Definite Integrals MCQ Quiz - Objective Question with Answer for Definite Integrals - Download Free PDF

Concept:

\(\rm \int {1\over {1 + x^2}} dx = tan ^ {-1} x\)

Calculation:

\(\rm \displaystyle\int_0^1 {1\over {1 + x^2}}dx = [tan ^{-1}x]_{0}^{1} = [tan ^{-1}1 - tan^{-1} 0] = {\pi\over 4}\)

Concept:

Integral properties: Consider a function f(x) defined on x.

• \(\mathop \smallint \limits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \left\{ {\begin{array}{*{20}{c}} {2\mathop \smallint \limits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}},\;\;\;f\left( {\rm{x}} \right) = f\left( { - x} \right)}\\ {0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;f\left( {\rm{x}} \right) = - f\left( { - x} \right)} \end{array}} \right.\)

Calculation:

Let f(x) = sin x – tan x

Checking the function is odd or even,

f(-x) = sin (-x) – tan (-x)

⇒ f(-x) = – sin x + tan x

⇒ f(-x) = –{sin x – tan x}

⇒ f(-x) = –f(x)

Hence, the function is odd.

And we know that, \(\mathop \smallint \limits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = 0\)  if f(x) is odd.

∴ \(\mathop \smallint \nolimits_{ - {\rm{\pi }}/4}^{{\rm{\pi }}/4} \left( {\sin {\rm{x}} - \tan {\rm{x}}} \right){\rm{dx}} = 0\)

\(I=\displaystyle\int_0^\pi\sqrt{1+4\sin^2(x/2)-4\sin(x/2)}dx\)

\(I=\displaystyle\int_0^\pi|2\sin(x/2)-1|dx\)

If x = 0, then \(2\sin (x/2)-1<0\)

And if x = \(\dfrac{\pi}{3}\), then \(2\sin (x/2)-1>0\)

\(\therefore I=\displaystyle\int_0^{\pi/3}-(2\sin(x/2)-1)dx+\displaystyle\int_{\pi/3}^{\pi}(2\sin(x/2)-1)dx\)

\(I=[4cos(x/2)+x]_0^{\pi/3}+[-4\cos (x/2)-x]_{\pi/3}^\pi\)

\(I=\dfrac{4\sqrt3}{2}+\dfrac{\pi}{3}-4+\Big(0-\pi+\dfrac{4\sqrt3}{2}+\dfrac{\pi}{3}\Big)\)

\(I=4\sqrt3-4-\dfrac{\pi}{3}\)

Calculation:

Given, I = \(\rm \int_{-1}^2\frac{|x|}{x}dx\)

= \(\rm \int_{-1}^0\frac{-x}{x}dx\) + \(\rm \int_{0}^2\frac{x}{x}dx\)

= \(\rm \int_{-1}^0(-1)dx\) + \(\rm \int_{0}^2(1) dx\)

= \(\rm -[x]_{-1}^0\) + \(\rm [x]_{0}^2\)

= - [0 - (- 1)] + [2 - 0]

= - 1 + 2

= 1

∴ The value of the integral is 1.

The correct answer is Option 2.

Calcualtion:

Let I = \(\rm \int_{\pi/6}^{\pi/4}cosec \ 2xdx\)

= \(\rm \frac{1}{2}\left[\log \tan x\right]_{\pi/6}^{\pi/4}\)

= \(\rm \frac{1}{2}\left[\log \tan \frac{\pi}{4}-\log \tan \frac{\pi}{6}\right]\)

= \(\rm \frac{1}{2}\left[\log 1-\log \frac{1}{\sqrt{3}}\right]\)

= \(\rm \frac{1}{2}\log \sqrt{3}\)

∴ The value of the integral is \(\rm \frac{1}{2}\log \sqrt{3}\).

The correct answer is Option 4.

