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

Concept:

Definite Integrals:

\(\rm \int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx\).

If f(x) = f(2a - x), then

\(\rm \int_0^{2a}f(x)\ dx=2\int_0^af(x)\ dx\).

A function f(x) is:

• Even, if f(-x) = f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx\).
• Odd, if f(-x) = -f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=0\).
• Periodic, if f(np ± x) = f(x), for some number p and n ∈ Z.

Calculation:

\(\rm\int_0^a[f(x)+f(-x)].dx=\int_{-a}^ag(x).dx \)

\(\rm =\int^0_{-a}g(x).dx+\int_0^ag(x).dx \)

\(\rm =\int_0^ag(-x).dx+\int_0^ag(x).dx \)

\(\rm \int_0^a[g(x)+g(-x)].dx \)

∴ f(x) = g(x)

Concept: 

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

Calculation: 

⇒ I =\( \int_{a}^{b}\)x f(x)dx ------------(1)

⇒ I = \( \int_{a}^{b}\)(a +b -x) f(a + b - x)dx---(2)

adding equation 1+2

⇒ 2I=\( \int_{a}^{b}\)x f(x)dx +\( \int_{a}^{b}\)(a + b -x)f(a + b - x)dx

given f(a + b -x)=f(x)

⇒ 2I = \( \int_{a}^{b}\)(b + a)f(x)dx

⇒ I = \({(b+a)\over2 } \int_{a}^{b}\)f(x)dx

Hence option 1 is correct

Calculation:

Given, I = \(\int_0^{\pi / 4}\) x2 cos2x dx

⇒ I = x²\(\int_0^{\pi / 4}\cos2x \ dx\) - \(\int_0^{\pi / 4}\left(\frac{d}{dx}x^2\int_0^{\pi/4}\cos2x \ dx\right)dx\)

⇒ I = \(\frac{x^2\sin2x}{2}\) - \(\int_0^{\pi / 4}x\sin2x \ dx\)

⇒ I = \(\frac{x^2\sin2x}{2}\) - (x\(\int_0^{\pi / 4}\sin2x \ dx\) - \(\int_0^{\pi / 4}\int_0^{\pi/4}\sin2x \ dx \ dx\))

⇒ I = \(\frac{x^2\sin2x}{2}\) - (\(-\frac{x\cos2x}{2}\) + \(\frac{\sin2x}{4}\))​​

⇒ I = \(\left[\frac{x^2\sin2x}{2}+\frac{x\cos2x}{2}-\frac{\sin2x}{4}\right]_0^{\pi/4}\)

= \(\left[\frac{\pi^2}{32}-\frac{1}{4}\right]-0\)

= \(\frac{\pi^2}{32}\) - \(\frac{1}{4}\)

∴ The value of integral is \(\frac{\pi^2}{32}\) - \(\frac{1}{4}\).

The correct answer is Option 2.

Explanation:

Let f(t) = \(ln (t+\sqrt{1+t^2}\)

f(-t) = \(ln (\sqrt{1+t^2} -t\)

= \(ln( \frac{1}{t+ \sqrt{1+t^2}})\)

= \(-ln(t+\sqrt{1+t^2})\) = -f(t)

∴  f(t) is an odd function

Now g(t) = tan f(t)

Then, g(–t) = tan f(–t)

= –tan (f(t)) = –g(t)

So g(t) is also an odd function

When  g(t) is an odd function, then

\(\rm \int_{-\pi}^\pi g(t)dt\) = 0

∴ Option (b) is correct

Explanation:

Let f(t) = \(ln (t+\sqrt{1+t^2}\)

f(-t) = \(ln (\sqrt{1+t^2} -t\)

= \(ln( \frac{1}{t+ \sqrt{1+t^2}})\)

= \(-ln(t+\sqrt{1+t^2})\) = -f(t)

∴  f(t) is an odd function

Now g(t) = tan f(t)

Then, g(–t) = tan f(–t)

= –tan (f(t)) = –g(t)

So g(t) is also an odd function

Hence, both statements I and II are correct.

