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

Last updated on Apr 22, 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:

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 ?

  1. \(\rm ​\frac 8 3 \;sq.\;units\)
  2. \(\rm ​\frac 1 3 \;sq.\;units\)
  3. \(\rm ​\frac 2 3 \;sq.\;units\)
  4. \(\rm ​\frac 4 3 \;sq.\;units\)
  5. None of the above

Answer (Detailed Solution Below)

Option 4 : \(\rm ​\frac 4 3 \;sq.\;units\)

Application of Integrals Question 1 Detailed Solution

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.

Application of Integrals Question 2:

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 2 Detailed Solution

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

Application of Integrals Question 3:

The following plot shows a function y which varies linearly with x. The value of the integral I = \(\mathop \smallint \limits_1^3 {y^2}dx\) 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 3 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 = \(\mathop \smallint \limits_1^3 {y^2}dx\)

I = \(\mathop \smallint \limits_1^3 {\left( {x + n} \right)^2}dx\)

I = \(\mathop \smallint \limits_1^3 \left[ {{{\rm{x}}^2} + {{\rm{n}}^2} + 2{\rm{xn}}} \right]{\rm{dx}}\)

n = 2

I = \(\mathop \smallint \limits_1^3 \left( {{{\rm{x}}^2} + 4 + 4{\rm{x}}} \right){\rm{dx}}\)

I = \(\left[ {\frac{{{{\rm{x}}^2}}}{3} + 4x + \frac{{4{{\rm{x}}^2}}}{3}} \right]_1^3\)

\(\begin{array}{l} {\rm{I}} = \left[ {\frac{{{3^3}}}{3} + 12 + 18 - \frac{1}{3} - 4 - 2} \right]\\ {\rm{I}} = 33 - \frac{1}{3} \end{array}\)

\({\rm{I}} = \frac{{98}}{3}\) = 32.67 units

Application of Integrals Question 4:

The area enclosed by the curves \(y = \sin x + \cos x\) and \(y = \left| \cos x - \sin x \right|\) over the interval \(\left[ 0, \frac{\pi}{2} \right]\) is

  1. \(4(\sqrt{2} - 1)\)
  2. \(2\sqrt{2}(\sqrt{2} - 1)\)
  3. \(2(\sqrt{2} + 1)\)
  4. \(2\sqrt{2}(\sqrt{2} + 1)\)
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : \(2\sqrt{2}(\sqrt{2} - 1)\)

Application of Integrals Question 4 Detailed Solution

Calculation

\(\cos x > \sin x, \forall x \in \left( 0, \frac{\pi}{4} \right)\) and \(\cos x < \sin x, \forall x \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right)\)

\(y_{1} = \sin x + \cos x\)

\(y_{2} = \left| \cos x - \sin x \right|\)

\(\Rightarrow \text{Area}\)

\(\int_{0}^{\pi/2} \left( y_{1} - y_{2} \right) dx\)

\(\int_{0}^{\pi/4} \left( (\sin x + \cos x) - (\cos x - \sin x) \right) dx + \int_{\pi/4}^{\pi/2} \left( (\sin x + \cos x) - (\sin x - \cos x) \right) dx\)

\(= 4 - 2\sqrt{2}\)

Hence option 2 is correct

Application of Integrals Question 5:

The area of the region bounded by the curve y2 = 4x and the line x = 3 is

  1. \(3\sqrt3\)
  2. \(3\sqrt8\)
  3. 8
  4. \(8\sqrt3\)
  5. 10

Answer (Detailed Solution Below)

Option 4 : \(8\sqrt3\)

Application of Integrals Question 5 Detailed Solution

Calculation

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Given:

Parabola: y² = 4x

Line: x = 3

Since y² = 4x, then y = \(2\sqrt{x}\) (in the first quadrant, y > 0)

Required Area = 2 × (Area of the region OCAO)

⇒ Area = \(2 \int_{0}^{3} y \, dx\)

⇒ Area = \(2 \int_{0}^{3} 2\sqrt{x} \, dx\)

⇒ Area = \(4 \int_{0}^{3} x^{\frac{1}{2}} \, dx\)

⇒ Area = \(4 \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_{0}^{3}\)

⇒ Area = \(4 \times \frac{2}{3} \left[ (3)^{\frac{3}{2}} - 0 \right]\)

⇒ Area = \(\frac{8}{3} (3\sqrt{3})\)

∴ Required area = \(8\sqrt{3}\) sq. units.

