Integrals - Exemplar Solutions

 CBSE Class–12 Mathematics

NCERT Exemplar
Chapter - 7
INTEGRALS - Short Answer Questions

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1. Integrate w.r.t. x

Sol. 

2. Evaluate 

Sol. Let              v=b2+c2x2$v=b^2+c^2{x}^2$      then 

Therefore, 

3. Verify the following using the concept of integration as an antiderivative.

Sol. 

Thus 

4. Evaluate 

Sol. Let 

where 

Put Therefore

Hence 

5. Evaluate .

Sol. Put and . Now

6. Evaluate 

Sol. 

7. Find∫x3dxx4+3x2+2$\int \frac{x^{3}dx}{x^4+3x^2+2}$

Sol. Put .

Now 

Consider 

Comparing coefficient, we get 

Then 

8. Find 

Sol. Dividing numerator and denominator by we have

Put so that . Then

I=∫dt2t2+5$I=\int \frac{dt}{2t^2+5}$

=12∫dtt2+(52√)2$=\frac{1}{2}\int \frac{dt}{t^2+{\left(\sqrt{\frac{5}{2}}\right)}^2}$

=12(25−−√)tan−1(t2√5√)+C=110√tan−1(2√tanx5√)+C$\frac{1}{2}\left(\sqrt{\frac{2}{5}}\right){\tan }^{-1}\left(\frac{t\sqrt{2}}{\sqrt{5}}\right)+C=\frac{1}{\sqrt{10}}{\tan }^{-1}\left(\frac{\sqrt{2}\tan x}{\sqrt{5}}\right)+C$

 

9. Evaluate as a limit of sums.

Sol. Here and i.e, .

Now, we have

Now that

Therefore,

10. Evaluate 

Sol. We have

Adding (1) and (2), we get

11. Find 

Sol. We have

I=∫28x√10−x√+x√dx$I=\underset{2}{\overset{8}{\int }}\frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}dx$....(2)

Adding (1) and (2), we get:

2I=∫2810−x√+x√10−x√+x√dx$2I=\underset{2}{\overset{8}{\int }}\frac{\sqrt{10-x}+\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}dx$

⇒I=∫281dx=(x)82=8−2=6$\Rightarrow I=\underset{2}{\overset{8}{\int }}1dx=(x{)}_2^8=8-2=6$

⇒$\Rightarrow$I=3$I=3$

12. Find 

Sol. We have

I = 1.

13. Find 

Sol. 

14. Find 

Sol. We have

Put then 

Therefore, 

 

Long Answer Questions

 

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15. Evaluate .

Sol. Let . Then

So 

Comparing coefficients, we get 

So 

Therefore,

16. Evaluate 

Sol. we have

Now 

Put so that . Therefore

Again, 

Put so that . Then

Thus 

17. Show that 

Sol. We have

Thus, we get 

Hence, 

18. Find 

Sol. 

Integrating by parts, we have

     I=(x22(tan−1x)2)10−∫012tan−1x11+x2x22dx$I={\left(\frac{x^2}{2}{\left({\tan }^{-1}x\right)}^2\right)}_0^1-\underset{0}{\overset{1}{\int }}2{\tan }^{-1}x\frac{1}{1+x^2}\frac{x^2}{2}dx$

 

I=π232−∫01tan−1xx21+x2dx$I=\frac{{\pi }^2}{32}-\underset{0}{\overset{1}{\int }}{\tan }^{-1}x\frac{x^2}{1+x^2}dx$

I=π232−I1whereI1=∫01x21+x2tan−1xdx$I=\frac{{\pi }^2}{32}-I_1^{}where{I}_1^{}=\underset{0}{\overset{1}{\int }}\frac{x^2}{1+x^2}{\tan }^{-1}xdx$

Now 

I1=∫01tan−1xdx−∫0111+x2tan−1xdx$I_1^{}=\underset{0}{\overset{1}{\int }}{\tan }^{-1}xdx-\underset{0}{\overset{1}{\int }}\frac{1}{1+x^2}{\tan }^{-1}xdx$

I1=I2−12[(tan−1x)2]10=I2−π232$I_1^{}=I_2^{}-\frac{1}{2}{\left[{({\tan }^{-1}x)}^2\right]}_0^1=I_2^{}-\frac{{\pi }^2}{32}$where I2=∫01tan−1xdx$I_2^{}=\underset{0}{\overset{1}{\int }}{\tan }^{-1}xdx$

