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

Last updated on Jun 14, 2025

Application of Integrals MCQs are important for assessing one's understanding of the practical uses of integration in various fields. Integration enables the calculation of areas, volumes, and accumulated quantities. Application of Integrals MCQs evaluate learners' knowledge of integration techniques, area under curves, volume of solids, and application-based problems. By answering such MCQs, individuals can enhance their comprehension of integration applications in physics, engineering, economics, and other disciplines. These Application of Integrals MCQs play a crucial role in strengthening individuals' grasp of integration concepts and their practical implementation.

Latest Application of Integrals MCQ Objective Questions

Application of Integrals Question 1:

Comprehension:

Consider the following for the two (02) items that follow:
The slope of the tangent to the curve y = f(x) at (x, f(x) ) is 4 for every real number x and the curve passes through the origin.

. What is the area bounded by the curve, the x-axis and the line x = 4?

  1. 8 square units 
  2. 16 square units 
  3. 32 square units 
  4. 64 square units 

Answer (Detailed Solution Below)

Option 3 : 32 square units 

Application of Integrals Question 1 Detailed Solution

Calculation: 

qImage6847eba9a36458e0b0f25d21

 

Given,

The equation of the curve is y = 4x , and the line x = 4 intersects the curve at the point (4, 16) . We need to find the area bounded by the curve, the x-axis, and the line x = 4 .

The region of interest is a right triangle with a base along the x-axis from x = 0 to x = 4  and a height of 16 units, corresponding to the point (4, 16) .

The area of the triangle is given by the formula:

Area=12×base×height

Substituting the values of the base (4 units) and the height (16 units):

Area=12×4×16=32square units

∴ The area is 32 square units.

Hence, the correct answer is option 3.

Application of Integrals Question 2:

Comprehension:

Consider the following for the two (02) items that follow:
The slope of the tangent to the curve y = f(x) at (x, f(x) ) is 4 for every real number x and the curve passes through the origin.

What is the nature of the curve?

  1. A straight line passing through (1,4)
  2. A straight line passing through (-14)
  3. A parabola with vertex at origin and focus at (2,0)
  4. A parabola with vertex at origin and focus at (1, 0)

Answer (Detailed Solution Below)

Option 1 : A straight line passing through (1,4)

Application of Integrals Question 2 Detailed Solution

Calculation:

Given,

The slope of the tangent to the curve y = f(x) at (x, f(x)) is 4 for every real number x , and the curve passes through the origin.

The slope of the tangent is the derivative of the function, so we have:

f(x)=4

Integrating f'(x) = 4  with respect to  x :

f(x)=4x+C

The curve passes through the origin, so when x = 0 , y = 0 . Substituting these values into the equation f(x) = 4x + C :

0=4(0)+CC=0

Therefore, the equation of the curve is:

f(x)=4x

This is the equation of a straight line with a slope of 4, passing through the origin.

 The curve is a straight line with a slope of 4, passing through the origin.

Hence, the correct answer is option 1.

Application of Integrals Question 3:

Find the area of the region bounded by the curves y = x22, the line x = 2, x  = 0 and the x - axis ?

  1. 83sq.units
  2. 13sq.units
  3. 23sq.units
  4. 43sq.units
  5. None of the above

Answer (Detailed Solution Below)

Option 4 : 43sq.units

Application of Integrals Question 3 Detailed Solution

Concept:

The area under the curve y = f(x) between x = a and x = b,is given by,  Area = x=ax=bf(x)dx

xndx=xn+1n+1+C

 

Calculation:

Here, we have to find the area of the region bounded by the curves y = x22, the line x = 2, x  = 0 and the x - axis

So, the area enclosed by the given curves = 02x22dx

As we know that, xndx=xn+1n+1+C

=02x22dx=[x36]02

=16(80)=43sq.units

Hence, option 4 is the correct answer.

Application of Integrals Question 4:

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

  1. 2 sq units
  2. 1 sq units
  3. 4 sq units
  4. 3 sq units
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : 2 sq units

Application of Integrals Question 4 Detailed Solution

Calculation

Area = 0π|cosx|dx

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

So, we split the integral:

Area = 0π2cosxdx+π2π(cosx)dx

Area = [sinx]0π2[sinx]π2π

Area = (sinπ2sin0)(sinπsinπ2)

Area = (10)(01)

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

Application of Integrals Question 5:

The following plot shows a function y which varies linearly with x. The value of the integral I = 13y2dx is

sbi po 1 1.26

  1. 27
  2.  32.67
  3. 35
  4. -32.67
  5. None of the above

Answer (Detailed Solution Below)

Option 2 :  32.67

Application of Integrals Question 5 Detailed Solution

Calculation:

Find the equation of line first intercept of y is 2

equation becomes:

y = x + 2

If intercept of y is n

Equation becomes

y = x + n

I = 13y2dx

I = 13(x+n)2dx

I = 13[x2+n2+2xn]dx

n = 2

I = 13(x2+4+4x)dx

I = [x23+4x+4x23]13

I=[333+12+181342]I=3313

I=983 = 32.67 units

Top Application of Integrals MCQ Objective Questions

What is the area of the parabola x2 = y bounded by the line y = 1?