Concept:

Definite Integral properties:

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{\;dx}}\)
Calculation:

Let f(x) = x(1 – x)⁹

Now using property, \(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{\;dx}}\)

\(\mathop \smallint \nolimits_0^1 {\rm{x}}{(1 - {\rm{x}})^9}{\rm{dx}} = \mathop \smallint \limits_0^1 \left( {1 - {\rm{x}}} \right){\left\{ {1 - \left( {1 - {\rm{x}}} \right)} \right\}^9}{\rm{dx}}\)

\(⇒ \mathop \smallint \limits_0^1 \left( {1 - {\rm{x}}} \right){{\rm{x}}^9}{\rm{dx}}\)

\(⇒ \mathop \smallint \limits_0^1 \left( {{{\rm{x}}^9} - {{\rm{x}}^{10}}} \right){\rm{dx}}\)

\(⇒\left[ {\frac{{{{\rm{x}}^{10}}}}{{10}} - \frac{{{{\rm{x}}^{11}}}}{{11}}} \right]_0^1\)

⇒ 1/10 – 1/11

⇒ 1/110

∴ The value of integral \(\mathop \smallint \nolimits_0^1 {\rm{x}}{(1 - {\rm{x}})^9}{\rm{dx}}\) is 1/110.

Concept:

\(\rm \int \frac{dx}{x^2+a^2} = \frac{1}{a}\; \tan^{-1}(\frac{x}{a}) + c\)

Calculation:

Let I = \(\displaystyle\int_0^2 \rm \dfrac{dx}{x^2+4}\)

= \(\displaystyle\int_0^2 \rm \dfrac{dx}{x^2+2^2}\)

\(= \rm [\frac{1}{2}\; \tan^{-1}(\frac{x}{2})] _{0}^{2}\)

\(= \rm [\frac{1}{2}\; \tan^{-1}1 -\frac{1}{2}\; \tan^{-1}0 ]\)

\(= \rm \frac{1}{2}\times \frac{\pi}{4} - 0\)

= \(\rm \dfrac{\pi}{8}\)

Concept:

\(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{dx}}\)

Calculation: 

Let I = \(\rm \int _{0}^{2π} \frac{\sin 2x}{a -b\cos x}dx\)         ----(1)

Using property f(a + b – x),

I = \(\rm \int _{0}^{2π} \frac{\sin 2(2π -x)}{a -b\cos (2π -x) }dx\)   

As we know,  sin (2π - x) = - sin x and cos (2π - x) = cos x

I = \(\rm \int _{0}^{2π} \frac{-\sin 2x}{a -b\cos x}dx\)         ----(2)       

I = -I

2I = 0

∴ I = 0

Concept:

\(\rm \int x^n dx = \frac{x^{n+1}}{n+1}+c\)

Calculation: 

I = \(\rm \int_{1}^{\infty} \frac{4}{x^4}dx\)

= \(\rm \int_{1}^{\infty}4{x^{-4}}dx\)

= \(\rm \left[\frac{4x^{-3}}{-3} \right ]_1^{\infty}\)

= \(\rm \frac{-4}{3}\left[\frac{1}{x^3} \right ]_1^{\infty}\)

= \(\rm \frac{-4}{3}\left[\frac{1}{\infty} - \frac{1}{1}\right ]\)

= \(\rm \frac{-4}{3}[0-1]\)

= \(\frac 4 3\)

Concept:

\(\rm \displaystyle\int_{a}^{b} f(x) dx = \displaystyle\int_{a}^{b} f(a+b-x) dx\)

Calculations:

Consider, I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx\)             ....(1)

I = \(\rm \displaystyle\int_{0}^{\pi/2} \dfrac{\sqrt{\sin (\dfrac{\pi}{2}-x)}}{\sqrt{\sin (\dfrac{\pi}{2}-x)}+ \sqrt{\cos (\dfrac{\pi}{2}-x)}}dx\)

I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos x}}{\sqrt{\cos x}+ \sqrt{\sin x}}dx\)                           ....(2)

