Integral Calculus MCQ Quiz - Objective Question with Answer for Integral Calculus - Download Free PDF

Concept:

The area under the curve y = f(x) between x = a and x = b,is given by,  Area = \(\rm \int_{x=a}^{x =b}f(x) \;dx\)

\(\smallint {{\rm{x}}^{\rm{n}}}{\rm{dx}} = \frac{{{{\rm{x}}^{{\rm{n}} + 1}}}}{{{\rm{n}} + 1}} + {\rm{C}}\)

Calculation:

Here, we have to find the area of the region bounded by the curves y = \(\rm \frac{x^2}{2}\), the line x = 2, x  = 0 and the x - axis

So, the area enclosed by the given curves = \(\rm \int_0^2 {\rm \frac{x^2}{2}}\;dx\)

As we know that, \(\smallint {{\rm{x}}^{\rm{n}}}{\rm{dx}} = \frac{{{{\rm{x}}^{{\rm{n}} + 1}}}}{{{\rm{n}} + 1}} + {\rm{C}}\)

\(=\rm \int_0^2 {\frac {x^2}{2}}\;dx = \left[ {\frac{{{x^3}}}{6}} \right]_0^2\)

\(\rm = \frac{1}{6}\;\left( {8- 0\;} \right) = \frac 4 3 \;sq.\;units\)

Hence, option 4 is the correct answer.

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.

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

Calculation:

Let I = \(\rm \int\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} d x\)

= \(\rm \int e^{x+\frac{1}{x}} d x+ \int x\left(1-\frac{1}{x^2}\right) e^{x+\frac{1}{x}} d x\)

= \(\rm \int e^{x+\frac{1}{x}} d x+ \int x\left(1-\frac{1}{x^2}\right) e^{x+\frac{1}{x}} d x\)

= \(\rm \int e^{x+\frac{1}{x}} d x+xe^{x+\frac{1}{x}} d x-\int\frac{d}{dx}(x)e^{x+\frac{1}{x}} d x\)

= \(\rm \int e^{x+\frac{1}{x}} d x+xe^{x+\frac{1}{x}} d x-\int e^{x+\frac{1}{x}} d x\) [∵ \(\rm \int \left(1-\frac{1}{x^2}\right) e^{x+\frac{1}{x}} d x = e^{x+\frac{1}{x}}\)]

= \(\rm x e^{x+\frac{1}{x}}+c \)

∴ The value of the integral is \(\rm x e^{x+\frac{1}{x}}+c\).

The correct answer is Option 2.

Calculation

Area = \(\int_{0}^{\pi} |\cos x| dx\)

cos x is positive from 0 to π/2 and negative from π/2 to π.

So, we split the integral:

Area = \(\int_{0}^{\frac{\pi}{2}} \cos x dx + \int_{\frac{\pi}{2}}^{\pi} (-\cos x) dx\)

Area = \(\left[ \sin x \right]_{0}^{\frac{\pi}{2}} - \left[ \sin x \right]_{\frac{\pi}{2}}^{\pi}\)

Area = \(\left( \sin \frac{\pi}{2} - \sin 0 \right) - \left( \sin \pi - \sin \frac{\pi}{2} \right)\)

Area = \((1 - 0) - (0 - 1)\)

Area = \(1 - (-1)\)

Area = \(1 + 1\)

Area = \(2\)

∴ The area bounded by the curve y = cos x, x = 0 and x = π is 2.

Hence option 1 is correct

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:

The area under the curve y = f(x) between x = a and x = b, is given by:

Area = \(\rm\int_{a}^{b}ydx\)

Similarly, the area under the curve y = f(x) between y = a and y = b, is given by:

Area = \(\rm\int_{a}^{b}xdy\)

Calculation:

Here, 

x² = y  and line y = 1 cut the parabola

∴ x² = 1

⇒ x = 1 and -1

\( \text{Area =}\int_{-1}^{1} y d x \)

Here, the area is symmetric about the y-axis, we can find the area on one side and then multiply it by 2, we will get the area,

\( \text{Area}_1 = \int_{0}^{1} y d x \)

\( \text{Area}_1 = \int_{0}^{1} x^{2} dx \)

\(= \rm\left[\frac{x^{3}}{3}\right]_{0}^{1}=\frac{1}{3}\)

This area is between y = x² and the positive x-axis.