Hence option 2 is correct

Top Application of Integrals MCQ Objective Questions

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

  1. \(\frac 1 3\) square unit
  2. \(\frac 2 3\) square unit
  3. \(\frac 4 3\) square units
  4. 2 square units

Answer (Detailed Solution Below)

Option 3 : \(\frac 4 3\) square units

Application of Integrals Question 6 Detailed Solution

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Concept:

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

Area = \(\rm\int_{a}^{b}ydx\)

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 = \(\rm\int_{a}^{b}xdy\)

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

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

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.

The area of the region bounded by the curve y = \(\rm \sqrt{16-x^2}\) 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

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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)

F6 Aman 15-1-2021 Swati D1

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

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

  1. \(\sqrt 2 - 1\)
  2. \(\sqrt 2 + 1\)
  3. \(2(\sqrt 2 - 1)\)
  4. \(2(\sqrt 2 + 1)\)

Answer (Detailed Solution Below)

Option 3 : \(2(\sqrt 2 - 1)\)

Application of Integrals Question 8 Detailed Solution

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Calculation:

F1 Tapesh 25.2.21 Pallavi D 3

Enclosed Area

\( = 2\mathop \smallint \nolimits_0^{\pi /4} \left( {\cos x - \sin x} \right)dx\)

\( = 2\left[ {\sin x + \cos x} \right]_0^{\pi /4}\)

\( = 2\left[ {\left( {\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right) - \left( {0 + 1} \right)} \right]\)

\( = 2\left( {\sqrt 2 - 1} \right)\)

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

  1. \( \frac{2}{32}\) Sq. unit
  2. \( \frac{32}{3}\) Sq. unit
  3. \( \frac{33}{2}\) Sq. unit
  4. None of these

Answer (Detailed Solution Below)

Option 2 : \( \frac{32}{3}\) Sq. unit

Application of Integrals Question 9 Detailed Solution

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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, \({\bf{A}} = \mathop \smallint \nolimits_{\bf{a}}^{\bf{b}} {\bf{xdy}} = \mathop \smallint \nolimits_{\bf{a}}^{\bf{b}} {\bf{f}}\left( {\bf{y}} \right){\bf{dy}}\)

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 \( = \mathop \int \nolimits_{-2}^2 {\rm{xdy}}\)

\(= \rm \int_{-2}^2 (4-y^2)dy\)

\(\rm = \left[4y- {\frac{{{{ {{\rm{y}} } }^3}}}{3}} \right]_{-2}^2\)

\(= \frac{32}{3}\) Sq. unit

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

  1. \(\mathop \smallint \limits_a^b \left( {{a^2} + {x^2}} \right)dx\)
  2. \(\mathop \smallint \limits_0^{2\pi } \sqrt {\left( {{a^2} - {x^2}} \right)} \;dx\)
  3. \(4 \times \mathop \smallint \limits_0^a \sqrt {\left( {{a^2} - {x^2}} \right)} \;dx\)
  4. \(\mathop \smallint \limits_0^a \sqrt {\left( {{a^2} - {x^2}} \right)} \;dx\)

Answer (Detailed Solution Below)

Option 3 : \(4 \times \mathop \smallint \limits_0^a \sqrt {\left( {{a^2} - {x^2}} \right)} \;dx\)

Application of Integrals Question 10 Detailed Solution

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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

\(y = \sqrt {x^2 - a^2}\)

Area of first Quadrant = \(\mathop \smallint \limits_0^a y\;dx\) = \(\mathop \smallint \limits_0^a \sqrt {{a^2} - {x^2}} \;dx\)