Adding (1) and (2), we have

2I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos x}+ \sqrt{sinx}}{\sqrt{\cos x}+ \sqrt{\sin x}}dx\)

2I = \(\rm \displaystyle\int_0^{\pi/2}dx\)

2I = \(\rm[x]^\frac{\pi}{2}_0\)

I = \(\dfrac{\pi}{4}\)

Concept:

\(\rm \dfrac{d(\ln x)}{dx} = \dfrac{1}{x}\)

Calculation:

Let I = \(\rm \int_{1}^{e}\frac{dx}{x\sqrt{2+\ln x}}\)

Let (2 + ln x) = t²

Differentiating with respect to x, we get

⇒ (0 + \(\rm \frac 1 x\))dx = 2tdt

⇒ \(\rm \frac 1 x\)dx = 2tdt

Now,

\(\rm I=\rm \int_{\sqrt 2}^{\sqrt 3}\frac{2tdt}{\sqrt{t^2}}\\=2\int_{\sqrt 2}^{\sqrt 3}\frac{tdt}{t}\\=2\int_{\sqrt 2}^{\sqrt 3}dt\\=2[t]_{\sqrt 2}^{\sqrt 3}\\=2(\sqrt 3- \sqrt 2)\)

Concept:

\(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{dx}}\)

Calculation: 

Let I = \(\rm \int _0^{\pi} \sin^6 x \cos^5 x\;dx\)             .... (1)

Using property f(a + b – x),

I = \(\rm \int _0^{\pi} \sin^6 (\pi -x) \cos^5 (\pi -x)\;dx\)

As we know,  sin (π - x) = sin x and cos (π - x) = -cos x

I = -\(\rm \int _0^{\pi} \sin^6 x \cos^5 x\;dx\)                .... (2)

I = -I

2I = 0

∴ I = 0

Concept:

f(x) = |x| will be equal to 

• x, if x > 0
• -x, if x < 0
• 0, if x = 0

∫ dx = x + C  (C is a constant)

∫ xⁿ dx = xⁿ⁺¹/n+1 + C

Calculation:

Let, \(I = \int^{-1}_{-2}\frac{x}{|x|}dx\)

Using the above concept, as x ∈ (-2, -1)

⇒ \(I=∫^{-1}_{-2}\frac{x}{-x}dx\)

⇒ \(I=-1∫^{-1}_{-2}(1)dx\)

⇒ \(I=-[x]^{-1}_{-2}\)  

⇒ I = -[-1 - (-2)] 

∴  \(\int^{-1}_{-2}\frac{x}{|x|}dx\) = -1

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C

Calculation:

I = \(\rm\int {2x+1\over\left({x^2+x+1}\right)^2}dx\)

Let x2 + x + 1 = t

⇒ (2x + 1) dx = dt

I = \(\rm \int {dt\over t^2}\)

I = \(\rm {t^{-1}\over{-1}}\)

I = \(\rm -1\over t\)

I = \(\rm -1\over x^2+x+1\)

Putting limits

I = \(\rm \left[-1\over x^2+x+1\right]_{-1}^1\)

I = \(\rm {-1\over 1^2+1+1} - \left({-1\over (-1)^2+(-1)+1}\right)\)

I = \(\rm 1-{1\over3}\) = \(\boldsymbol{2\over3}\)

Concept:

\(\mathop \smallint \nolimits_{\rm{a}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = 0\)

Calculation:

We know, 

\(\mathop \smallint \nolimits_{\rm{a}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = 0\)

Here, limit of integration is same (i.e., π/4)

\(\therefore \mathop \smallint \nolimits_{\frac{{\rm{\pi }}}{4}}^{\frac{{\rm{\pi }}}{4}} \frac{{\cos \left( {{{\rm{e}}^{3{\rm{x}}}}} \right)}}{{{{\rm{x}}^4} + {{\rm{x}}^3} + 1}} = 0\)

Hence, option (3) is correct.