Ex 7.1 integration ncert solution maths class 12

06/02/2023 by Team Gmath

EXERCISE 7.1 (Integration)

Find an anti derivative (or integral) of the following functions by the method of inspection.(Ex 7.1 integration ncert solution maths class 12)

Question 1: $\sin 2x$

Solution: Using method of inspection

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

$\Rightarrow -\frac{1}{2}\frac{d}{dx}\cos 2x = \sin2x$

$\therefore \sin 2x = \frac{d}{dx}(-\frac{1}{2}\cos 2x)$

Thus, the anti derivative of $\sin 2x$ is $-\frac{1}{2}\cos 2x$

Question 2: $\cos 3x$

Solution: Using method of inspection

$\frac{d}{dx}\sin 3x= 3\cos 3x$

$\Rightarrow \frac{1}{3}\frac{d}{dx}\sin 3x = \cos3x$

$\therefore \cos 3x = \frac{d}{dx}(\frac{1}{3}\sin 3x)$

Thus, the anti derivative of $\cos 3x$ is $\frac{1}{3}\sin 3x$

Question 3: $e^{2x}$

Solution: Using method of inspection

$\frac{d}{dx}e^{2x}= 2e^{2x}$

$\Rightarrow \frac{1}{2}\frac{d}{dx}e^{2x} = e^{2x}$

$\therefore e^{2x} = \frac{d}{dx}(\frac{1}{2}e^{2x})$

Thus, the anti derivative of $e^{2x}$ is $\frac{1}{2}e^{2x}$

Question 4: $(ax+b)^2$

Solution : Using method of inspection

$\frac{d}{dx}(ax+b)^3= 3a(ax+b)^2$

$\Rightarrow \frac{1}{3a}\frac{d}{dx}(ax+b)^3= (ax+b)^2$

$\therefore (ax+b)^2 = \frac{d}{dx}(\frac{1}{3a}(ax+b)^3)$

Thus, the anti derivative of $e^{2x}$ is $\frac{1}{3a}(ax+b)^3$

Question 5: $\sin 2x-4e^{3x}$

Solution: Using method of inspection

Anti derivative of $\sin 2x = -\frac{1}{2}\cos 2x$

And antiderivative of $4e^{3x}=\frac{4}{3}e^{3x}$

$\therefore \frac{d}{dx}( -\frac{1}{2}\cos 2x-\frac{4}{3}e^{3x})=\sin 2x-4e^{3x}$

Thus the anitiderivative of $\sin 2x-4e^{3x}= -\frac{1}{2}\cos 2x-\frac{4}{3}e^{3x}$

Find the following integrals in Exercises 6 to 20:

Question 6: $\int (4e^{3x}+1)$

Solution: Let I = $\int (4e^{3x}+1)$

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

$=4\frac{e^{3x}}{3}+x+C$

$=\frac{4}{3}e^{3x}+x+C$

Question 7: $\int x^2(1-\frac{1}{x^2})dx$

Solution: Let I = $\int x^2(1-\frac{1}{x^2})dx$

$=\int (x^2-1)dx$

$=\frac{x^3}{3}-x+C$

Question 8: $\int (ax^2+bx+c)$

Solution: Let I = $\int (ax^2+bx+c)$

$=a\int x^2dx+b\int xdx+c\int 1dx$

$=a(\frac{x^3}{3})+b(\frac{x^2}{2})+cx+d$

$=\frac{1}{3}ax^3+\frac{1}{2}bx^2+cx+d$

Question 9: $\int(2x^2+e^x)dx$

Solution: Let I = $\int(2x^2+e^x)dx$

$= 2\int x^2dx+\int e^xdx$

$=2(\frac{x^3}{3})+e^x+C$

$= \frac{2}{3}x^3+e^x+C$

Question 10: $\int \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2dx$

Solution: Let I = $\int \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2dx$