To get the area of the shaded region, we have to subtract this area from the area of square i.e.

\((1 \times 1)-\frac{1}{3}=\frac{2}{3}\)

\(Total\;Area=2\times{\frac{2}{3}} =\frac{4}{3}\) square units.

Concept:

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

Calculation: 

I = \(\rm \int_0^1 x \sqrt{x^2 + 4}\;dx\)

Let x² + 4 = t

Differentiating with respect to x, we get

⇒ 2xdx = dt

⇒ xdx = \(\rm \frac {dt}{2}\)

Now,

I = \(\rm \frac{1}{2}\int_4^5 \sqrt{t}\;dt\)

= \(\rm \frac{1}{2} \left[\frac{t^{3/2}}{\frac{3}{2}} \right ]_4^5\)

= \(\rm \frac{1}{3} \left[5^{3/2} - 4^{3/2} \right ]\)

= \(\rm \frac{1}{3}[5\sqrt 5 - 8]\)

Concept:

1 + cos 2x = 2cos² x

1 - cos 2x = 2sin² x

\(\rm \int \cos x\;dx = \sin x + c\)

Calculation:

I = \(\rm \int cos^2 x\;dx\)

= \(\rm \int \frac{1+\cos 2x}{2}\;dx\)

= \(\rm \frac{1}{2}\int (1+\cos 2x)\;dx\)

= \(\rm \frac{1}{2} \left[x+\frac{\sin 2x}{2} \right ] + c\)

= \(\rm \frac{x}{2}+\frac{\sin 2x}{4} + c\)

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: 

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

Function y = √f(x) is defined for f(x) ≥ 0. Therefore y can not be negative.

Calculation:

Given: 

y = \(\rm √{16-x^2}\) and x-axis

At x-axis, y will be zero

y = \(\rm √{16-x^2}\)

⇒ 0 = \(\rm √{16-x^2}\)

⇒ 16 - x² = 0

⇒ x2 = 16

∴ x = ± 4

So, the intersection points are (4, 0) and (−4, 0)

Since the curve is y = \(\rm √{16-x^2}\)

So, y ≥ o [always]

So, we will take the circular part which is above the x-axis

Area of the curve, A \(\rm =\int_{-4}^{4}√{16-x^2}\;dx\)

We know that,

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

= \( \rm [ \frac x 2 √{(4^2- x^2) }+ \frac {16}{2}sin^{-1} \frac x4]_{-4}^{4 }\) 

= \( \rm [ \frac x 2 √{(4^2- 4^2) }+ \frac {16}{2}sin^{-1} \frac 44]- \rm [ \frac x 2 √{(4^2- (-4)^2) }+ \frac {16}{2}sin^{-1} \frac {4}{-4})]\)

= 8 sin^-1 (1) + 8 sin-1 (1)

= 16 sin^-1 (1)

= 16 × π/2

= 8π sq units

Concept:

\(\rm \int \frac{1}{\sqrt{a^2-x^2}}dx= \sin^{-1 } \left(\frac{x}{a} \right ) + c\)

Calculation:

I = \(\rm \int \frac {1}{\sqrt{16-25x^2}}dx\)

= \(\rm \int \frac {1}{\sqrt{16-(5x)^2}}dx\)

Let 5x = t

Differentiating with respect to x, we get

⇒ 5dx = dt

⇒ dx = \(\rm \frac {dt}{5}\)

Now,

I = \(\rm \frac {1}{5}\int \frac {1}{\sqrt{4^2-t^2}} dt\)

= \(\rm \frac 1 5 \sin^{-1} \left(\frac t 4 \right)\) + c

= \(\rm \frac 1 5 \sin^{-1} \left(\frac {5x} {4} \right)\) + c