Home » Integrals MCQ with Solutions

Integrals MCQ with Solutions

MCQs
Integrals MCQ

Integrals: Chapter 7 Links

Current page: Integrals MCQ with Solutions. Go to other exercises and content from this chapter :

7.1  |  7.2  |  7.3  |  7.4  |  7.5  |  7.6  |  7.7  |  7.8  |  7.9  |  7.10  |  Misc.

Integrals MCQ with answers and solutions from NCERT Class 12 Maths. Revise all 23 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 Application of Derivatives MCQs.

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

Integrals MCQ#1 (NCERT Exercise 7.1 – Question 21)

📍The anti-derivative of \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right) equals

Options:

  • (A) \dfrac{1}{3}x^{3/2}+2x^{1/2}+C
  • (B) \dfrac{2}{3}x^{3/2}+\dfrac{1}{2}x^{2}+C
  • (C) \dfrac{2}{3}x^{3/2}+2x^{1/2}+C
  • (D) \dfrac{3}{2}x^{3/2}+\dfrac{1}{2}x^{1/2}+C

Answer: ✅ (C)

Explanation:
\sqrt{x}+\dfrac{1}{\sqrt{x}}=x^{1/2}+x^{-1/2}

\int x^{1/2}dx=\dfrac{2}{3}x^{3/2}, \int x^{-1/2}dx=2x^{1/2}

Hence,
\int\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)dx =\dfrac{2}{3}x^{3/2}+2x^{1/2}+C

Watch Video Solution on YouTube

Integrals MCQ#2 (NCERT Exercise 7.1 – Question 22)

📍If \dfrac{d}{dx}f(x)=4x^{3}-\dfrac{3}{x^{4}} such that f(2)=0, then f(x) is

Options:

  • (A) x^{4}+\dfrac{1}{x^{3}}-\dfrac{129}{8}
  • (B) x^{3}+\dfrac{1}{x^{4}}+\dfrac{129}{8}
  • (C) x^{4}+\dfrac{1}{x^{3}}+\dfrac{129}{8}
  • (D) x^{3}+\dfrac{1}{x^{4}}-\dfrac{129}{8}

Answer: ✅ (A)

Explanation:
f(x)=\int\left(4x^{3}-\dfrac{3}{x^{4}}\right)dx =x^{4}+\dfrac{1}{x^{3}}+C

Using f(2)=0,
0=16+\dfrac{1}{8}+C \Rightarrow C=-\dfrac{129}{8}

Therefore,
f(x)=x^{4}+\dfrac{1}{x^{3}}-\dfrac{129}{8}

Watch Video Solution on YouTube

Let me tell you all of these 23 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 Integrals MCQ 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.

Integrals MCQ#3 (NCERT Exercise 7.2 – Question 38)

📍Evaluate \displaystyle \int \frac{10x^9 + 10^x \log_e 10}{x^{10} + 10^x}\,dx

Options:

  • (A) 10^x - x^{10} + C
  • (B) 10^x + x^{10} + C
  • (C) (10^x - x^{10})^{-1} + C
  • (D) \log(10^x + x^{10}) + C

Answer: ✅ (D)

Explanation:
Let f(x)=x^{10}+10^x

Then,
f'(x)=10x^9+10^x\log_e 10

Hence,
\int \frac{f'(x)}{f(x)}\,dx =\log f(x)+C =\log(x^{10}+10^x)+C

Watch Video Solution on YouTube

Integrals MCQ#4 (NCERT Exercise 7.2 – Question 39)

📍Evaluate \displaystyle \int \frac{dx}{\sin^2 x \cos^2 x}

Options:

  • (A) \tan x + \cot x + C
  • (B) \tan x - \cot x + C
  • (C) \tan x \cot x + C
  • (D) \tan x - \cot 2x + C

Answer: ✅ (B)

Explanation:
Using 1=\sin^2 x+\cos^2 x ,

So, \int \frac{dx}{\sin^2 x \cos^2 x} = \int \frac{\sin^2 x+\cos^2 x}{\sin^2 x \cos^2 x}\,dx

= \int \left(\frac{1}{\cos^2 x}+\frac{1}{\sin^2 x}\right)dx

= \int (\sec^2 x+\csc^2 x)\,dx = \tan x-\cot x + C

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 23 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.

