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Ex 7.1 integration ncert solution maths class 12

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)

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

Team Gmath

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Chapter 7

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