RD Sharma Solutions Class 12 Mathematics Chapter 3 VSA

RD Sharma Solutions Class 12 Mathematics Chapter 19 VSA

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RD Sharma class 12th exercise VSA is an extremely important book for all students who are in class 12. RD Sharma solutions is a trusted and famous name among numerous other NCERT solutions. Students and teachers highly recommend this book especially for its Mathematics solutions. The RD Sharma class 12 chapter 19 exercise VSA, especially can be an excellent guidebook for students who want to study and home well before their board exams..

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RD Sharma Class 12 Solutions Chapter19 VSA Definite Integrals - Other Exercise
Definite Integrals Excercise:VSA
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter19 VSA Definite Integrals - Other Exercise

Definite Integrals Excercise:VSA

Definite Integrals exercise Very short answer type question 1

Answer: $\frac{π}{4}$
Hint: You must know the integration rule of trigonometric functions with its limits.
Given: $\int_{0}^{\frac{π}{2}} \sin^{2} x d x$
Solution:
$\begin{aligned} = \int_{0}^{\frac{π}{2}} \frac{1 - \cos 2 x}{2} d x \\ = \frac{1}{2} \int_{0}^{\frac{π}{2}} 1 . d x - \frac{1}{2} \int_{0}^{\frac{π}{2}} \cos 2 x d x \end{aligned}$
$\begin{aligned} = \frac{1}{2} [x]_{0}^{\frac{π}{2}} - \frac{1}{2} {[\frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} \\ = \frac{1}{2} [\frac{π}{2} - 0] - \frac{1}{4} [\sin (2 \frac{π}{2}) - \sin 0] \end{aligned}$
$\begin{aligned} = \frac{π}{4} - \frac{1}{4} [0] \\ = \frac{π}{4} \end{aligned}$

Definite Integrals exercise Very short answer type question 2

Answer: $\frac{π}{4}$
Hint: You must know the integration rule of trigonometric functions.
Given: $\int_{0}^{\frac{π}{2}} \cos^{2} x d x$
Solution: $\int_{0}^{\frac{π}{2}} \cos^{2} x d x$
$\begin{aligned} = \int_{0}^{\frac{π}{2}} \frac{1 + \cos 2 x}{2} d x \\ = \frac{1}{2} \int_{0}^{\frac{π}{2}} 1 . d x + \frac{1}{2} \int_{0}^{\frac{π}{2}} \cos 2 x d x \end{aligned}$
$\begin{aligned} = \frac{1}{2} [x]_{0}^{\frac{π}{2}} + \frac{1}{2} {[\frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} \\ = \frac{1}{2} [\frac{π}{2} - 0] + \frac{1}{4} [\sin \frac{2 π}{2} - \sin 0] \\ = \frac{π}{4} \end{aligned}$

Definite Integrals exercise Very short answer type question 3

Answer: $\frac{π}{2}$
Hint: You must know the integration rules of trigonometric function with its limits
Given: $\int_{\frac{- π}{2}}^{\frac{π}{2}} \sin^{2} x d x$
Solution: $\int_{\frac{- π}{2}}^{\frac{π}{2}} \sin^{2} x d x$
$\begin{aligned} = \int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{1 - \cos 2 x}{2} d x \\ = \frac{1}{2} \int_{\frac{- π}{2}}^{\frac{π}{2}} 1 \cdot d x - \frac{1}{2} \int_{\frac{- π}{2}}^{\frac{π}{2}} \cos 2 x d x \end{aligned}$
$= \frac{1}{2} [x]_{\frac{- π}{2}}^{\frac{π}{2}} - \frac{1}{2} {[\frac{\sin 2 x}{2}]}_{\frac{- π}{2}}^{\frac{π}{2}}$
$\begin{aligned} = \frac{1}{2} [\frac{π}{2} - (- \frac{π}{2})] - \frac{1}{4} [\sin \frac{2 (π)}{2} - \sin \frac{2 (- π)}{2}] \\ = \frac{1}{2} [\frac{2 π}{2}] - \frac{1}{4} [0 + 0] \\ = \frac{π}{2} \end{aligned}$

Definite Integrals exercise Very short answer type question 4
Answer: $\frac{π}{2}$

Hint: You must know the integration rules of trigonometric function with its limits
Given: $\int_{\frac{- π}{2}}^{\frac{π}{2}} \cos^{2} x d x$
Solution: $\int_{\frac{- π}{2}}^{\frac{π}{2}} \cos^{2} x d x$

\begin{aligned} = \int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{1 + \cos 2 x}{2} d x \\ = \frac{1}{2} \int_{\frac{- π}{2}}^{\frac{π}{2}} 1 . d x + \frac{1}{2} \int_{\frac{- π}{2}}^{\frac{π}{2}} \cos 2 x d x \end{aligned}

\begin{aligned} = \frac{1}{2} [x]_{\frac{- π}{2}}^{\frac{π}{2}} + \frac{1}{2} {[\frac{\sin 2 x}{2}]}_{\frac{π}{2}}^{\frac{π}{2}} \\ = \frac{1}{2} [\frac{π}{2} - (- \frac{π}{2})] + \frac{1}{4} [\sin \frac{2 (π)}{2} + \sin \frac{2 (- π)}{2}] \end{aligned}

\begin{aligned} = \frac{1}{2} [\frac{2 π}{2}] + \frac{1}{4} [0 + 0] \\ = \frac{π}{2} \end{aligned}