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

$= \frac{x^2}{2}+\log x-2x+C$

Solution 11: $\int \frac{x^3+5x^2-4}{x^2}dx$

Solution: Let $I = \int \frac{x^3+5x^2-4}{x^2}dx$

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

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

$= \int xdx+5\int 1dx-4\int x^{-2}dx$

$= \frac{x^2}{2}+5x-4\frac{x^{-2+1}}{-2+1}+C$

$= \frac{1}{2}x^2+5x+\frac{4}{x}+C$

Question 12: $\int \frac{x^3+3x+4}{\sqrt{x}}dx$

Solution: Let $I =\int \frac{x^3+3x+4}{\sqrt{x}}dx$

$= \int (\frac{x^3}{\sqrt{x}})+3(\frac{x}{\sqrt{x}})+(\frac{4}{\sqrt{x}})dx$

$= \int x^{(3-1/2)}dx+3\int x^{(1-1/2)}dx+4\int x^{-1/2}dx$

$=\int x^{5/2}dx+3\int x^{1/2}dx+4\int x^{-1/2}dx$

$=\frac{x^{7/2}}{7/2}+3(\frac{x^{3/2}}{3/2})+4(\frac{x^{1/2}}{1/2})+C$

$= \frac{2}{7}x^{7/2}+2x^{3/2}+8x^{1/2}+C$

Question 13: $\int \frac{x^3-x^2+x-1}{x-1}dx$

Solution: Let I = $\int \frac{x^3-x^2+x-1}{x-1}dx$

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

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

$= \int x^2+1dx$

$= \frac{x^3}{3}+x+C$

Question 14: $\int(1-x)\sqrt{x}dx$

Solution: Let I = $\int(1-x)\sqrt{x}dx$

$= \int (\sqrt{x}-x^{(1+1/2)})dx$

$=\int (x^{1/2}-x^{3/2})dx$

$= \frac{x^{3/2}}{3/2}-\frac{x^5/2}{5/2}+C$

$=\frac{2}{3}x^{3/2}-\frac{2}{5}x^{5/2}+C$

Question 15: $\int \sqrt{x}(3x^2+2x+3)dx$

Solution: Let I = $\int \sqrt{x}(3x^2+2x+3)dx$

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

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

$= 3(\frac{x^{7/2}}{7/2})+2(\frac{x^{5/2}}{5/2})+3(\frac{x^{3/2}}{3/2})+C$

$=\frac{6}{7}x^{7/2}+\frac{4}{5}x^{5/2}+2x^{3/2}+C$

Question 16: $\int(2x-3\cos x+e^x)dx$

Solution: Let I= $\int(2x-3\cos x+e^x)dx$

$=2\int xdx-3\int \cos xdx+\int e^xdx$

$= 2(\frac{x^2}{2})-3\sin x+e^x+C$

$= x^2-3\sin x+e^x+C$

Question 17: $\int(2x^2-3\sin x+5\sqrt{x})dx$

Solution: Let I = $\int(2x^2-3\sin x+5\sqrt{x})dx$

$=2\int x^2dx-3\int\sin xdx+5\int x^{1/2}dx$

$=2(\frac{x^3}{3})+3\cos x+5(\frac{x^{3/2}}{3/2})+C$

$=\frac{2}{3}x^3+3\cos x+\frac{10}{3}x^{3/2}+C$

Question 18: $\int \sec x(\sec x+\tan x)dx$

Solution: Let I = $\int \sec x(\sec x+\tan x)dx$

$= \int (\sec^2x+\sec x\tan x)dx$

$= \tan x+\sec x+C$

Question 19: $\int\frac{\sec^2x}{\operatorname{cosec^2}x}dx$

Solution: I = $\int\frac{\sec^2x}{\operatorname{cosec^2}x}dx$

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

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

$= \int \tan^2 x$

$=\int (\sec^2x-1)dx$

$=\tan x-x+C$

Question 20: $\int \frac{2-3\sin x}{\cos^2x}dx$

Solution: Let I = $\int \frac{2-3\sin x}{\cos^2x}dx$

$=\int (\frac{2}{\cos^2x}-3\frac{\sin x}{\cos^2x})dx$

$=\int (2\sec^2x-\tan x\sec x)$

$= 2\tan x-\sec x+C$

Choose the correct answer in Exercises 21 and 22.

Question 21: The antiderivative of $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$ equals

(A) $\frac{1}{3}x^{1/3}+2x^{1/2}+C$

(B) $\frac{25}{3}x^{2/3}+\frac{1}{2}x^{2}+C$

(C) $\frac{2}{3}x^{3/2}+2x^{1/2}+C$

(D) $\frac{3}{2}x^{3/2}+\frac{1}{2}x^{1/2}+C$

Solution: Answer is option (C)

$\int \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$

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

$= \frac{x^{3/2}}{3/2}+\frac{x^{1/2}}{1/2}+C$

$=\frac{2}{3}x^{3/2}+2x^{1/2}+C$

Question 22: If $\frac{d}{dx}f(x)=4x^3-\frac{3}{x^4}$ such that $f(2) = 0$ . Then $f(x)$ is

(A) $x^4+\frac{1}{x^3}-\frac{1}{129}$

(B) $x^3+\frac{1}{x^4}+\frac{129}{8}$

(C) $x^4+\frac{1}{x^3}+\frac{129}{8}$

(D) $x^3+\frac{1}{x^4}-\frac{129}{8}$

Solution: Answer is option (A)

Since, $\frac{d}{dx}f(x)=4x^3-\frac{3}{x^4}$

$df(x)=(4x^3-\frac{3}{x^4})dx$

Integrating both side

$\int df(x)=\int (4x^3-\frac{3}{x^4})dx$

$\Rightarrow f(x) = 4(\frac{x^4}{4})-3\frac{x^{-4+1}}{(-4+1)}+C$

$\Rightarrow f(x) = x^4+\frac{x^{-3}}{3}+C$

$\Rightarrow f(x)= x^4+\frac{1}{x^3}+C ---(i)$

$\Rightarrow f(2) = (2)^4+\frac{1}{(2)^3}+C$

$\Rightarrow 0= 16+\frac{1}{8}+C$

$\Rightarrow 0=\frac{129}{8}+C$

$\Rightarrow C = -\frac{129}{8}$

Putting in (i)

$f(x) = x^4+\frac{1}{x^3}-\frac{129}{8}$

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