Application of Integrals MCQ with Solutions

Integrals: Chapter 8 Links

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8.1 | Misc.

Application of Integrals MCQ with answers and solutions from NCERT Class 12 Maths. Revise all 4 important questions step by step with clear explanations. You can also watch the video solutions for better understanding. This is in continuation of the previous topic on Integrals MCQs.

Ready? Okay, let’s begin with the first question!

Application of Integrals MCQ#1 (NCERT Exercise 8.1 – Question 3)

📍Area lying in the first quadrant and bounded by the circle $x^{2}+y^{2}=4$ and the lines $x=0$ and $x=2$ is

Options:

(A) $\pi$
(B) $\dfrac{\pi}{2}$
(C) $\dfrac{\pi}{3}$
(D) $\dfrac{\pi}{4}$

Answer:
✅ Correct option: (A)

Explanation:
From $x^{2}+y^{2}=4$ , we get
$y=\sqrt{4-x^{2}}$

Required area in first quadrant:
$\displaystyle \int_{0}^{2}\sqrt{4-x^{2}}\,dx$

This represents one-fourth of a circle of radius 2.

Area of full circle $=\pi r^{2}=\pi(2)^{2}=4\pi$

Hence required area $=\dfrac{1}{4}\times 4\pi=\pi$ .

Alternative Method (Using Integration Formula):

From $x^{2}+y^{2}=4$ , we get
$y=\sqrt{4-x^{2}}$

Required area:
$\displaystyle \int_{0}^{2}\sqrt{4-x^{2}}\,dx$

We use the standard result:
$\displaystyle \int \sqrt{a^{2}-x^{2}}\,dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\left(\frac{x}{a}\right)+C$

Here $a=2$ .

So,
$\displaystyle \int_{0}^{2}\sqrt{4-x^{2}}\,dx=\left[\frac{x}{2}\sqrt{4-x^{2}}+2\sin^{-1}\left(\frac{x}{2}\right)\right]_{0}^{2}$

Now substitute limits:

At $x=2$ :
$\frac{2}{2}\sqrt{4-4}+2\sin^{-1}(1)=0+2\cdot \frac{\pi}{2}=\pi$

At $x=0$ :
$0+2\sin^{-1}(0)=0$

Hence,
$\displaystyle \int_{0}^{2}\sqrt{4-x^{2}}\,dx=\pi$ .

Therefore, required area is $\pi$ .

Watch Video Solution on YouTube

Let me tell you all of these 4 Application of Integrals MCQ are very simple and easy to understand and therefore scoring too. But you must practice them regularly.

So, my suggestion to you is to solve these NCERT MCQs a good number of times and feel confident. And if you ever need any clarification, just drop a comment — I’ll be more than happy to help.

Application of Integrals MCQ#2 (NCERT Exercise 8.1 – Question 4)

📍Area of the region bounded by the curve $y^{2}=4x$ , y-axis and the line $y=3$ is

Options:

(A) $2$
(B) $\dfrac{9}{4}$
(C) $\dfrac{9}{3}$
(D) $\dfrac{9}{2}$

Answer:
✅ Correct option: (B)

Explanation:
Given $y^{2}=4x$

Express x in terms of y:
$x=\dfrac{y^{2}}{4}$

Area bounded by curve and y-axis from $y=0$ to $y=3$ is
$\displaystyle \int_{0}^{3}\dfrac{y^{2}}{4}\,dy$

$=\dfrac{1}{4}\int_{0}^{3}y^{2}dy$

i.e. Area $=\dfrac{1}{4}\left[\dfrac{y^{3}}{3}\right]_{0}^{3}$

$=\dfrac{1}{4}\times \dfrac{27}{3}=\dfrac{9}{4}$

Hence, required area is $\dfrac{9}{4}$ .

Watch Video Solution on YouTube

Did you know, there are about 100 MCQs in NCERT textbooks for Class 12 (both Part 1 and Part 2)? I shall be taking one Chapter at a time and show you the explanations for each of the questions.