Integrals MCQ#5 (NCERT Exercise 7.3 – Question 23)

📍Evaluate \displaystyle \int \frac{\sin^2 x-\cos^2 x}{\sin^2 x \cos^2 x}\,dx

Options:

  • (A) \tan x + \cot x + C
  • (B) \tan x + \csc x + C
  • (C) -\tan x + \cot x + C
  • (D) \tan x + \sec x + C

Answer: ✅ (A)

Explanation:
Consider, \int \frac{\sin^2 x-\cos^2 x}{\sin^2 x \cos^2 x}\,dx = \int \left(\frac{1}{\cos^2 x}-\frac{1}{\sin^2 x}\right)dx

= \int (\sec^2 x-\csc^2 x)\,dx

= \tan x+\cot x+C

Watch Video Solution on YouTube

Integrals MCQ#6 (NCERT Exercise 7.3 – Question 24)

📍Evaluate \displaystyle \int \frac{e^x(1+x)}{\cos^2(e^xx)}\,dx

Options:

  • (A) -\cot (ex^x)+C
  • (B) \tan (xe^x)+C
  • (C) \tan (e^x)+C
  • (D) \cot (e^x)+C

Answer: ✅ (B)

Explanation:
Let t=xe^x \Rightarrow dt=(xe^x + e^x)dx

Therefore, \int \frac{e^x(1+x)}{\cos^2(xe^x)}\,dx = \int \frac{dt}{\cos^2 t}

= \int \sec^2 t\,dt = \tan t + C

= \tan(xe^x)+C

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 set of Integrals MCQ in this part of the series.

Integrals MCQ#7 (NCERT Exercise 7.4 – Question 24)

📍Evaluate \displaystyle \int \frac{dx}{x^2+2x+2}

Options:

  • (A) x\tan^{-1}(x+1)+C
  • (B) \tan^{-1}(x+1)+C
  • (C) (x+1)\tan^{-1}x+C
  • (D) \tan^{-1}x+C

Answer: ✅ (B)

Explanation:
x^2+2x+2=(x+1)^2+1

\int \frac{dx}{(x+1)^2+1} = \tan^{-1}(x+1)+C

Watch Video Solution on YouTube

Integrals MCQ#8 (NCERT Exercise 7.4 – Question 25)

📍Evaluate \displaystyle \int \frac{dx}{\sqrt{9x-4x^2}}

Options:

  • (A) \frac{1}{9}\sin^{-1}\!\left(\frac{9x-8}{8}\right)+C
  • (B) \frac{1}{2}\sin^{-1}\!\left(\frac{8x-9}{9}\right)+C
  • (C) \frac{1}{3}\sin^{-1}\!\left(\frac{9x-8}{8}\right)+C
  • (D) \frac{1}{2}\sin^{-1}\!\left(\frac{9x-8}{9}\right)+C

Answer: ✅ (B)

Explanation:
Rewrite the expression under the square root:

9x-4x^2 = -4\left(x^2-\frac{9}{4}x\right)

Complete the square:

x^2-\frac{9}{4}x = \left(x-\frac{9}{8}\right)^2-\left(\frac{9}{8}\right)^2

So,

9x-4x^2 = 4\left[\left(\frac{9}{8}\right)^2-\left(x-\frac{9}{8}\right)^2\right]

Hence,

\int \frac{dx}{\sqrt{9x-4x^2}} = \frac{1}{2}\int \frac{dx}{\sqrt{\left(\frac{9}{8}\right)^2-\left(x-\frac{9}{8}\right)^2}}

Using the standard result,

\int \frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\!\left(\frac{u}{a}\right)+C

We get,

\int \frac{dx}{\sqrt{9x-4x^2}} = \frac{1}{2}\sin^{-1}\!\left(\frac{8x-9}{9}\right)+C

Watch Video Solution on YouTube

Following are the two MCQs of Exercise 7.5 of NCERT book for Class 12.