Area of circle = 4 × \(\mathop \smallint \limits_0^a \sqrt {{a^2} - {x^2}} \;dx\)

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

  1. \(16 \sqrt{2}\)
  2. \(\frac{16}{3} \sqrt{3} \)
  3. \(\frac{16}{3}\)
  4. \(\frac{16}{3} \sqrt{2}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{16}{3} \sqrt{2}\)

Application of Integrals Question 11 Detailed Solution

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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

\(\rm A=\int_a^b(y_1-y_2)\ dx\)

Calculation:

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

then 3x2 = x2 + 4

⇒ x2 = 2

⇒ x = ± √ 2

Then the area is 

\(\rm A=\int_{-\sqrt2}^{\sqrt2}(x^2+4-3x^2) \ dx\)

\(\rm A=\int_{-\sqrt2}^{\sqrt2}(4-2x^2) \ dx\)

\(\rm A=[4x-2\frac{x^3}{3}]_{-\sqrt2}^{\sqrt2}\)

\(\rm A=4[\sqrt2-(-\sqrt2)]-\frac{2}{3}[{\sqrt2}^3-{(-\sqrt2)}^3]\)

\(\rm A=8\sqrt2-\frac{8}{3}\sqrt2\)

\(\rm A=\frac{16\sqrt2}{3}\) 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

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Concept:

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

A = \(\rm \int_{x_1}^{x_2}f(x) dx\)

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) = \(\rm \left|\int_{-2}^3 4x^3dx\right|\)

⇒ A = \(\rm \left|\int_{-2}^0 4x^3dx\right| + \left|\int_0^3 4x^3dx\right|\)

⇒ A = \(\rm \left|4\left[x^4\over4\right]_{-2}^0\right| + \left|4\left[x^4\over4\right]_0^3\right|\)

⇒ A = \(\rm \left|\left[0- 2^4\right]\right| + \left|\left[3^4 - 0\right]\right| \)

⇒ A = \(\rm \left|-16\right| + \left|81\right|\)

⇒ A = 97

Additional Information

Integral property:

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

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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,

\(Required~Area~=~∫_{-4}^4(16-x^2)~dx \)

By using Integral property we have,

 \(A=~2∫_{0}^4(16-x^2)~dx \)

\(= 2\left[ {16x - \frac{{{x^3}}}{3}} \right]_0^4\)

\(= 2\left[ {16x - \frac{{{x^3}}}{3}} \right]_0^4\)

\( = 2\left[ {64 - \frac{{64}}{3}} \right] \)

\(= 2 \times 64 \times \frac{2}{3}\;\)

 \(A=\frac{256}{3}~sq.units\)

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:

\(Area~=~2\int_0^{16}x~dy\)

\(Area~=~2\int_0^{16}\sqrt{y}~dy\)

\(Area~=~2~\times~\frac{2}{3}~\times~[y^{\frac{3}{2}}]_0^{16} \)

\(Area~=~2\times \frac{2}{3}\times [16^{\frac{3}{2}}-0]\)

Area = \(\frac{256}{3}~sq.unit\)

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

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

 

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 \(\rm \mathop \int \nolimits_{-1}^2{x^2}\;dx\)

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

Area = \(\rm \mathop \int \nolimits_{-1}^2{x^2}\;dx\)

\( \rm \left[ {\frac{{{x^3}}}{3}} \right]_{-1}^2\)

\(\left[\frac 83 - \frac {-1}{3}\right] = \frac 93=3\)

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

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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 = \(\int_{-2}^{4}\left [y+1-\left ( \frac{y^{2}}{2}-3 \right ) \right ]dy \)

\(\int_{-2}^{4}\left (4+y-\frac{y^{2}}{2} \right )dy \)

\(\left [ 4y+\frac{y^{2}}{2} -\frac{y^{3}}{6}\right ]_{-2}^{4}\)

\(16+8-\frac{32}{3}-\left ( -8+2+\frac{4}{3} \right )\)

∴ A = 18

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