I2=(xtan−1x)10−∫0111+x2xdx$I_2^{}=(x{\tan }^{-1}x{)}_0^1-\underset{0}{\overset{1}{\int }}\frac{1}{1+x^2}xdx$

I2=π4−12[log∣∣1+x2∣∣]10=π4−12log2$I_2^{}=\frac{\pi }{4}-\frac{1}{2}{\left[\log \left|1+x^2\right|\right]}_0^1=\frac{\pi }{4}-\frac{1}{2}\log 2$

Thus,I1=π4−12log2−π232$I_1^{}=\frac{\pi }{4}-\frac{1}{2}\log 2-\frac{{\pi }^2}{32}$

Therefore,  I=π232−π4+12log2+π232=π216−π4+12log2$I=\frac{{\pi }^2}{32}-\frac{\pi }{4}+\frac{1}{2}\log 2+\frac{{\pi }^2}{32}=\frac{{\pi }^2}{16}-\frac{\pi }{4}+\frac{1}{2}\log 2$

 I=π2−4π16+log2–√$I=\frac{{\pi }^2-4\pi }{16}+\log \sqrt{2}$

 

19. Evaluate 

Sol. We can redefine as 

Therefore, 

 

Objective Questions

 

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Choose the correct answer from the given four options in each of the Examples from 20 to 30.

20. is equal to

(A) 

(B) 

(C) 

(D) 

Sol. (A) is the correct answer since∫ex(f(x)+f′(x))dx=exf(x)+C$\int e^x(f(x)+f^{\prime }(x))dx=e^{x}f(x)+C$

. Here .

So, =excosx+C$e^x\cos x+C$

21. is equal to

(A) 

(B) 

(C) 

(D) 

Sol. (C) is the correct answer, since

22. If , then

(A) 

(B) 

(C) 

(D) 

Sol. (C) is the correct answer, since differentiating both sides, we have

Giving . Comparing coefficients on both sides, we get

. This verifies 

23. is equal to

(A) 

(B) 

(C) 

(D) 

Sol. (B) is the correct answer, since by putting we get

24. If are continuous in [0, 1] satisfying and thenthen is equal to

(A) 

(B) 

(C) 

(D) 

Sol. (B) is the correct answer. Since 

Or 

25. Ifx=∫0y11+9t2√dtandd2ydx2=ay$Ifx=\underset{0}{\overset{y}{\int }}\frac{1}{\sqrt{1+9t^2}}dtand\frac{d^{2}y}{d{x}^2}=ay$then is equal to

(A) 3

(B) 6

(C) 9

(D) 1

Sol. (C) is the correct answer, sincex=∫0y11+9t2√dt$x=\underset{0}{\overset{y}{\int }}\frac{1}{\sqrt{1+9t^2}}dt$⇒dxdy=11+9y2√⇒dydx=1+9y2−−−−−−√$\Rightarrow \frac{dx}{dy}=\frac{1}{\sqrt{1+9y^2}}\Rightarrow \frac{dy}{dx}=\sqrt{1+9y^2}$

which gives 

So, a = 9

26. is equal to

(A) log 2

(B) 2 log 2

(C) log 2

(D) 4 log 2

Sol. (B) is the correct answer, since 

[odd function + even function]

27. If then is equal to

(A) 

(B) 

(C) 

(D) 

Sol. (B) is the correct answer, since 

Therefore, 

28. is equal to

(A) 

(B) 

(C) 

(D) 

Sol. (A) is the correct answer, since 

(Since∫xcosπxdx=xsinπxπ+cosπxπ2$\int x\cos \pi xdx=\frac{x\sin \pi x}{\pi }+\frac{\cos \pi x}{{\pi }^2}$)

Fill in the blanks in each of the Examples 29 to 32.

29. ______________.

Sol. Since, ∫sin6xcos6xsec2xdx=∫tan6x.sec2xdxlettanx=t⇒sec2xdx=dt$\begin{array}{l}\int \frac{{\sin }^{6}x}{{\cos }^{6}x}{\sec }^{2}xdx=\int {\tan }^{6}x.{\sec }^{2}xdx\\ let\tan x=t\Rightarrow {\sec }^{2}xdx=dt\end{array}$

⇒$\Rightarrow$

30. if is an ______________ function.

Sol. Odd.

31. 

Sol. 

32. _________________.

Sol. .