Definite Integrals exercise Very short answer type question 5

Answer: 0
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{- π}^{\frac{π}{2}} \sin^{3} x d x$
Solution: $\int_{- π}^{\frac{π}{2}} \sin^{3} x d x$
$\begin{aligned} = \int_{\frac{- π}{2}}^{\frac{π}{2}} \sin^{2} x \sin x d x \\ = \int_{\frac{- π}{2}}^{\frac{π}{2}} (1 - \cos^{2} x) \sin x d x \end{aligned}$

= \int_{\frac{- π}{2}}^{\frac{π}{2}} \sin x d x + \int_{\frac{- π}{2}}^{\frac{π}{2}} \cos^{2} x \sin x d x

We know that

\int \sin x d x = - \cos x and \int \cos x d x = \sin x

\begin{aligned} = [- \cos x]_{\frac{- π}{2}}^{\frac{π}{2}} + {[\frac{- \cos^{3} x}{3}]}_{\frac{- π}{2}}^{\frac{π}{2}} \\ = - \cos \frac{π}{2} - (- \cos \frac{- π}{2}) + [\frac{\cos^{3} (\frac{π}{2})}{3} - \frac{\cos^{3} (\frac{- π}{2})}{3}] \\ = 0 + 0 + 0 - 0 \\ = 0 \end{aligned}

Definite Integrals exercise Very short answer type question 6

Answer: 0
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{\frac{- π}{2}}^{\frac{π}{2}} x \cos^{2} x d x$
Solution: $I = \int_{\frac{- π}{2}}^{\frac{π}{2}} x \cos^{2} x d x$

\begin{aligned} f (x) = x \cos^{2} x \\ f (- x) = (- x) \cos^{2} (- x) \\ = - x \cos^{2} x \\ = - f (x) \end{aligned}

Hence, f(x) is an odd function.
Since,

\int_{- a}^{a} f (x) d x = 0

if f(x) is an odd

∴ \int_{\frac{- π}{2}}^{\frac{π}{2}} x \cos^{2} x d x = 0

Definite Integrals exercise Very short answer type question 7

Answer: $1 - \frac{π}{4}$
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{\frac{π}{4}} \tan^{2} x d x$
Solution: $\int_{0}^{\frac{π}{4}} \tan^{2} x d x$
$\begin{aligned} = \int_{0}^{\frac{π}{4}} (\sec^{2} x - 1) d x \\ = \int_{0}^{\frac{π}{4}} \sec^{2} x d x - \int_{0}^{\frac{π}{4}} 1 d x \end{aligned}$
$\begin{aligned} = [\tan x]_{0}^{\frac{π}{4}} - [x]_{0}^{\frac{π}{4}} \\ = \tan \frac{π}{4} - \tan 0 - \frac{π}{4} + 0 \end{aligned}$
$\begin{aligned} = 1 - 0 - \frac{π}{4} + 0 \\ = 1 - \frac{π}{4} \end{aligned}$

Definite Integrals exercise Very short answer type question 8

Answer: $\frac{π}{4}$
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{1} \frac{1}{x^{2} + 1} d x$
Solution: $\int_{0}^{1} \frac{1}{x^{2} + 1} d x$
$\begin{aligned} = [\tan^{- 1} 1 - \tan^{- 1} 0] \\ = \frac{π}{4} - 0 \\ = \frac{π}{4} \end{aligned}$

Definite Integrals exercise Very short answer type question 9

Answer: $- 1$
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{- 2}^{1} \frac{| x |}{x} d x$
Solution: $\int_{- 2}^{1} \frac{| x |}{x} d x$

\begin{aligned} = \int_{- 2}^{0} \frac{- x}{x} d x + \int_{0}^{1} \frac{x}{x} d x \\ = \int_{- 2}^{0} - 1 d x + \int_{0}^{1} 1 d x \end{aligned}

\begin{aligned} = - [x]_{- 2}^{0} + [x]_{0}^{1} \\ = - [0 + 2] + [1 + 0] \\ = - 2 + 1 \\ = - 1 \end{aligned}