  1. 13 square unit
  2. 23 square unit
  3. 43 square units
  4. 2 square units

Answer (Detailed Solution Below)

Option 3 : 43 square units

Application of Integrals Question 6 Detailed Solution

Download Solution PDF

Concept:

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

Area = abydx

F7 5f3573a3f346800d0e2814b3 Aman.K 20-08-2020 Savita Dia

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

Area = abxdy

Calculation:

Here, 

x2 = y  and line y = 1 cut the parabola

∴ x2 = 1

⇒ x = 1 and -1

F5 5f3574b68881b70d100bb46f Aman.K 20-8-2020 Savita Dia

Area =11ydx

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,

Area1=01ydx

Area1=01x2dx

=[x33]01=13

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.

(1×1)13=23

TotalArea=2×23=43 square units.

The area of the region bounded by the curve y = 16x2 and x-axis is 

  1. 8π sq.units
  2. 20π sq. units 
  3. 16π sq. units
  4. None of these

Answer (Detailed Solution Below)

Option 1 : 8π sq.units

Application of Integrals Question 7 Detailed Solution

Download Solution PDF

Concept: 

a2x2dx=x2x2a2+a22sin1xa+c 

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

Calculation:

Given: 

y = 16x2 and x-axis

At x-axis, y will be zero

y = 16x2

⇒ 0 = 16x2

⇒ 16 - x2 = 0

⇒ x2 = 16

∴ x = ± 4

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

F6 Aman 15-1-2021 Swati D1

Since the curve is y = 16x2

So, y ≥ o [always]

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

Area of the curve, A =4416x2dx

We know that,

a2x2dx=x2x2a2+a22sin1xa+c

[x2(42x2)+162sin1x4]44 

[x2(4242)+162sin144][x2(42(4)2)+162sin144)]

= 8 sin-1 (1) + 8 sin-1 (1)

= 16 sin-1 (1)

= 16 × π/2

= 8π sq units

The area enclosed between the curves y = sin x, y = cos x, 0 ≤ x ≤ π/2 is

  1. 21
  2. 2+1
  3. 2(21)
  4. 2(2+1)

Answer (Detailed Solution Below)

Option 3 : 2(21)

Application of Integrals Question 8 Detailed Solution

Download Solution PDF

Calculation:

F1 Tapesh 25.2.21 Pallavi D 3

Enclosed Area

=20π/4(cosxsinx)dx

=2[sinx+cosx]0π/4

=2[(12+12)(0+1)]

=2(21)

The area bounded by the parabola x = 4 - y2 and y-axis, in square units, is

  1. 232 Sq. unit
  2. 323 Sq. unit
  3. 332 Sq. unit
  4. None of these

Answer (Detailed Solution Below)

Option 2 : 323 Sq. unit

Application of Integrals Question 9 Detailed Solution

Download Solution PDF

Concept:

Area under a Curve by Integration

F1 A.K 12.5.20 Pallavi D3

Find the area under this curve by summing vertically.

  • In this case, we find the area is the sum of the rectangles, height x = f(y) and width dy.
  • If we are given y = f(x), then we need to re-express this as x = f(y) and we need to sum from the bottom to top.


So, A=abxdy=abf(y)dy

Calculation:

Given Curve: x = 4 - y2

⇒ y2 = 4 - x
⇒ y2 = - (x - 4)           

The above curve is the equation of the Parabola,

We know that at y-axis; x = 0

⇒ y2 = 4 - x

⇒ y2 = 4 - 0 = 4

⇒ y = ± 2

 (0, 2) or (0, -2) are Point of intersection.

F1 SachinM Madhuri 01.03.2022 D2

Area under the curve =22xdy

=22(4y2)dy

=[4yy33]22

=323 Sq. unit

The area of a circle of radius ‘a’ can be found by following integral

  1. ab(a2+x2)dx
  2. 02π(a2x2)dx
  3. 4×0a(a2x2)dx
  4. 0a(a2x2)dx

Answer (Detailed Solution Below)

Option 3 : 4×0a(a2x2)dx

Application of Integrals Question 10 Detailed Solution

Download Solution PDF

Explanation:

F1 Ateeb 19.3.21 Pallavi D12

Equation of circle is given by x2 + y2 = a2

Let's take the strip along a y-direction and integrate it from 0 to 'a' this will give the area of the first quadrant and in order to find out the area of a circle multiply by 4

y=x2a2

Area of first Quadrant = 0aydx = 0aa2x2dx

Area of circle = 4 × 0aa2x2dx

The area bound by the parabolas y = 3x2 and x- y + 4 = 0 is:

  1. 162
  2. 1633
  3. 163
  4. 1632

Answer (Detailed Solution Below)

Option 4 : 1632

Application of Integrals Question 11 Detailed Solution

Download Solution PDF

Given:

The parabolas y = 3x2 and x- y + 4 = 0

Concept:

Apply concept of area between two curves y1 and y2 between x = a and x = b

A=ab(y1y2) dx

Calculation:

The parabolas y = 3x2 and x- y + 4 = 0

then 3x2 = x2 + 4

⇒ x2 = 2

⇒ x = ± √ 2

Then the area is 

A=22(x2+43x2) dx

A=22(42x2) dx

A=[4x2x33]22

A=4[2(2)]23[23(2)3]

A=82832

A=1623 sq unit.