Integrals MCQ#9 (NCERT Exercise 7.5 – Question 22)

📍Evaluate \displaystyle \int \frac{x\,dx}{(x-1)(x-2)}

Options:

  • (A) \log\left|\frac{(x-1)^2}{x-2}\right|+C
  • (B) \log\left|\frac{(x-2)^2}{x-1}\right|+C
  • (C) \log\left|\left(\frac{x-1}{x-2}\right)^2\right|+C
  • (D) \log|(x-1)(x-2)|+C

Answer: ✅ (B)

Explanation:
Resolve into partial fractions:

\frac{x}{(x-1)(x-2)}=\frac{-1}{x-1}+\frac{2}{x-2}

So, \int \frac{x\,dx}{(x-1)(x-2)} = \int\!\left(\frac{-1}{x-1}+\frac{2}{x-2}\right)dx

= -\log|x-1|+2\log|x-2|+C = \log\left|\frac{(x-2)^2}{x-1}\right|+C

Watch Video Solution on YouTube

Integrals MCQ#10 (NCERT Exercise 7.5 – Question 23)

📍Evaluate \displaystyle \int \frac{dx}{x(x^2+1)}

Options:

  • (A) \log|x|-\frac{1}{2}\log(x^2+1)+C
  • (B) \log|x|+\frac{1}{2}\log(x^2+1)+C
  • (C) -\log|x|+\frac{1}{2}\log(x^2+1)+C
  • (D) \frac{1}{2}\log|x|+\log(x^2+1)+C

Answer: ✅ (A)

Explanation:
Multiply and divide the integrand by x ,

\int \frac{dx}{x(x^2+1)} = \int \frac{x\,dx}{x^2(x^2+1)}

So, \displaystyle \int \frac{dx}{x(x^2+1)} = \frac{1}{2}\int \frac{2x\,dx}{x^2(x^2+1)}

Let u=x^2 , so du=2x\,dx . Then,

= \frac{1}{2}\int\!\left(\frac{1}{u}-\frac{1}{u+1}\right)du

= \frac{1}{2}\log|u|-\frac{1}{2}\log(u+1)+C = \log|x|-\frac{1}{2}\log(x^2+1)+C

Watch Video Solution on YouTube

Now, following are the two MCQs of Exercise 7.6 of NCERT book for Class 12.

Integrals MCQ#11 (NCERT Exercise 7.6 – Question 23)

📍Evaluate \displaystyle \int x^2 e^{x^3}\,dx

Options:

  • (A) \frac{1}{3}e^{x^3}+C
  • (B) \frac{1}{3}e^{x^2}+C
  • (C) \frac{1}{2}e^{x^3}+C
  • (D) \frac{1}{2}e^{x^2}+C

Answer: ✅ (A)

Explanation:
Let u=x^3 , then du=3x^2dx

\int x^2 e^{x^3}dx = \frac{1}{3}\int e^u du = \frac{1}{3}e^{x^3}+C

Watch Video Solution on YouTube

Integrals MCQ#12 (NCERT Exercise 7.6 – Question 24)

📍Evaluate \displaystyle \int e^x \sec x (1+\tan x)\,dx

Options:

  • (A) e^x\cos x + C
  • (B) e^x\sec x + C
  • (C) e^x\sin x + C
  • (D) e^x\tan x + C

Answer: ✅ (B)

Explanation:
Recall the standard result (from integration by parts):

\int e^x\,[f(x)+f'(x)]\,dx = e^x f(x)+C

Here, take
f(x)=\sec x
so that
f'(x)=\sec x\tan x

Hence the given integral becomes,

\int e^x\sec x(1+\tan x)\,dx = \int e^x[\sec x+\sec x\tan x]\,dx

Applying the standard result,

\int e^x\sec x(1+\tan x)\,dx = e^x\sec x + C

Watch Video Solution on YouTube

You know, solving these integrals require a lot of practice, and at the same time you must remember the types. Like the following two of Exercise 7.7 are based on some standard types of integrals.