Definite Integrals exercise Very short answer type question 10

Answer: 1
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{\infty} e^{- x} d x$
Solution: ${[\frac{e^{- x}}{- 1}]}_{0}^{\infty}$
$\begin{aligned} = - [\frac{1}{e^{\infty}} - \frac{1}{e^{0}}] \\ = 0 + 1 \\ = 1 \end{aligned}$

Definite Integrals exercise Very short answer type question 11

Answer: $\frac{π}{2}$
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{4} \frac{1}{\sqrt{16 - x^{2}}} d x$
Solution: $\int_{0}^{4} \frac{1}{\sqrt{16 - x^{2}}} d x$
$= \int_{0}^{4} \frac{1}{\sqrt{(4)^{2} - x^{2}}} d x$
$\begin{aligned} Put x = 4 \sin θ θ = \sin^{- 1} \frac{x}{4} \\ d x = 4 \cos θ d θ \end{aligned}$
$\begin{aligned} = \int_{0}^{4} \frac{4 \cos θ}{\sqrt{16 - 16 \sin^{2} θ}} d θ \\ = \int_{0}^{4} \frac{4 \cos θ}{\sqrt{16 (1 - \sin^{2} θ)}} d θ \end{aligned}$
$\begin{aligned} = \int_{0}^{4} \frac{4 \cos θ}{4 \sqrt{\cos^{2} θ}} d θ \\ = \int_{0}^{4} \frac{4 \cos θ}{4 \cos θ} d θ \end{aligned}$
$\begin{aligned} = \int_{0}^{4} 1 d θ \\ = [θ]_{0}^{4} = {[{Sin}^{- 1} \frac{x}{4}]}_{0}^{4} \end{aligned}$
$\begin{aligned} = [\sin^{- 1} \frac{4}{4} - \sin^{- 1} \frac{0}{4}] \\ = \sin^{- 1} 1 \\ = \frac{π}{2} \end{aligned}$

Definite Integrals exercise Very short answer type question 12

Answer: $\frac{π}{12}$
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{3} \frac{1}{x^{2} + 9} d x$
Solution: $\int_{0}^{3} \frac{1}{x^{2} + 9} d x$
Use the formula

\int \frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} \tan^{- 1} \frac{x}{a} + c

\begin{aligned} = \frac{1}{3} {[\tan^{- 1} \frac{x}{3}]}_{0}^{3} \\ = \frac{1}{3} [\tan^{- 1} 1 - \tan^{- 1} 0] \end{aligned}

\begin{aligned} = \frac{1}{3} [\frac{π}{4}] \\ = \frac{π}{12} \end{aligned}

Definite Integrals exercise Very short answer type question 13

Answer: $\sqrt{2}$
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{\frac{π}{2}} \sqrt{1 - \cos 2 x} d x$
Solution: $\int_{0}^{\frac{π}{2}} \sqrt{1 - \cos 2 x} d x$
$\begin{aligned} = \int_{0}^{\frac{π}{2}} \sqrt{2 \sin^{2} x} d x \\ = \int_{0}^{\frac{π}{2}} \sqrt{2} \sin x d x \end{aligned}$
$\begin{aligned} = \sqrt{2} [- \cos x]_{0}^{\frac{π}{2}} \\ = \sqrt{2} [- \cos \frac{π}{2} + \cos 0] \\ = \sqrt{2} \end{aligned}$

Definite Integrals exercise Very short answer type question 14

Answer: 0
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{\frac{π}{2}} \log \tan x d x$
Solution: $\int_{0}^{\frac{π}{2}} \log \tan x d x$ .................(i)

I = \int_{0}^{\frac{π}{2}} \log \tan (\frac{π}{2} - x) d x

= \int_{0}^{\frac{π}{2}} \log \cot x d x

..................(ii)
Adding (i) and (ii)

2 I = \int_{0}^{\frac{π}{2}} \log (\tan x) d x + \int_{0}^{\frac{π}{2}} \log (\cot x) d x

= \int_{0}^{\frac{π}{2}} (\log \tan x + \log \cot x) d x

= \int_{0}^{\frac{π}{2}} (\log \tan x \cdot \cot x) d x [∵ \log m + \log n = \log m n]

= \int_{0}^{\frac{π}{2}} (\log 1) d x [∵ \tan x \cdot \cot x = 1]

\begin{aligned} = \int_{0}^{\frac{π}{2}} (0) d x [∵ \log 1 = 0] \\ = 0 \end{aligned}

Definite Integrals exercise Very short answer type question 15

Answer: 0
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{\frac{π}{2}} \log (\frac{3 + 5 \cos x}{3 + 5 \sin x}) d x$ .............(i)
Solution: $\int_{0}^{\frac{π}{2}} \log (\frac{3 + 5 \cos x}{3 + 5 \sin x}) d x$
Property:

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

I = \int_{0}^{\frac{π}{2}} \log \frac{3 + 5 \cos (\frac{π}{2} - x)}{3 + 5 \sin (\frac{π}{2} - x)} d x

\begin{aligned} = \int_{0}^{\frac{π}{2}} \log \frac{3 + 5 \sin x}{3 + 5 \cos x} d x \\ = - \int_{0}^{\frac{π}{2}} \log \frac{3 + 5 \cos x}{3 + 5 \sin x} d x \end{aligned}

................(ii)
Adding (i) and (ii),

\begin{aligned} 21 = \int_{0}^{\frac{π}{2}} \log (\frac{3 + 5 \cos x}{3 + 5 \sin x}) d x + [- \int_{0}^{\frac{π}{2}} \log \frac{3 + 5 \cos x}{3 + 5 \sin x} d x] \\ 21 = 0 \\ I = 0 \end{aligned}

Definite Integrals exercise Very short answer type question 16

Answer: $\frac{π}{4}$
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{\frac{π}{2}} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} d x$
Solution: $\int_{0}^{\frac{π}{2}} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} d x$ ....................(i)
Property:

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

= \int_{0}^{\frac{π}{2}} \frac{\sin^{n} (\frac{π}{2} - x)}{\sin^{n} (\frac{π}{2} - x) + \cos^{n} (\frac{π}{2} - x)} d x

= \int_{0}^{\frac{π}{2}} \frac{\cos^{n} x}{\sin^{n} x + \cos^{n} x} d x

..................(ii)
Add (i) and (ii)

\begin{aligned} 2 l = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} x + \cos^{n} x}{\sin^{n} x + \cos^{n} x} d x \\ 2 I = \int_{0}^{\frac{π}{2}} 1 \cdot d x \end{aligned}

\begin{aligned} 21 = [x]_{0}^{\frac{π}{2}} \\ 2 I = [\frac{π}{2} - 0] \\ I = \frac{π}{4} \end{aligned}

Definite Integrals exercise Very short answer type question 17

Answer: 0
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{π} \cos^{5} x d x$
Solution: Using property

\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x

I = \int_{0}^{π} \cos^{5} x d x = \int_{0}^{π} \cos x \cos^{4} x d x

= \int_{0}^{π} \cos x {(1 - \sin^{2} x)}^{2} d x

\begin{aligned} Put \sin x = t \\ \cos x d x = d t \end{aligned}

\begin{aligned} I = \int_{0}^{π} {(1 - t^{2})}^{2} d x \\ = \int_{0}^{π} (1 + t^{4} - 2 t^{2}) d x \end{aligned}

= {[t + \frac{t^{5}}{5} - \frac{2 t^{3}}{3}]}_{0}^{π}

\begin{aligned} = {[\sin x + \frac{\sin^{5} x}{5} - \frac{2 \sin^{3} x}{3}]}_{0}^{π} [\sin π = 0, \sin 0 = 0] \\ I = 0 \end{aligned}

Definite Integrals exercise Very short answer type question 18

Answer: 0
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{\frac{- π}{2}}^{\frac{π}{2}} \log (\frac{a - \sin θ}{a + \sin θ}) d θ$

Solution: $I = \int_{\frac{- π}{2}}^{\frac{π}{2}} \log (\frac{a - \sin θ}{a + \sin θ}) d θ$ .............(i)
Let

f (x) = \log (\frac{a - \sin θ}{a + \sin θ})

f (- x) = \log (\frac{a - \sin (- θ)}{a + \sin (- θ)})

= \log (\frac{a + \sin θ}{a - \sin θ}) [∵ \sin (- x) = - \sin x]

\begin{aligned} = - \log (\frac{a - \sin θ}{a + \sin θ}) \\ = - f (x) \end{aligned}

Hence, f(x) is an odd function.

Since,

\int_{- a}^{a} f (x) d x = 0

if f(x) is an odd.

∴ \int_{\frac{- π}{2}}^{\frac{π}{2}} \log (\frac{a - \sin θ}{a + \sin θ}) d θ = 0

Definite Integrals exercise Very short answer type question 19

Answer: 0
Hint: you must know the rule of integration for function of x

Given: $\int_{- 1}^{1} x | x | d x$
Solution: $I = \int_{- 1}^{1} x | x | d x$
$\begin{aligned} = \int_{- 1}^{0} x (- x) d x + \int_{0}^{1} x (x) d x \\ = \int_{- 1}^{0} - x^{2} d x + \int_{0}^{1} x^{2} d x \end{aligned}$
$\begin{aligned} = - \frac{1}{3} {[x^{3}]}_{- 1}^{0} + \frac{1}{3} {[x^{3}]}_{0}^{1} \\ = - \frac{1}{3} [0 - (- 1)] + \frac{1}{3} [1 - 0] \end{aligned}$
$\begin{aligned} = - \frac{1}{3} + \frac{1}{3} \\ = 0 \end{aligned}$

Definite Integrals exercise Very short answer type question 20

Answer: $\frac{b - a}{2}$
Hint: you must know the rule of integration for function of x