∴ Option (c) is correct

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:

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

CALCULATION:

Here, we have to find the value of the integral \(\mathop \smallint \nolimits_0^{\frac{\pi }{2}} \frac{{\sqrt {tanx} }}{{1 + \sqrt {tanx} }}\)

Let \(I = \mathop \smallint \nolimits_0^{\frac{\pi }{2}} \frac{{\sqrt {tanx} }}{{1 + \sqrt {tanx} }}dx\)---------(1)

As we know that, \(\mathop \smallint \nolimits_a^b f\left( x \right)\;dx = \;\mathop \smallint \nolimits_a^b f\left( {a + b - x} \right)\;dx\)

⇒ \(I = \mathop \smallint \nolimits_0^{\frac{\pi }{2}} \frac{{\sqrt {cotx} }}{{1 + \sqrt {cotx} }} dx\)----------(2)

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

⇒ \(2I = \;\mathop \smallint \nolimits_0^{\frac{\pi }{2}\;} \left[ {\frac{{\sqrt {\tan x} }}{{1 + \sqrt {\tan x} }} + \frac{{\sqrt {\cot x} }}{{1 + \sqrt {\cot x} }}} \right] dx\)

⇒ \(2I = \;\mathop \smallint \limits_0^{\frac{\pi }{2}} dx = \frac{\pi }{2}\;\)

⇒ \(I = \frac{\pi}{4}\)

Hence, option B is the correct answer.

Concept:

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

Calculation:

Consider I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\)      ----(i)

⇒ I =  \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 (π/2 -x)}}{\sqrt{\sin^8 (π/2 -x)}+ \sqrt{\cos^8 (π/2 -x)}}dx\)

⇒ I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\)      ----(ii)

Add (i) and (ii), we get

⇒ 2I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos^8 x}+ \sqrt{sin^8x}}{\sqrt{\cos^8 x}+ \sqrt{\sin^8 x}}dx\)

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

⇒ 2I = \(\rm[x]^\frac{\pi}{2}_0\)

⇒ I = \(\dfrac{\pi}{4}\)

Concept:

Definite Integral:

If ∫ f(x) dx = g(x) + C, then \(\rm \int_a^b f(x)\ dx = \left[ g(x)\right]_a^b\) = g(b) - g(a).

Properties:

• For even functions: f(-x) = f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx\).
• For odd functions: f(-x) = -f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=0\).
• \(\rm \int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx\).
• If f(x) = f(2a - x), then \(\rm \int_0^{2a}f(x)\ dx=2\int_0^af(x)\ dx\).

Calculation:

It can be observed that sin (-θ) = -sin θ.

Let f(x) = |sin x|

Put x = -x

⇒ f(-x) = |sin -x| = |-sin x| = |sin x| = f(x)

∴ |sin x| is an even function.

Therefore, using the properties of definite integrals, we get:

\(\rm \int_{-\pi/2}^{\pi/2}|\sin x|\ dx\)

= 2\(\rm \int_{0}^{\pi/2}\sin x\ dx\)

= 2\(\rm \left[-\cos x\right]_{0}^{\pi/2}\)

= -2\(\rm \left[\cos \pi/2-\cos 0\right]\)

= -2[0 - 1]

= 2.

Concept:

Integral property:

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

Calculation:

I = \(\rm \int_{-2}^2\) (sin x + cos x)dx​

​Let f₁(x) = sinx and f₂(x) = cosx

f₁(x) = sinx

f₁(-x) = sin(-x) = -sinx = -f₁(x)

f₂(x) = cosx

f₂(-x) = cox(-x) = cosx = f₂(x)

By integration property f₁(x) = 0

I  = \(\rm 2\mathop \smallint \nolimits_0^2 cosx\;dx\)

= 2 

= 2[sin 2 - sin 0]

= 2sin 2

Concept:

If f(x)  is odd function then f(-x) = -f(x)

Properties of definite integral

If f(x)  is even function then \(\mathop \smallint \nolimits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = 2\mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}}\)

If f(x)  is odd function then \(\mathop \smallint \nolimits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} =0\)

Calculation:

Let I = \(\displaystyle\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \rm \dfrac{sin^5 \ x \ cos^3 \ x}{x^4}dx\)

Let f(x) = \(\rm \dfrac{sin^5 \ x \ cos^3 \ x}{x^4}\)

Replaced x by -x, 

⇒ f(-x) = \(\rm \dfrac{sin^5 \ (-x) \ cos^3 \ (-x)}{(-x)^4}\)

As we know sin (-θ) = - sin θ and cos (-θ) = cos θ

= \(\rm \dfrac{-sin^5 \ x \ cos^3 \ x}{x^4}\)

⇒ f(-x) = -f(x)      

So, f(x) is odd function

Therefore, I = 0     

Concept:

Definite Integral properties:

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

Calculation :

Let , I = \(\rm\int_{0}^{π/2}\) log cot x dx             ....(i)

Now using property, \(\mathop \smallint \limits_{\rm{0}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{0}}^{\rm{a}} {\rm{f}}\left( {{\rm{a}} - {\rm{x}}} \right){\rm{\;dx}}\)

