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

Last updated on Mar 24, 2025

Latest Definite Integrals MCQ Objective Questions

Definite Integrals Question 1:

The value of the integral 0π2sinθcos5θdθ will be

  1. 2231
  2. -64
  3. 1231
  4. 64231
  5. 131

Answer (Detailed Solution Below)

Option 4 : 64231

Definite Integrals Question 1 Detailed Solution

Analysis:

Consider I=0π/2sinθcos5θdθ

Put sin θ = t

cos θ dθ = dt

If θ = 0 to θ=π2 then t = 0 to t = 1

Now, I=θ=0π/2sinθcosθ(1sin2θ)2dθ

I=t=01t(1t2)2dt

I=t=01t(1+t42t2)dt

I=(t3232+t1121122t7272)01

I=23+21147

I=64231

Definite Integrals Question 2:

The value of integral xcos2x is equal to

  1. x tan x
  2. log cos x
  3. x tan x + log (cos x)
  4. x tan x - log cos x
  5. log sin x

Answer (Detailed Solution Below)

Option 3 : x tan x + log (cos x)

Definite Integrals Question 2 Detailed Solution

Concept:

Trigonometric Ratio Fundamental Identities

secx=1cosx

Integration by parts

when u and v are functions of x.

u×v dx=uvdx[dudxvdx]dx

Integral of standard function 

sec2(ax+b)dx=tan(ax+b)

tanxdx=log|secx|=log cosx

The derivative of standard function

ddx(xn)=nxn1

ddx(x)=1

Calculation:

Given:

we have xcos2x

where u = x and v=1cos2x=sec2x

Integration by parts

when u and v are functions of x.

Integration by parts

when u and v are functions of x.

u×v dx=uvdx[dudxvdx]dx

x×sec2x dx=xsec2x dx[ddxxsec2x dx]dx

x×sec2x dx=xtanx dxtanxdx

∫x × sec2 x dx = x tan x - ∫ tan x dx

∫x × sec2 x dx = x tan x - (-log (cos x))

∫ x × sec2 x dx = x tan x + log (cos x)

Definite Integrals Question 3:

The length of the curve y=23x3/2 between x = 0 and x = 1 is

  1. 0.27
  2. 0.67
  3. 1
  4. 1.22
  5. 2

Answer (Detailed Solution Below)

Option 4 : 1.22

Definite Integrals Question 3 Detailed Solution

Concept:

Arc length or curve length is the distance between two points along the section of the curve. Determining the length of an irregular section of the arc is termed as rectification of the curve.

The length of the curve y = f(x) from x = a to x = b is given as:

l=x=ax=b1+(dydx)2dx

or,

If the curve is parametrized in the form x = f(t) and y = g(t) with the parameter t going from a to b then

l=t=at=b(dxdt)2+(dydt)2dt

Calculation:

y=23x3/2

dydx=23×32x12

(dydx)2=x

Now, the arc length(l) is

l=x=0x=11+xdx

=|(1+x)3232|01

=(23×232)(23)=1.21

Definite Integrals Question 4:

The integral 2dxxlogx

  1. diverges to ∞ 
  2. diverges to -∞ 
  3. Converges to 2
  4. Converges to -3
  5. Converges to 0

Answer (Detailed Solution Below)

Option 1 : diverges to ∞ 

Definite Integrals Question 4 Detailed Solution

Explanation:

The given integral is an improper integral of 1st kind.

I=2(1/x)logxdx

I=[log(logx)]2

I = log [log (∞)] – log [log (2)]

I =

Given integral is divergent and diverges to ∞

Additional Information

An improper integral of first kind is when integral limits have -∞ or +∞ or both.

An improper integral of second kind is when integral limits are finite but function is infinite at some value between those limits.