Given: $\int_{a}^{b} \frac{f (x)}{f (x) + f (a + b - x)} d x$
Solution: $I = \int_{a}^{b} \frac{f (x)}{f (x) + f (a + b - x)} d x$ ...............(i)
Using property

\int_{a}^{b} f (x) = \int_{a}^{b} f (a + b - x) d x

I = \int_{a}^{b} \frac{f (a + b - x)}{f (a + b - x) + f (a + b - (a + b - x))} d x

I = \int_{a}^{b} \frac{f (a + b - x)}{f (a + b - x) + f (x)} d x

................(ii)
Add (i) and (ii)

\begin{aligned} 2 I = \int_{a}^{b} \frac{f (x) + f (a + b - x)}{f (x) + f (a + b - x)} d x \\ 2 I = \int_{a}^{b} 1 . d x \end{aligned}

\begin{aligned} 2 l = [x]_{a}^{b} \\ 2 l = b - a \\ I = \frac{b - a}{2} \end{aligned}

Definite Integrals exercise Very short answer type question 21

Answer: $\frac{π}{4}$
Hint: you must know the rule of integration for function of x
Given: $\int_{0}^{1} \frac{1}{1 + x^{2}} d x$
Solution: $\int_{0}^{1} \frac{1}{1 + x^{2}} d x$
$\begin{aligned} = {[\tan^{- 1} x]}_{0}^{1} \\ = [\tan^{- 1} 1 - \tan^{- 1} 0] \\ = \frac{π}{4} \end{aligned}$

Definite Integrals exercise Very short answer type question 22

Answer: $\frac{1}{2} \log 2$
Hint: you must know the rule of integration for function of x
Given: $\int_{0}^{\frac{π}{4}} \tan x d x$
Solution: $\int_{0}^{\frac{π}{4}} \tan x d x$
$\begin{aligned} = \log [\sec x]_{0}^{\frac{π}{4}} \\ = \log (\sec \frac{π}{4}) - \log (\sec 0) \\ = \log | \sqrt{2} | - \log | 1 | \end{aligned}$
$\begin{aligned} = \log \sqrt{2} - 0 \\ = \log \sqrt{2} \\ = \log (2)^{\frac{1}{2}} \\ = \frac{1}{2} \log 2 \end{aligned}$

Definite Integrals exercise Very short answer type question 23
Answer: $\log \frac{3}{2}$

Hint: you must know the rule of integration for function of x
Given: $\int_{2}^{3} \frac{1}{x} d x$
Solution: $\int_{2}^{3} \frac{1}{x} d x$
$\begin{aligned} = [\log | x |]_{2}^{3} \\ = \log 3 - \log 2 \\ = \log \frac{3}{2} \end{aligned}$

Definite Integrals exercise Very short answer type question 24

Answer: $π$
Hint: you must know the rule of integration
Given: $\int_{0}^{2} \sqrt{4 - x^{2}}$
Solution: $\int_{0}^{2} \sqrt{4 - x^{2}}$
$\begin{aligned} Put x = 2 \sin θ \\ d x = 2 \cos θ d θ \end{aligned}$
$\begin{aligned} = \int_{0}^{\frac{π}{2}} \sqrt{4 - 4 \sin^{2} θ} \cdot 2 \cos θ d θ \\ = \int_{0}^{\frac{π}{2}} 4 \cos^{2} θ d θ \end{aligned}$
$\begin{aligned} = \int_{0}^{\frac{π}{2}} 4 \cdot \frac{1}{2} (1 + \cos 2 θ) d θ \\ = 2 {[θ + \frac{\sin 2 θ}{2}]}_{0}^{\frac{π}{2}} \end{aligned}$
$\begin{aligned} = 2 [\frac{π}{2} + \frac{\sin π}{2} - 0 - 0] \\ = π \end{aligned}$

Definite Integrals exercise Very short answer type question 25

Answer: $\log_{e} 2$
Hint: you must know the rule of integration for function of x
Given: $\int_{0}^{1} \frac{2 x}{1 + x^{2}} dx$
Solution: $\int_{0}^{1} \frac{2 x}{1 + x^{2}} dx$
Put

1 + x^{2} = t

2 x d x = d t

\begin{aligned} = \int_{0}^{1} \frac{d t}{t} \\ = [\log t]_{0}^{1} \end{aligned}

\begin{aligned} = {[\log (1 + x^{2})]}_{0}^{1} \\ = [\log (1 + 1) - \log (1 + 0)] \\ = \log_{e} 2 \end{aligned}

