Given integral: $I={\int }_1^2\frac{5x^2}{x^2+4x+3}$

So, we can rewrite the integral as;

$I={\int }_1^2\frac{5x^2}{x^2+4x+3}={\int }_1^2(5-\frac{20x+15}{x^2+4x+3})dx$

$={\int }_1^{2}5dx-{\int }_1^2\frac{20x+15}{x^2+4x+3}dx$

$=[5x{]}_1^2-{\int }_1^2\frac{20x+15}{x^2+4x+3}dx$

$I=5-I_1$ where $I={\int }_1^2\frac{20x+15}{x^2+4x+3}dx$ . ................(1)

Now, consider $I={\int }_1^2\frac{20x+15}{x^2+4x+3}dx$

Take numerator $20x+15=A\frac{d}{dx}(x^2+4x+3)+B$

$=2Ax+(4A+B)$

We now equate the coefficients of x and constant term, we get

A=10 and B=-25

$\Rightarrow I_1=10{\int }_1^2\frac{2x+4}{x^2+4x+3}dx-25{\int }_1^2\frac{dx}{x^2+4x+3}$

Now take denominator $x^2+4x+3=t$

Then we have $(2x+4)dx=dt$

$\Rightarrow I_1=10\int \frac{dt}{t}-25\int \frac{dx}{(x+2{)}^2-1^2}$

$=10\log t-25[\frac{1}{2}\log \left(\frac{x+2-1}{x+2+1}\right)]$

$=[10\log (x^2+4x+3){]}_1^2-25{[\frac{1}{2}\log \left(\frac{x+1}{x+3}\right)]}_1^2$

$=[10\log 15-10\log 8]-25[\frac{1}{2}\log \frac{3}{5}-\frac{1}{2}\log \frac{2}{4}]$

$=[10\log 5+10\log 3-10\log 4-10\log 2]-\frac{25}{2}[\log 3-\log 5-\log 2+\log 4]$ $=(10+\frac{25}{2})\log 5+(-10-\frac{25}{2})\log 4+(10-\frac{25}{2})\log 3+(-10+\frac{25}{2})\log 2$ $=\frac{45}{2}\log 5-\frac{45}{2}\log 4-\frac{5}{2}\log 3+\frac{5}{2}\log 2$

$=\frac{45}{2}\log \frac{5}{4}-\frac{5}{2}\log \frac{3}{2}$

Then substituting the value of $I_1$ in equation (1), we get

$I=5-(\frac{45}{2}\log \frac{5}{4}-\frac{5}{2}\log \frac{3}{2})$

$=5-\frac{5}{2}(9\log \frac{5}{4}-\log \frac{3}{2})$

Question 17: Evaluate the definite integral

${\int }_0^{\frac{\pi }{4}}(2{\sec }^{2}x+x^3+2)dx$

Answer:

Given integral: ${\int }_0^{\frac{\pi }{4}}(2{\sec }^{2}x+x^3+2)dx$

Consider the integral $\int (2{\sec }^{2}x+x^3+2)dx$

$\int (2{\sec }^{2}x+x^3+2)dx=2\tan x+\frac{x^4}{4}+2x$

So, we have the function of $x$ , $f(x)=2\tan x+\frac{x^4}{4}+2x$

Now, by Second fundamental theorem of calculus, we have

$I=f(\frac{\pi }{4})-f(0)$

$=\{(2\tan \frac{\pi }{4}+\frac{1}{4}{\left(\frac{\pi }{4}\right)}^4+2\frac{\pi }{4})-(2\tan 0+0+0)\}$

$=2\tan \frac{\pi }{4}+\frac{{\pi }^4}{4^5}+\frac{\pi }{2}$

$+2+\frac{\pi }{2}+\frac{{\pi }^4}{1024}$

Question 18: Evaluate the definite integral

${\int }_0^{\pi }({\sin }^2\frac{x}{2}-{\cos }^2\frac{x}{2})dx$

Answer:

Given integral: ${\int }_0^{\pi }({\sin }^2\frac{x}{2}-{\cos }^2\frac{x}{2})dx$

Consider the integral $\int ({\sin }^2\frac{x}{2}-{\cos }^2\frac{x}{2})dx$

can be rewritten as: $-\int ({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2})dx=-{\int }_0^{\pi }\cos xdx$

$=\sin x$

So, we have the function of $x$ , $f(x)=\sin x$

Now, by Second fundamental theorem of calculus, we have

$I=f(\pi )-f(0)$

$\Rightarrow \sin \pi -\sin 0$

$=0-0$

$=0$

Question 19: Evaluate the definite integral

${\int }_0^2\frac{6x+3}{x^2+4}$

Answer:

Given integral: ${\int }_0^2\frac{6x+3}{x^2+4}$

Consider the integral $\int \frac{6x+3}{x^2+4}$

can be rewritten as: $\int \frac{6x+3}{x^2+4}=3\int \frac{2x+1}{x^2+4}dx$

$=3\int \frac{2x}{x^2+4}dx+3\int \frac{1}{x^2+4}dx$

$=3\log (x^2+4)+\frac{3}{2}{\tan }^{-1}\frac{x}{2}$

So, we have the function of $x$ , $f(x)=3\log (x^2+4)+\frac{3}{2}{\tan }^{-1}\frac{x}{2}$