Integrals MCQ#13 (NCERT Exercise 7.7 – Question 10)

📍Evaluate \displaystyle \int \sqrt{1+x^2}\,dx

Options:

  • (A) \frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}\log\left|x+\sqrt{1+x^2}\right|+C
  • (B) \frac{2}{3}(1+x^2)^{3/2}+C
  • (C) \frac{2}{3}x(1+x^2)^{3/2}+C
  • (D) \frac{x^2}{2}\sqrt{1+x^2}+\frac{1}{2}x^2\log\left|x+\sqrt{1+x^2}\right|+C

Answer: ✅ (A)

Explanation:
Using the standard result,

\int \sqrt{x^2+a^2}\,dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right|+C

Here a=1 , hence

\int \sqrt{1+x^2}\,dx = \frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\log\left|x+\sqrt{1+x^2}\right|+C

Watch Video Solution on YouTube

Integrals MCQ#14 (NCERT Exercise 7.7 – Question 11)

📍Evaluate \displaystyle \int \sqrt{x^2-8x+7}\,dx

Options:

  • (A) \frac{1}{2}(x-4)\sqrt{x^2-8x+7}+9\log\left|x-4+\sqrt{x^2-8x+7}\right|+C
  • (B) \frac{1}{2}(x+4)\sqrt{x^2-8x+7}+9\log\left|x+4+\sqrt{x^2-8x+7}\right|+C
  • (C) \frac{1}{2}(x-4)\sqrt{x^2-8x+7}-3\sqrt{2}\log\left|x-4+\sqrt{x^2-8x+7}\right|+C
  • (D) \frac{1}{2}(x-4)\sqrt{x^2-8x+7}-\frac{9}{2}\log\left|x-4+\sqrt{x^2-8x+7}\right|+C

Answer: ✅ (D)

Explanation:
Complete the square,

x^2-8x+7=(x-4)^2-9 = (x-4)^2-3^2

Using the standard result,

\int \sqrt{u^2-a^2}\,du = \frac{u}{2}\sqrt{u^2-a^2} - \frac{a^2}{2}\log\left|u+\sqrt{u^2-a^2}\right|+C

Here u=x-4,\ a=3 , hence

\int \sqrt{x^2-8x+7}\,dx = \frac{1}{2}(x-4)\sqrt{x^2-8x+7} - \frac{9}{2}\log\left|x-4+\sqrt{x^2-8x+7}\right|+C

Watch Video Solution on YouTube

Further, here we have the next set of two definite integrals of Exercise 7.8 of NCERT book for Class 12.

Integrals MCQ#15 (NCERT Exercise 7.8 – Question 21)

📍Evaluate \displaystyle \int_{1}^{\sqrt{3}} \frac{dx}{1+x^2}

Options:

  • (A) \frac{\pi}{3}
  • (B) \frac{2\pi}{3}
  • (C) \frac{\pi}{6}
  • (D) \frac{\pi}{12}

Answer: ✅ (D)

Explanation:
Using the standard result,

\int \frac{dx}{1+x^2} = \tan^{-1}x + C

Applying limits,

\int_{1}^{\sqrt{3}} \frac{dx}{1+x^2} = \left[\tan^{-1}x\right]_{1}^{\sqrt{3}} = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1)

= \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}

Watch Video Solution on YouTube

Integrals MCQ#16 (NCERT Exercise 7.8 – Question 22)

📍Evaluate \displaystyle \int_{0}^{\frac{2}{3}} \frac{dx}{4+9x^2}

Options:

  • (A) \frac{\pi}{6}
  • (B) \frac{\pi}{12}
  • (C) \frac{\pi}{24}
  • (D) \frac{\pi}{4}

Answer: ✅ (C)

Explanation:
Rewrite the denominator as a sum of squares,

4+9x^2 = 2^2 + (3x)^2

Using the standard result,

\int \frac{dx}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) + C

\int \frac{dx}{4+9x^2} = \frac{1}{6}\tan^{-1}\!\left(\frac{3x}{2}\right)

Applying limits,

\left[\frac{1}{6}\tan^{-1}\!\left(\frac{3x}{2}\right)\right]_{0}^{\frac{2}{3}} = \frac{1}{6}\left(\tan^{-1}1 - 0\right) = \frac{1}{6}\cdot\frac{\pi}{4} = \frac{\pi}{24}

Watch Video Solution on YouTube

You need to really practice hard, some special questions of definite integrals also like the following two Integrals MCQ of Exercise 7.9 of NCERT. In fact, I would like to add here, that a few questions like below, need to be remembered also. Because looking at them in the first go, you may not get an idea how to start them even!