Moreover, I have also covered these solutions in the video form and they are freely available on my YouTube channel, @Mathsbetter. And here also, in each of the 4 Application of Integrals MCQ on this page, I have provided the direct links to each and every MCQ for you to understand the solutions in a very simple and easy manner.

Application of Integrals MCQ#3 (NCERT Misc. Exercise Chapter 8 – Question 4)

📍Area bounded by the curve $y=x^{3}$ , the x-axis and the ordinates $x=-2$ and $x=1$ is

Options:

(A) $-9$
(B) $-\dfrac{15}{4}$
(C) $\dfrac{15}{4}$
(D) $\dfrac{17}{4}$

Answer:
✅ Correct option: (D)

Explanation:
Since area is required, we take modulus where curve lies below x-axis.

$y=x^{3}$ is negative on $[-2,0]$ and positive on $[0,1]$ .

Required area:
$=\int_{-2}^{0}(-x^{3})dx+\int_{0}^{1}x^{3}dx$

$=-\left[\dfrac{x^{4}}{4}\right]_{-2}^{0}+\left[\dfrac{x^{4}}{4}\right]_{0}^{1}$

And after putting the limits, A $=-\left(0-\dfrac{16}{4}\right)+\dfrac{1}{4}$

$=4+\dfrac{1}{4}=\dfrac{17}{4}$

Hence, required area is $\dfrac{17}{4}$ .

Watch Video Solution on YouTube

If you have been following this website as a resource for some of the important questions. For example, how to Integrate Square Root of tan x, or may be a complete guide to look at using Matrix Method in solving a system of linear equations.

Then trust me, similarly, this series on NCERT Class 12 MCQs is going to help you in many ways.

Moving onto the next Application of Integrals MCQ in this part of the series.

Application of Integrals MCQ#4 (NCERT Misc. Exercise Chapter 8 – Question 5)

📍The area bounded by the curve $y=x|x|$ , x-axis and the ordinates $x=-1$ and $x=1$ is

Options:

(A) $0$
(B) $\dfrac{1}{3}$
(C) $\dfrac{2}{3}$
(D) $\dfrac{4}{3}$

Answer:
✅ Correct option: (C)

Explanation:
Given hint:
$y=x^{2} \text{ if } x>0$
$y=-x^{2} \text{ if } x<0$

Area is symmetric about y-axis.

Required area:

$=\int_{-1}^{0}(-x^{2})dx+\int_{0}^{1}x^{2}dx$

$=-\left[\dfrac{x^{3}}{3}\right]_{-1}^{0}+\left[\dfrac{x^{3}}{3}\right]_{0}^{1}$

And after putting the limits, we get, Area $=-\left(0-\left(-\dfrac{1}{3}\right)\right)+\dfrac{1}{3}$

$=\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}$

Hence, required area is $\dfrac{2}{3}$ .

Watch Video Solution on YouTube

Quick Answer Key

Q1 → (A)
Q2 → (B)
Q3 → (D)
Q4 → (C)

Closing Note

That completes all the important NCERT Class 12 Maths Application of Integrals MCQ from chapter 8.
I hope the explanations and video solutions made things clearer and gave you more confidence. Stay tuned for the next chapter, where we’ll cover all the important Differential Equations MCQ (Chapter 9 from NCERT) with the same clarity and video support.

👉 Make sure to practice these questions again to strengthen your concepts.
👉 Watch the video solutions for a clearer and faster revision.

Keep practicing, and you’ll master the application of integrals easily! 🚀

Integrals: Chapter 8 Links

Application of Integrals MCQ#1 (NCERT Exercise 8.1 – Question 3)

Application of Integrals MCQ#2 (NCERT Exercise 8.1 – Question 4)

Application of Integrals MCQ#3 (NCERT Misc. Exercise Chapter 8 – Question 4)

Application of Integrals MCQ#4 (NCERT Misc. Exercise Chapter 8 – Question 5)

Quick Answer Key

Closing Note

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