I = \(\rm\int_{0}^{π/2}\) log cot ( \(\rm \frac{\pi}{2}\) - x ) dx 

I = \(\rm\int_{0}^{π/2}\) log tan x dx                    .... (ii)

Adding eq. (i) and (ii), we get 

⇒ 2I = \(\rm\int_{0}^{π/2}\) (log cot x + log tan x ) dx 

⇒ 2I = \(\rm\int_{0}^{π/2}\) log ( tan x × cot x ) dx              [∵ log m + log n = log mn]

⇒ 2I = \(\rm\int_{0}^{π/2}\) log 1 dx 

⇒ 2I = 0                                                      [ ∵ log 1 = 0 ]   

⇒  I = 0 

The correct option is 1. 

Concept:

Property of definite integrals: \(\mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {{\rm{a}} - {\rm{x}}} \right){\rm{dx}}\)

Calculation:

I = \(\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{1}{{1 + \tan {\rm{x}}}}{\rm{dx}}\)                            …. (1)

Using the property of definite integrals: 

I = \(\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{1}{{1 + \tan \left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right)}}{\rm{dx}}\)

I \(= \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{1}{{1 + \cot {\rm{x}}}}{\rm{dx}}\)                          ..... (2)

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

2I = \(\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{1}{{1 + \tan {\rm{x}}}}{\rm{dx}} + \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{1}{{1 + \cot {\rm{x}}}}{\rm{dx}}\)

\(=\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \left[ {\frac{1}{{1 + \tan {\rm{x}}}} + \frac{1}{{1 + \cot {\rm{x}}}}} \right]{\rm{dx}} \)

\(= \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \left[ {\frac{1}{{1 + \tan {\rm{x}}}} + \frac{{\tan {\rm{x}}}}{{1 + \tan {\rm{x}}}}} \right]{\rm{dx}} \)

\(= \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \left[ {\frac{{1 + \tan {\rm{x}}}}{{1 + \tan {\rm{x}}}}} \right]{\rm{dx}}\)

\( = \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} 1{\rm{dx}}\)

\( = {\rm{\;}}\left[ x \right]_0^{\frac{{\rm{\pi }}}{2}} \)

\(= \frac{{\rm{\pi }}}{2}\)

∴ I \(= \frac{{\rm{\pi }}}{4}\)

Concept:

Properties of definite integral:

\(\rm \int\limits_a^b f(x) dx\) = \(\rm \int\limits_a^b f(b+a -x) dx\)

Calculation:
I = \(\rm \int\limits_2^8 {\frac{{\sqrt {10 - x} }}{{\sqrt x + \sqrt {10 - x} }}} dx\)       ...(i)

I = \(\rm\int\limits_2^8 {\frac{{\sqrt {10 - (8+2-x)} }}{{\sqrt{8+2- x} + \sqrt {10 - (8+2-x)} }}} dx\)

I = \(\rm\int\limits_2^8 {\frac{{\sqrt { x} }}{{\sqrt {10 - x} +\sqrt x }}} dx\)    ...(ii)

Adding (i) and (ii), we get

2I = \(\rm\int\limits_2^8 {\frac{{\sqrt {10 - x}+\sqrt x }}{{\sqrt x + \sqrt {10 - x} }}} dx\)

2I = \(\rm\int\limits_2^8 dx\)

2I = \(\rm\left[x\right]_2^8\)

2I = 8 - 2 = 6

I = 3

Concept:

Properties of definite integral:

\(\rm \int\limits_a^b f(x) dx\) = \(\rm \int\limits_a^b f(b+a -x) dx\)

Calculation:
I = \(\rm \int\limits_0^5 {\frac{{\sqrt {5- x} }}{{\sqrt x + \sqrt {5- x} }}} dx\)       ...(i)

I = \(\rm\int\limits_0^5 {\frac{{\sqrt {5 - (5+0-x)} }}{{\sqrt{5+0- x} + \sqrt {5- (5+0-x)} }}} dx\)

I = \(\rm\int\limits_0^5 {\frac{{\sqrt { x} }}{{\sqrt {5 - x} +\sqrt x }}} dx\)    ...(ii)

Adding (i) and (ii), we get

2I = \(\rm\int\limits_0^5 {\frac{{\sqrt {5- x}+\sqrt x }}{{\sqrt x + \sqrt {5- x} }}} dx\)

2I = \(\rm\int\limits_0^5dx\)

2I = \(\rm\left[x\right]_0^5\)

2I = 5 - 0 = 5

I = 2.5