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

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

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

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

$= 2$

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

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

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

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

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

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

$= \frac{64}{3}$

Question 4: Evaluate the definite integral

$\int_0^\frac{\pi}{4}\sin 2x dx$

Answer:

Given integral: $\int_0^\frac{\pi}{4}\sin 2x dx$

Consider the integral $\int \sin 2x dx$

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

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

$\int_0^\frac{\pi}{2}\cos 2x dx$

Answer:

Given integral: $\int_0^\frac{\pi}{2}\cos 2x dx$

Consider the integral $\int \cos 2x dx$

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

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

Now, by Second fundamental theorem of calculus, we have

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

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

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

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

$\int^\frac{\pi}{4}_0 \tan x dx$

Answer:

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

Consider the integral $\int \tan x dx$

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

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

Now, by Second fundamental theorem of calculus, we have

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

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

$= -\log\left | \cos \frac{1}{\sqrt2} \right | +\log|1|$

$= -\log\left | \frac{1}{\sqrt2} \right | + 0 = -\log (2)^{-\frac{1}{2}}$

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

Question 8: Evaluate the definite integral

$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \text{cosec}\,x\,dx$

Answer:

Given integral: $\int_\frac{\pi}{6}^\frac{\pi}{4}\textup{cosec} x dx$

Consider the integral $\int\textup{cosec} x dx$

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

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

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

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

$= \log|\sqrt2 -1 | - \log|2 -\sqrt3 |$

$= \log \left ( \frac{\sqrt2 -1}{2-\sqrt3} \right )$

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

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

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

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

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

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

Question 12: Evaluate the definite integral

$\int_0^\frac{\pi}{2}\cos^2 x dx$

Answer:

Given integral: $\int_0^\frac{\pi}{2}\cos^2 x dx$

Consider the integral $\int \cos^2 x dx$

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

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

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

Now, by Second fundamental theorem of calculus, we have

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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:

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

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

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

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

$= \frac{1}{5}\log(5x^2+1) +\frac{3}{5}\times \frac{1}{\frac{1}{\sqrt5}} \tan^{-1}\frac{x}{\frac{1}{\sqrt5}}$

$= \frac{1}{5}\log(5x^2+1) +\frac{3}{\sqrt5}\tan^{-1}(\sqrt5 x )$

So, we have the function of $x$ , $f(x) = \frac{1}{5}\log(5x^2+1) +\frac{3}{\sqrt5}\tan^{-1}(\sqrt5 x )$

Now, by Second fundamental theorem of calculus, we have

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

$= \left \{ \frac{1}{5}\log(1+5)+\frac{3}{\sqrt5}\tan^{-1}(\sqrt5) \right \} - \left \{ \frac{1}{5}\log(1)+\frac{3}{\sqrt5}\tan^{-1}(0) \right \}$

$= \frac{1}{5}\log 6 +\frac{3}{\sqrt 5}\tan^{-1}{\sqrt5}$

Question 15: Evaluate the definite integral

$\int_0^1xe^{x^2}dx$

Answer:

Given integral: $\int_0^1xe^{x^2}dx$

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

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

As, $x\rightarrow0 ,t \rightarrow0$ and as $x\rightarrow1 ,t \rightarrow1$ .

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;

$I = \int_1^2 \frac{5x^2}{x^2 + 4x +3}= \int_1^2 \left ( 5 - \frac{20x+15}{x^2 + 4x +3} \right ) dx$

$= \int_1^2 5 dx - \int_1^2 \frac{20x+15}{x^2+4x+3}dx$

$= [5x]_1^2 - \int_1^2 \frac{20x+15}{x^2+4x+3}dx$

$I = 5-I_1$ 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 $20x+15 = A \frac{d}{dx}\left ( x^2+4x+3 \right )+B$

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

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 $(2x+4)dx =dt$

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

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

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

$= \left [ 10\log15 -10\log 8 \right ] -25 \left [ \frac{1}{2}\log\frac{3}{5} -\frac{1}{2}\log\frac{2}{4} \right ]$

$= \left [ 10\log5 +10\log3 -10\log4-10\log2 \right ] -\frac{25}{2}\left [ \log3 -\log5-\log2+\log4 \right ]$ $= \left ( 10+\frac{25}{2} \right )\log5 + \left ( -10-\frac{25}{2} \right )\log 4 + \left ( 10-\frac{25}{2} \right )\log 3 + \left ( -10+\frac{25}{2} \right )\log 2$ $= \frac{45}{2}\log5 -\frac{45}{2}\log4 - \frac{5}{2}\log3 +\frac{5}{2}\log2$

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

Then substituting the value of $I_{1}$ in equation (1), we get

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

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

Question 17: Evaluate the definite integral

$\int_0^\frac{\pi}{4}(2\sec^2x + x^3 + 2)dx$

Answer:

Given integral: $\int_0^\frac{\pi}{4}(2\sec^2x + x^3 + 2)dx$

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

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

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

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

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

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

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

Question 18: Evaluate the definite integral

$\int^\pi_0(\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx$

Answer:

Given integral: $\int^\pi_0(\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx$

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

can be rewritten as: $-\int (\cos^2\frac{x}{2} - \sin^2\frac{x}{2})dx = -\int_0^{\pi} \cos x dx$

$= \sin x$

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

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

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

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

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

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

Question 20: Evaluate the definite integral

$\int_0^1 \left( x e^x + \text{sin}\,\frac{\pi x}{4} \right) dx$

Answer:

Given integral: $\int_0^1 \left( xe^x + \text{sin}\, \frac{\pi x}{4} \right) dx$

Consider the integral $\int \left( xe^x + \text{sin}\, \frac{\pi x}{4} \right) dx$

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

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

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

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

Now, by Second fundamental theorem of calculus, we have

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

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

$= e-e -\frac{4}{\pi}\left ( \frac{1}{\sqrt2} \right )+1+\frac{4}{\pi}$

Question 21: Choose the correct answer

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

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

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

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

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

Answer:

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

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

$\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2} = f(\sqrt3) - f(1)$

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

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

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

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

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

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

$\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2} = f(\frac{2}{3}) - f(0)$

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

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

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

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