Now, by Second fundamental theorem of calculus, we have

$I=f(2)-f(0)$

$=\{3\log (2^2+4)+\frac{3}{2}{\tan }^{-1}\left(\frac{2}{2}\right)\}-\{3\log (0+4)+\frac{3}{2}{\tan }^{-1}\left(\frac{0}{2}\right)\}$ $=3\log 8+\frac{3}{2}{\tan }^{-1}1-3\log 4-\frac{3}{2}{\tan }^{-1}0$

$=3\log 8+\frac{3}{2}\times \frac{\pi }{4}-3\log 4-0$

$=3\log \frac{8}{4}+\frac{3\pi }{8}$

or we have $=3\log 2+\frac{3\pi }{8}$

Question 20: Evaluate the definite integral

${\int }_0^1(x{e}^x+\text{sin}\frac{\pi x}{4})dx$

Answer:

Given integral: ${\int }_0^1(x{e}^x+\text{sin}\frac{\pi x}{4})dx$

Consider the integral $\int (x{e}^x+\text{sin}\frac{\pi x}{4})dx$

can be rewritten as: $x\int e^{x}dx-\int \{\left(\frac{d}{dx}x\right)\int e^{x}dx\}dx+\left\{\frac{-\cos \frac{\pi x}{4}}{\frac{\pi }{4}}\right\}$

$=x{e}^x-\int e^{x}dx-\frac{4\pi }{\pi }\cos \frac{x}{4}$

$=x{e}^x-e^x-\frac{4\pi }{\pi }\cos \frac{x}{4}$

So, we have the function of $x$ , $f(x)=x{e}^x-e^x-\frac{4\pi }{\pi }\cos \frac{x}{4}$

Now, by Second fundamental theorem of calculus, we have

$I=f(1)-f(0)$

$=(1.e^t-e^t-\frac{4}{\pi }\cos \frac{\pi }{4})-(0.e^0-e^0-\frac{4}{\pi }\cos 0)$

$=e-e-\frac{4}{\pi }\left(\frac{1}{\sqrt{2}}\right)+1+\frac{4}{\pi }$

Question 21: Choose the correct answer

${\int }_1^{\sqrt{3}}\frac{dx}{1+x^2}$ equals

(A) $\frac{\pi }{3}$

(B) $\frac{2\pi }{3}$

(C) $\frac{\pi }{6}$

(D) $\frac{\pi }{12}$

Answer:

Given definite integral ${\int }_1^{\sqrt{3}}\frac{dx}{1+x^2}$

Consider $\int \frac{dx}{1+x^2}={\tan }^{-1}x$

we have then the function of x, as $f(x)={\tan }^{-1}x$

By applying the second fundamental theorem of calculus, we will get

${\int }_1^{\sqrt{3}}\frac{dx}{1+x^2}=f(\sqrt{3})-f(1)$

$={\tan }^{-1}\sqrt{3}-{\tan }^{-1}1$

$=\frac{\pi }{3}-\frac{\pi }{4}$

$=\frac{\pi }{12}$

Therefore the correct answer is $\frac{\pi }{12}$ .

Question 22: Choose the correct answer

${\int }_0^{\frac{2}{3}}\frac{dx}{4+9x^2}$ equals

(A) $\frac{\pi }{6}$

(B) $\frac{\pi }{12}$

(C) $\frac{\pi }{24}$

(D) $\frac{\pi }{4}$

Answer:

Given definite integral ${\int }_0^{\frac{2}{3}}\frac{dx}{4+9x^2}$

Consider $\int \frac{dx}{4+9x^2}=\int \frac{dx}{2^2+(3x{)}^2}$

Now, putting $3x=t$

we get, $3dx=dt$

Therefore we have, $\int \frac{dx}{2^2+(3x{)}^2}=\frac{1}{3}\int \frac{dt}{2^2+t^2}$

$=\frac{1}{3}(\frac{1}{2}{\tan }^{-1}\frac{t}{2})=\frac{1}{6}{\tan }^{-1}\left(\frac{3x}{2}\right)$

we have the function of x , as $f(x)=\frac{1}{6}{\tan }^{-1}\left(\frac{3x}{2}\right)$

So, by applying the second fundamental theorem of calculus, we get

${\int }_0^{\frac{2}{3}}\frac{dx}{4+9x^2}=f(\frac{2}{3})-f(0)$

$=\frac{1}{6}{\tan }^{-1}(\frac{3}{2}.\frac{2}{3})-\frac{1}{6}{\tan }^{-1}0$

$=\frac{1}{6}{\tan }^{-1}1-0$

$=\frac{1}{6}\times \frac{\pi }{4}=\frac{\pi }{24}$

Therefore the correct answer is $\frac{\pi }{24}$.