Definite Integrals exercise Very short answer type question 26

Answer: $\frac{1}{2} (e - 1)$
Hint: you must know the rule of integration for function of x
Given: $\int_{0}^{1} x e^{x^{2}} d x$
Solution:
Put

\begin{aligned} x^{2} = t \\ 2 x d x = d t \\ x d x = \frac{d t}{2} \end{aligned}

\begin{aligned} = \frac{1}{2} \int_{0}^{1} e^{t} d t \\ = \frac{1}{2} {[e^{t}]}_{0}^{1} \end{aligned}

\begin{aligned} = \frac{1}{2} {[e^{x^{2}}]}_{0}^{1} \\ = \frac{1}{2} [e^{1} - e^{0}] \\ = \frac{1}{2} [e - 1] \end{aligned}

Definite Integrals exercise Very short answer type question 27

Answer: $\frac{1}{2}$
Hint: you must know the rule of integration
Given: $\int_{0}^{\frac{π}{4}} \sin 2 x d x$
Solution: $\int_{0}^{\frac{π}{4}} \sin 2 x d x$
$\begin{aligned} = \frac{1}{2} (- \cos 2 x)_{0}^{\frac{π}{4}} \\ = - \frac{1}{2} (\cos 2 x)_{0}^{\frac{π}{4}} \end{aligned}$
$\begin{aligned} = \frac{- 1}{2} [\cos 2 (\frac{π}{4}) - \cos 2 (0)] \\ = \frac{- 1}{2} [\cos \frac{π}{2} - \cos 0] \end{aligned}$
$\begin{aligned} = \frac{- 1}{2} [0 - 1] \\ = \frac{1}{2} \end{aligned}$

Definite Integrals exercise Very short answer type question 28

Answer: $\log_{e} 2$
Hint: you must know the rule of integration
Given: $\int_{e}^{e^{2}} \frac{1}{x \log x} d x$
Solution: Put

\log x = t

$\frac{d x}{x} = dt$
$\begin{aligned} = \int_{e}^{e^{2}} \frac{1}{t} d t \\ = [\log ∣ t]_{- e}^{e^{2}} \end{aligned}$
$\begin{aligned} = [\log [\log x]]_{e}^{e^{2}} \\ = \log (\log e^{2}) - \log (\log e) \\ = \log (2 \log e) - \log (1) \\ = \log 2 \end{aligned}$

Definite Integrals exercise Very short answer type question 29

Answer: 1
Hint: you must know the rule of integration
Given: $\int_{0}^{\frac{π}{2}} e^{x} (\sin x - \cos x) d x$
Solution: $\int_{0}^{\frac{π}{2}} e^{x} (\sin x - \cos x) d x$
$= - \int_{0}^{\frac{π}{2}} e^{x} (\cos x - \sin x) d x$
$\begin{aligned} f (x) = \cos x \\ f^{'} (x) = - \sin x \\ \int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + c \end{aligned}$
$\begin{aligned} we get, I = - {[e^{x} \cos x]}_{0}^{\frac{π}{2}} \\ = - e^{\frac{π}{2}} \cos \frac{π}{2} + e^{0} \cos (0) \\ = 0 + (1) (1) \\ = 1 \end{aligned}$

Definite Integrals exercise Very short answer type question 30

Answer: $\frac{1}{2} \log (\frac{17}{5})$
Hint: you must know the rule of integration
Given: $\int_{2}^{4} \frac{x}{x^{2} + 1} d x$
Solution: Put

x^{2} + 1 = t

\begin{aligned} 2 x d x = d t \\ x d x = \frac{d t}{2} \end{aligned}

\begin{aligned} I = \frac{1}{2} \int_{2}^{4} \frac{d t}{t} \\ = \frac{1}{2} [\log | t |]_{2}^{4} \end{aligned}

\begin{aligned} = \frac{1}{2} {[\log | x^{2} + 1 |]}_{2}^{4} \\ = \frac{1}{2} [\log (16 + 1) + \log (4 + 1)] \\ = \frac{1}{2} \log (\frac{17}{5}) \end{aligned}

Definite Integrals exercise Very short answer type question 31

Answer: 2
Hint: you must know the rule of integration
Given: $If \int_{0}^{1} (3 x^{2} + 2 x + k) d x = 0 find k$
Solution: $\int_{0}^{1} (3 x^{2} + 2 x + k) d x = 0$
$\begin{aligned} {[\frac{3 x^{3}}{3} + \frac{2 x^{2}}{2} + k x]}_{0}^{1} = 0 \\ {[x^{3} + x^{2} + k x]}_{0}^{1} = 0 \end{aligned}$
$\begin{aligned} [1 + 1 + k - 0] = 0 \\ k = - 2 \end{aligned}$