Integrals MCQ#17 (NCERT Exercise 7.9 – Question 9)

📍Evaluate \displaystyle \int_{1/3}^{1} \frac{(x-x^{3})^{1/3}}{x^{4}}\,dx

Options:

  • (A) 6
  • (B) 0
  • (C) 3
  • (D) 4

Answer: ✅ (A)

Explanation:
x-x^3=x^3\!\left(\frac{1}{x^2}-1\right)

(x-x^3)^{1/3}=x\!\left(\frac{1}{x^2}-1\right)^{1/3}

Hence,
\int_{1/3}^{1}\frac{(x-x^3)^{1/3}}{x^4}\,dx =\int_{1/3}^{1}\frac{\left(\frac{1}{x^2}-1\right)^{1/3}}{x^3}\,dx

Put u=\frac{1}{x^2}-1 , then
du=-\frac{2}{x^3}\,dx \Rightarrow \frac{dx}{x^3}=-\frac{1}{2}d

When x=\frac{1}{3},\ u=8 and when x=1,\ u=0 .

Therefore,
\int_{1/3}^{1}\frac{(x-x^3)^{1/3}}{x^4}\,dx =\frac{1}{2}\int_{0}^{8} u^{1/3}\,du

=\frac{1}{2}\left[\frac{3}{4}u^{4/3}\right]_{0}^{8} =\frac{3}{8}\times 16 =6

Watch Video Solution on YouTube

Integrals MCQ#18 (NCERT Exercise 7.9 – Question 10)

📍If f(x)=\int_{0}^{x} t\sin t\,dt , then f'(x) is

Options:

  • (A) \cos x + x\sin x
  • (B) x\sin x
  • (C) x\cos x
  • (D) \sin x + x\cos x

Answer: ✅ (B)

Explanation:
Using integration by parts,

\int t\sin t\,dt = -t\cos t + \int \cos t\,dt = -t\cos t + \sin t + C

Now apply limits from 0 \text{ to } x :

f(x) =\left[-t\cos t+\sin t\right]_{0}^{x} = (-x\cos x+\sin x)-(0+0)

So,
f(x)=-x\cos x+\sin x

Differentiate w.r.t. x :

f'(x) = \frac{d}{dx}(-x\cos x+\sin x) = -\cos x + x\sin x + \cos x = x\sin x

Watch Video Solution on YouTube

I just love the following types of integrals, which may be otherwise difficult or quite lengthy to evaluate. But by using properties of definite integrals, such questions can be damn easy to solve. See it yourself, the following two Integrals MCQ of Exercise 7.10 of NCERT.

Integrals MCQ#19 (NCERT Exercise 7.10 – Question 20)

📍Evaluate \displaystyle \int_{-\pi/2}^{\pi/2} (x^3+x\cos x+\tan^5x+1)\,dx

Options:

  • (A) 0
  • (B) 2
  • (C) \pi
  • (D) 1

Answer: ✅ (C)

Explanation:
x^3,\;x\cos x,\;\tan^5x are odd functions, hence their integrals over [-\tfrac{\pi}{2},\tfrac{\pi}{2}] are zero.

Only the constant term contributes:

\int_{-\pi/2}^{\pi/2}1\,dx=\pi

Watch Video Solution on YouTube

Integrals MCQ#20 (NCERT Exercise 7.10 – Question 21)

📍Evaluate \displaystyle \int_{0}^{\pi/2}\log\!\left(\frac{4+3\sin x}{4+3\cos x}\right)\,dx

Options:

  • (A) 2
  • (B) \frac{3}{4}
  • (C) 0
  • (D) -2

Answer: ✅ (C)

Explanation:
Let
I=\int_{0}^{\pi/2}\log\!\left(\frac{4+3\sin x}{4+3\cos x}\right)dx

Replacing x by \tfrac{\pi}{2}-x ,

I=\int_{0}^{\pi/2}\log\!\left(\frac{4+3\cos x}{4+3\sin x}\right)dx

Adding both,
2I=\int_{0}^{\pi/2}\log(1)\,dx=0

Hence,
I=0

Watch Video Solution on YouTube

Last but not the least, here we have a set of 3 important Integrals MCQ of Miscellaneous Exercise of Chapter 7 of NCERT.