Definite Integrals Question 5:

The value of the Integral I = 0π/2x2sinxdx is

  1. (x + 2)/2
  2. 2/(π – 2) 
  3. π - 2
  4. π + 2
  5. π - 4

Answer (Detailed Solution Below)

Option 3 : π - 2

Definite Integrals Question 5 Detailed Solution

Concept:

Using Integration by parts using ILATE

f(x)g(x)dx=f(x)g(x)dx[(f(x).g(x)dx]dx

Calculation:

I=0π/2x2sinxdx

f(x)=x2andg(x)=sinx

I=[x2cosx(2xcosx)]0π/2

I=[x2cosx+2×(xsinxsinx)]0π/2

I=[x2cosx+2×(xsinx+cosx)]0π/2

I=[0+2(π2(1)+0)][0+2(0+1)]

I=π2

Top Definite Integrals MCQ Objective Questions

The value of the definite integral 1exln(x)dx is

  1. 49e3+29
  2. 29e349
  3. 29e3+49
  4. 49e329

Answer (Detailed Solution Below)

Option 3 : 29e3+49

Definite Integrals Question 6 Detailed Solution

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Concept:

We know that,

By Parts method

uvdx=uvdx[dudxvdx]dx

Where, u, v should follow the ILATE sequence.[I= Inverse, L= Logarithmic, A= Algebraic, T= Trigonometric, E= Exponential terms]

Calculation:

Given:

From the given Equation, 1exln(x)dx

u = ln(x), v = √x

Now,

uvdx=uvdx[dudxvdx]dx

1elnxxdx=lnx1exdx1e[dudxxdx]dx  

1elnxxdx=[ln(x)×x3232]1e[1x×x3232]dx

 1elnxxdx=[ln(x)×x32×2349×x32]1e

∴  1exln(x)dx  =29e3+49     

The value (round off to one decimal place) of  11xe|x|dx ______

Answer (Detailed Solution Below) 0

Definite Integrals Question 7 Detailed Solution

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Explanation

Given,

 Function f(x) = x e|x|

 Integral is -1 to 1.

If f(-x) = f(x) then the function is said to be even function

 If f(-x) = - f(x) then the function is said to be odd function.

 f(-x) = -x e|-x| = -x e|x| = - f(x)

 ∴ The given function is an odd function.

 For an odd function:

 aaxf(x)dx = 0

For a even function

 aaxf(x)dx = 2 × 0axf(x)dx

 Now, as the function is odd

 11xe|x|dx = 0

The value of I=11e|x|dx is equal to

  1. (e - 1)
  2. 2(e - 1)
  3. 3(e - 1)
  4. 2(1 - e)

Answer (Detailed Solution Below)

Option 2 : 2(e - 1)

Definite Integrals Question 8 Detailed Solution

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Explanation:

I=11e|x|dx.

I=10exdx+01exdx

I=[ex]10+[ex]01

I = [-e-0 + e1] + [e1 - e0]

I = -1 + e1 + e - 1

I = 2 (e - 1)

A parametric curve defined by x=cos(πu2),y=sin(πu2)in the range of 0 ≤ u ≤ 1 is rotated about the X – axis by 360 degrees. Area of the surface generated is

  1. π2
  2. π

Answer (Detailed Solution Below)

Option 3 : 2π

Definite Integrals Question 9 Detailed Solution

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Concept:

x=cos(πu2),y=sin(πu2)

x2+y2=cos2(πu2)+sin2(πu2)=1

So it represents an equation of circle in x-y plane.

Given 0 ≤ u ≤ 1

So, 0 ≤ x ≤ 1,  0 ≤ y ≤ 1

i.e., 0 ≤ θ ≤ π/2

30

So we get a quarter circle in x-y plane and by revolving it 360°, we get a hemisphere.

Area of hemi-sphere = 2π(1)2 = 2π

A parabola x = y2 with 0 ≤ x ≤ 1 is shown in the figure. The volume of the solid of rotation obtained by rotating the shaded area by 360° around the x-axis is

GATE ME 2019 Shift 1 Solution writing 8Qs images Vivek  D 1

  1. π4
  2. π2
  3. π

Answer (Detailed Solution Below)

Option 2 : π2

Definite Integrals Question 10 Detailed Solution

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Concept:

Volume of the solid of rotation obtained by rotating the shaded area by 360° around the x-axis is asked,

there is a direct relation for this;

 1=πy2dx…1)

Calculation:

Given area;

F1 V.S. N.J. 13.09.2019 D 1

Using (1); Volume of solid of rotation can be calculated by:

1=01πxdx

=π01xdx=π{x22}01

=π2{10}

1=π2units;

Key points:

In the given problem, the volume is generated by revolving the area by 360° about the x-axis.