Definite Integrals exercise Very short answer type question 32

Answer: $a = 2$
Hint: you must know the rule of integration
Given: $If \int_{0}^{a} 3 x^{2} d x = 8, find a$
Solution: $\int_{0}^{a} 3 x^{2} d x = 8$

\begin{aligned} {[\frac{3 x^{3}}{3}]}_{0}^{a} = 8 \\ {[x^{3}]}_{0}^{a} = 8 \end{aligned}

\begin{aligned} a^{3} - 0 = 8 \\ a = \sqrt[3]{8} \\ a = 2 \end{aligned}

Definite Integrals exercise Very short answer type question 33

Answer: $f^{'} (x) = x \sin x$
Hint: you must know the rule of integration
Given: $f (x) = \int_{0}^{x} t \sin t d t$
Solution: $f (x) = \int_{0}^{x} t \sin t d t$
$= {[t \int \sin t d t - \int {\frac{d t}{d t} \int \sin t d t} d t]}_{0}^{x}$
$\begin{aligned} = {[t (- \cos t) + \int \cos t d t]}_{0}^{x} \\ = [- t \cos t + \sin t]_{0}^{x} \end{aligned}$
$= - x \cos x + \sin x - (- 0 \cos 0 + \sin 0)$
$\begin{aligned} f (x) = \sin x - x \cos x \\ f^{'} (x) = \cos x - (x (- \sin x) + \cos x) \end{aligned}$
$\begin{aligned} = \cos x + x \sin x - \cos x \\ = x \sin x \end{aligned}$

Definite Integrals exercise Very short answer type question 34

Answer: $a = 2$
Hint: you must know the rule of integration
Given: $- \int_{0}^{a} \frac{1}{4 + x^{2}} d x = \frac{π}{8}, find a$
Solution:
$\begin{aligned} \int_{0}^{a} \frac{1}{4 + x^{2}} d x = \frac{π}{8} \\ \int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \tan^{- 1} \frac{x}{a} \end{aligned}$
$\begin{aligned} ∴ \int_{0}^{a} \frac{1}{2^{2} + x^{2}} d x = \frac{π}{8} \\ \Rightarrow \frac{1}{2} \tan^{- 1} \frac{a}{2} - 0 = \frac{π}{8} \end{aligned}$
$\begin{aligned} \Rightarrow \tan^{- 1} \frac{a}{2} = \frac{π}{4} \\ \Rightarrow \frac{a}{2} = \tan^{- 1} \frac{π}{4} \\ \Rightarrow \frac{a}{2} = 1 \end{aligned}$
$\Rightarrow a = 2$

Definite Integrals exercise Very short answer type question 35

Answer:
Hint: Separate the terms of x and y and then integrate them.
Given:
Solution:
Integrating both sides

Definite Integrals exercise Very short answer type question 36

Answer: $\frac{18}{\log_{e} 3}$
Hint: you must know the rule of integration
Given: $I = \int_{2}^{3} 3^{x} d x$
Solution:
Let

y = 3^{x}

..............(i)
Taking logarithm on both sides

\begin{aligned} \log y = \log 3^{x} \\ \log y = x \log 3 \\ y = e^{x \log 3} \end{aligned}

........(ii)
From eqn (i) and (ii) we can write

y = 3^{x} = e^{x \log 3}

This means instead of integrating

3^{x}

we will integrate

e^{x \log 3}

\begin{aligned} I = \int_{2}^{3} 3^{x} d x \\ = \int_{2}^{3} e^{x \log 3} d x \end{aligned}

\begin{aligned} = {[\frac{e^{x \log 3}}{\log 3}]}_{2}^{3} \\ = \frac{1}{\log 3} [e^{3 \log 3} - e^{2 \log 3}] \end{aligned}

\begin{aligned} = \frac{1}{\log 3} [e^{\log 3^{3}} - e^{\log 3^{2}}] \\ = \frac{1}{\log 3} [3^{3} - 3^{2}] \end{aligned}

\begin{aligned} = \frac{(27 - 9)}{\log 3} \\ = \frac{18}{\log 3} \end{aligned}

Definite Integrals exercise Very short answer type question 37

Answer: 1
Hint: you must know the rule of integration
Given: $\int_{0}^{2} [x] d x$
Solution: $\int_{0}^{2} [x] d x$
$\begin{aligned} = \int_{0}^{1} 0 d x + \int_{1}^{2} 1 d x \\ = 0 + [x]_{1}^{2} \\ = 2 - 1 \\ = 1 \end{aligned}$

Definite Integrals exercise Very short answer type question 38

Answer: $\frac{1}{2}$
Hint: you must know the rule of integration
Given: $\int_{0}^{1.5} [x] d x$
Solution: $\int_{0}^{1.5} [x] d x$
$\begin{aligned} = \int_{0}^{1} 0 d x + \int_{1}^{1.5} 1 d x \\ = [x]_{1}^{1.5} \end{aligned}$
$\begin{aligned} = [1.5 - 1] \\ = 0.5 \\ = \frac{1}{2} \end{aligned}$