Hence option (4) is correct.

Find the area of the curve y = 4x3 between the end points x = [-2, 3]

  1. 97
  2. 65
  3. 70
  4. 77

Answer (Detailed Solution Below)

Option 1 : 97

Application of Integrals Question 12 Detailed Solution

Download Solution PDF

Concept:

The area of the curve y = f(x) is given by:

A = x1x2f(x)dx

where x1 and x2 are the endpoints between which the area is required.

Imp. Note: The net area will be the addition of the area below the x-axis and the area above the x-axis.

Calculation:

The f(x) = y = 4x3

Given the end points x1 = -2, x2 = 3

Area of the curve (A) = |234x3dx|

⇒ A = |204x3dx|+|034x3dx|

⇒ A = |4[x44]20|+|4[x44]03|

⇒ A = |[024]|+|[340]|

⇒ A = |16|+|81|

⇒ A = 97

Additional Information

Integral property:

  • ∫ xn dx = xn+1n+1+ C ; n ≠ -1
  • 1xdx=lnx + C
  • ∫ edx = ex+ C
  • ∫ adx = (ax/ln a) + C ; a > 0,  a ≠ 1
  • ∫ sin x dx = - cos x + C
  • ∫ cos x dx = sin x + C 

The area of the region bounded by the curve y = x2 and the line y = 16 is

  1. 32/3
  2. 256/3
  3. 64/3
  4. 128/3

Answer (Detailed Solution Below)

Option 2 : 256/3

Application of Integrals Question 13 Detailed Solution

Download Solution PDF

Explanation:

Given equation of curves are

y = x2    ---(1) and y = 16    ---(2)

By solving both equation (1) and (2) we have:

x2 = 16

x = 4, -4.

∴ Points of intersection are (4, 16) and (-4, 16).

F1 Shraddha Shubham 18.12.2020 D1

From the figure we have,

Required Area = 44(16x2) dx

By using Integral property we have,

 A= 204(16x2) dx

=2[16xx33]04

=2[16xx33]04

=2[64643]

=2×64×23

 A=2563 sq.units

Alternate Method 

There is another method also by which we can solved the problem,

By considering horizontal strip and by the condition of symmetry we have:

Area = 2016x dy

Area = 2016y dy

Area = 2 × 23 × [y32]016

Area = 2×23×[16320]

Area = 2563 sq.unit

The area under the curve y = x2 and the lines x = -1, x = 2 and x-axis is:

  1. 3 sq. units.
  2. 5 sq. units.
  3. 7 sq. units.
  4. 9 sq. units.

Answer (Detailed Solution Below)

Option 1 : 3 sq. units.

Application of Integrals Question 14 Detailed Solution

Download Solution PDF

Concept:

The area under a Curve by Integration:

F1 Aman.K 10-07-2020 Savita D1

Find the area under this curve is by summing horizontally.

In this case, we find the area is the sum of the rectangles, heights y = f(x) and width dx.

We need to sum from left to right.

∴ Area =  abydx=abf(x)dx

 

Calculation: 

Here, we have to find the area of the region bounded by the curves y = x2, x-axis and ordinates x = -1 and x = 2

F1 Aman.K 14-12-20 Savita D2

So, the area enclosed by the given curves is given by 12x2dx

As we know that, xndx=xn+1n+1+C

Area = 12x2dx

[x33]12

[8313]=93=3

Area = 3 sq. units.

The area enclosed by the curves y = x - 1 and y2 = 2x + 6 is:

  1. 21
  2. 24
  3. 18
  4. 20

Answer (Detailed Solution Below)

Option 3 : 18

Application of Integrals Question 15 Detailed Solution

Download Solution PDF

Explanation:

Given curves are y = x - 1 and y2 = 2x + 6 

F1 Vinanti Defence 31.12.22 D1

On solving, we get, 

y2 = 2(y + 1) + 6

⇒ y2 - 2y - 8 = 0

⇒ (y - 4)(y + 2) = 0

⇒ y = -2, 4

Now, we can find the area by

A = 24[y+1(y223)]dy

24(4+yy22)dy

[4y+y22y36]24

16+8323(8+2+43)

∴ A = 18

Get Free Access Now
Hot Links: teen patti master 2023 teen patti master apk teen patti gold online teen patti rules teen patti star login