But if rotation/revolution is about the y-axis, then the volume of solid of rotation is calculated by:

2=πx2dy …2)

So, always be careful about which axis rotation is asked.

Depending upon that, you should use either 1) or 2).

Let f be a real-valued function of a real variable defined as f(x) = x – [x], where [x] denotes the largest integer less than or equal to x. The value of 0.251.25f(x)dx is _______ (up to 2 decimal places).

Answer (Detailed Solution Below) 0.49 - 0.51

Definite Integrals Question 11 Detailed Solution

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f(x) = x – [x]

for 0.25 < x < 1, f(x) = x

for 1 < x < 1.25, f(x) = x – 1

0.251.25f(x)dx=0.251xdx+11.25(x1)dx

=[x22]0.251+[x22x]11.25

=12[1116]+[(253254)(121)]

=12[1516]+[253254+12]

=15+2540+1632=0.5

If for non-zero x, if af(x)+bf(1x)=1x25 where a ≠ b then 12f(x)dx is

  1. 1a2b2[a(ln225)+47b2]
  2. 1a2b2[a(2ln225)47b2]
  3. 1a2b2[a(2ln225)+47b2]
  4. 1a2b2[a(ln225)47b2]

Answer (Detailed Solution Below)

Option 1 : 1a2b2[a(ln225)+47b2]

Definite Integrals Question 12 Detailed Solution

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Given x as non- zero,

af(x)+bf(1x)=1x25       ---(1)

Consider x as 1/x

af(1x)+bf(11x)=x25

af(1x)+bf(x)=x25       ---(2)

Multiply equation 1 by a and 2 by b and subtract both

a2f(x)+abf(1x)abf(1x)b2f(x)=ax25abx+25b

a2f(x)b2f(x)=ax25abx+25b

f(x)=ax(a2b2)bxa2b225(ab)a2b2

12f(x)dx=1a2b2[a(ln225)+47b2]

Solve:

0π2f(x)f(x)+f(π2x)dx = ?

  1. π/2
  2. 1
  3. 0
  4. π/4

Answer (Detailed Solution Below)

Option 4 : π/4

Definite Integrals Question 13 Detailed Solution

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Concept:

abf(x)dx=f(a+bx)dx

Calculation:

⇒ Let, I = 0π2f(x)f(x)+f(π2x)dx    ----- equation(1)

⇒ I = 0π2f(π2x)f(π2x)+f(π2(π2x))dx

⇒ I = 0π2f(π2x)f(x)+f(π2x)dx     ---- equation(2)

On adding equation (1) and (2)

⇒ 2I = 0π2f(x)+f(π2x)f(x)+f(π2x)dx

⇒ 2I = 0π2dx

⇒ 2I = π2

⇒ I = π4

∴ The value of 0π2f(x)f(x)+f(π2x)dx is π4

If 02π|xsinx|dx=kπ, then the value of k is equal to ______.

Answer (Detailed Solution Below) 4

Definite Integrals Question 14 Detailed Solution

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Concept:

Following steps to solve the equation 02π|xsinx|dx=kπ,

  1. To remove the modulus
  2. To keep sin x positive in the interval 0 to π to 2π and to keep the sin x negative in the interval.This is because x in the above equation is always positive but the value sinx changes in the two mentioned intervals.

 

Explanation:

Solving the equation as per the steps,

02π|xsinx|dx=0πxsinxdx+(π2πxsinxdx)

=0πxsinxdxπ2πxsinxdx

Keeping u = x, du = dx, dv = sinxdx, so v = - cosx,

=0πudv=uv0πvdu

=0πxsinxdx=[xcosx]0π+0πcosxdx

=π+[sinx]0π

Now, repeating the same with π2πxsinxdx, we get -3

Hence, π - (-3π) = 4π

Therefore, k = 4

The value of the following definite integral is ______ (round off to three decimal places)

e1(xlnx)dx

Answer (Detailed Solution Below) 2.090 - 2.104

Definite Integrals Question 15 Detailed Solution

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Concept:

Use integration by parts for solving this problem.

Calculation:

1e(logx x)dx

={logx.x22}1e1e1x.x22dx 

={x2logx2}1e121exdx 

={x2logx2}1e{x24}1e 

={e2loge2log12}{e2414} 

=e22e24+14 

e2+14=2.097

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