Definite Integrals exercise Very short answer type question 39

Answer: $\frac{1}{2}$
Hint: you must know the rule of integration
Given: $\int_{0}^{1} {x} d x$
Solution: $\int_{0}^{1} {x} d x$
$\begin{aligned} = \int_{0}^{1} (x - [x]) d x \\ = \int_{0}^{1} x d x - \int_{0}^{1} [x] d x \end{aligned}$
$\begin{aligned} = & {[\frac{x^{2}}{2}]}_{0}^{1} - \int_{0}^{1} 0 d x \\ = & [\frac{1}{2} - 0] - 0 \end{aligned}$
$= \frac{1}{2}$

Definite Integrals exercise Very short answer type question 40

Answer: $e - 1$
Hint: you must know the rule of integration
Given: $\int_{0}^{1} e^{{x}} d x$
Solution: $\int_{0}^{1} e^{{x}} d x$
Integration of exponential function is same as the function

∴ {[e^{{x}}]}_{0}^{1} [∵ \int_{a}^{b} e^{x} d x = {[e^{x}]}_{a}^{b}]

\begin{aligned} = e^{1} - e^{0} \\ = e - 1 \end{aligned}

Definite Integrals exercise Very short answer type question 41
Answer: $\frac{3}{2}$

Hint: you must know the rule of integration
Given: $\int_{0}^{2} x [x] d x$
Solution: $\int_{0}^{2} x [x] d x$
$= \int_{0}^{1} x [0] d x + \int_{1}^{2} x (1) d x$
$\begin{aligned} = {[0 + \frac{x^{2}}{2}]}_{1}^{2} \\ = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \end{aligned}$

Definite Integrals exercise Very short answer type question 42

Answer: $\frac{1}{\log_{e} 2}$
Hint: you must know the rule of integration
Given: $I = \int_{0}^{1} 2^{x - [x]} d x$
Solution:
We know the values of greatest integer function on

0 < x < 1, [x] = 0

\begin{aligned} I = \int_{0}^{1} 2^{x - 0} d x \\ = \int_{0}^{1} 2^{x} d x \\ = {[\frac{2^{x}}{\log 2}]}_{0}^{1} \end{aligned}

\begin{aligned} = [\frac{2^{1}}{\log 2} - \frac{2^{0}}{\log 2}] \\ = \frac{(2 - 1)}{\log 2} \\ = \frac{1}{\log 2} \end{aligned}

Definite Integrals exercise Very short answer type question 43

Answer: 0
Hint: you must know the rule of integration
Given: $\int_{1}^{2} \log_{e} [x] d x$
Solution: $\int_{1}^{2} \log_{e} [x] d x$
Using rule of greatest integer

\begin{aligned} \int_{0}^{1} 0 d x + \int_{1}^{2} 1 d x + \int_{2}^{3} 2 d x \\ I = & \int_{1}^{2} \log [x] d x = \int_{0}^{1} \log (0) d x + \int_{1}^{2} \log 1 d x \\ = & 0 \end{aligned}

Definite Integrals exercise Very short answer type question 44

Answer: $\sqrt{2} - 1$
Hint: The values of greatest integer function when

0 < x < 1, [x] = 0

And when

1 < x < 2, [x] = 1

Given: $\int_{0}^{\sqrt{2}} [x^{2}] d x$
Solution: $\int_{0}^{\sqrt{2}} [x^{2}] d x$
$= \int_{0}^{1} [x^{2}] d x + \int_{1}^{\sqrt{2}} [x^{2}] d x$ [ using the value of greatest integer function ]

\begin{aligned} = 0 + \int_{1}^{\sqrt{2}} 1 d x \\ = [x]_{1}^{\sqrt{2}} \\ = \sqrt{2} - 1 \end{aligned}

Note: Final answer is not matching with the book.

Definite Integrals exercise Very short answer type question 45

Answer: $\frac{\sqrt{2} - 1}{\sqrt{2}}$
Hint: you must know the rule of integration
Given: $\int_{0}^{\frac{π}{4}} \sin {x} d x$
Solution: $\int_{0}^{\frac{π}{4}} \sin {x} d x$
$\begin{aligned} = - [\cos {x}]_{0}^{\frac{π}{4}} \\ = - [\cos \frac{π}{4} - \cos 0] \end{aligned}$
$\begin{aligned} = - [\frac{1}{\sqrt{2}} - 1] \\ = \frac{\sqrt{2} - 1}{\sqrt{2}} \end{aligned}$

The RD Sharma class 12 solutions Definite Integrals VSA can be an excellent free study material for students who will appear for their board exams. Chapter 19 of the NCERT maths book deals with Definite Integral, Evaluating Definite Integrals, Definite Integral & Riemann integral Formulas, etc. Exercise VSA has 45 questions that require short and simple answers. The RD Sharma class 12th exercise VSA will guide you while you attempt to solve these questions.

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RD Sharma Chapter wise Solutions

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RD Sharma Solutions Class 12 Mathematics Chapter 19 VSA

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