Integrals MCQ#21 (NCERT Misc. Exercise Chapter 7 – Question 38)

📍Evaluate \displaystyle \int \frac{dx}{e^x+e^{-x}}

Options:

  • (A) \tan^{-1}(e^x)+C
  • (B) \tan^{-1}(e^{-x})+C
  • (C) \log(e^x-e^{-x})+C
  • (D) \log(e^x+e^{-x})+C

Answer: ✅ (A)

Explanation:
Multiply numerator and denominator by e^x ,

\int \frac{dx}{e^x+e^{-x}} = \int \frac{e^x\,dx}{e^{2x}+1}

Let u=e^x \Rightarrow du=e^x dx ,

= \int \frac{du}{u^2+1} = \tan^{-1}(u)+C = \tan^{-1}(e^x)+C

Watch Video Solution on YouTube

Integrals MCQ#22 (NCERT Misc. Exercise Chapter 7 – Question 39)

📍Evaluate \displaystyle \int \frac{\cos 2x}{(\sin x+\cos x)^2}\,dx

Options:

  • (A) -\frac{1}{\sin x+\cos x}+C
  • (B) \log|\sin x+\cos x|+C
  • (C) \log|\sin x-\cos x|+C
  • (D) \frac{1}{(\sin x+\cos x)^2}+C

Answer: ✅ (B)

Explanation:
Using identity \cos 2x=(\cos x-\sin x)(\cos x+\sin x) ,

\int \frac{\cos 2x}{(\sin x+\cos x)^2}\,dx = \int \frac{\cos x-\sin x}{\sin x+\cos x}\,dx

Let u=\sin x+\cos x \Rightarrow du=(\cos x-\sin x)dx ,

= \int \frac{du}{u} = \log|u|+C = \log|\sin x+\cos x|+C

Watch Video Solution on YouTube

Integrals MCQ#23 (NCERT Misc. Exercise Chapter 7 – Question 40)

📍If f(a+b-x)=f(x) , then evaluate \displaystyle \int_a^b x f(x)\,dx

Options:

  • (A) \frac{a+b}{2}\int_a^b f(b-x)\,dx
  • (B) \frac{a+b}{2}\int_a^b f(b+x)\,dx
  • (C) \frac{b-a}{2}\int_a^b f(x)\,dx
  • (D) \frac{a+b}{2}\int_a^b f(x)\,dx

Answer: ✅ (D)

Explanation:
Let I=\int_a^b x f(x)\,dx .

Using substitution x=a+b-t ,

I=\int_a^b (a+b-t)f(a+b-t)\,dt

Given f(a+b-t)=f(t) , so

So, I=(a+b)\int_a^b f(t)\,dt - I

\Rightarrow 2I=(a+b)\int_a^b f(x)\,dx

\therefore I=\frac{a+b}{2}\int_a^b f(x)\,dx

Watch Video Solution on YouTube

Quick Answer Key

  • Q1 → (C)
  • Q2 → (A)
  • Q3 → (D)
  • Q4 → (B)
  • Q5 → (A)
  • Q6 → (B)
  • Q7 → (B)
  • Q8 → (B)
  • Q9 → (B)
  • Q10 → (A)
  • Q11 → (A)
  • Q12 → (B)
  • Q13 → (A)
  • Q14 → (D)
  • Q15 → (D)
  • Q16 → (C)
  • Q17 → (A)
  • Q18→ (B)
  • Q19 → (C)
  • Q20 → (C)
  • Q21 → (A)
  • Q22 → (B)
  • Q23 → (D)

Closing Note

That completes all the important NCERT Class 12 Maths Integrals MCQ from chapter 7.
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 Application of Integrals MCQ (Chapter 8 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 these integrals easily! 🚀

Related Posts

Leave a Comment

Search
YouTube
YouTube
Scroll to Top

Discover more from Maths Better

Subscribe now to keep reading and get access to the full archive.

Continue reading

Discover more from Maths Better

Subscribe now to keep reading and get access to the full archive.

Continue reading