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NCERT Solutions for Exercise 7.10 Class 12 Maths Chapter 7 - Integrals

NCERT Solutions for Exercise 7.10 Class 12 Maths Chapter 7 - Integrals

Edited By Komal Miglani | Updated on Apr 24, 2025 08:49 AM IST | #CBSE Class 12th

Derivatives measure the rate of change, like the speed of your bike, while integrals sum up all the little changes over time to find the total accumulation, like the total distance travelled by you. Integrals are one of the fundamental concepts in calculus, which play a vital role in solving real-world problems involving areas and volumes. After learning about the definite integrals, we can now look forward to the topic of evaluating definite integrals. In exercise 7.9 of the chapter Integrals, we will learn about the evaluation of definite integrals by substitution. This exercise will help the students in applying a suitable substitution to change some complex expressions into more manageable ones and evaluate them easily. This article on the NCERT Solutions for Exercise 7.9 Class 12 Maths Chapter 7 - Integrals, offers detailed and easy-to-understand solutions to help students clear their doubts and get a clear idea about the logic behind these solutions. For syllabus, notes, and PDF, refer to this link: NCERT.

This Story also Contains
  1. Class 12 Maths Chapter 7 Exercise 7.9 Solutions: Download PDF
  2. Integrals Class 12 Chapter 7 Exercise: 7.9
  3. Topics covered in Chapter 7, Integrals: Exercise 7.9
  4. NCERT Solutions Subject Wise
  5. NCERT Exemplar Solutions Subject Wise
LiveCBSE 10th, 12th Results 2025 Live: When will CBSE declare results? Date, time, latest updatesMay 7, 2025 | 10:54 PM IST

CBSE will assign grades to Class 10 students based on the following criteria:

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Class 12 Maths Chapter 7 Exercise 7.9 Solutions: Download PDF

Download PDF

Integrals Class 12 Chapter 7 Exercise: 7.9

Question 1: Evaluate the integral using substitution.

∫01xx2+1dx

Answer:

∫01xx2+1dx
let x2+1=t⇒xdx=dt/2
when x = 0 then t = 1 and when x =1 then t = 2
∴∫o1xx2+1dx=12∫12dtt
=12[log⁡|t|]12=12log⁡2


Question 2: Evaluate the integral using substitution.

∫0π2sin⁡ϕcos5⁡ϕdϕ

Answer:

∫0π2sin⁡ϕcos5⁡ϕdϕ
let sin⁡ϕ=t⇒cos⁡ϕdϕ=dt
when ϕ=0,t→0 and ϕ=π/2,t→1

using the above substitution we can evaluate the integral as

∴∫01t(1−t2)dt

=∫01t12(1+t4−2t2)dt

=∫01t12dt+∫01t92dt−2∫01t52dt

=[2t3/23+2t11/211−4t7/27]01

=64231


Question 3: Evaluate the integral using substitution.

∫01sin−1⁡(2x1+x2)dx

Answer:

∫01sin−1⁡(2x1+x2)dx
let x=tan⁡θ⇒dx=sec2⁡θdθ
when x = 0 then θ=0 and when x = 1 then θ=π/4

=∫0π/4sin−1⁡(2tan⁡θ1+tan⁡θ)sec2⁡θdθ=∫0π/4sin−1⁡(sin⁡2θ)sec2⁡θdθ=∫0π/42θsec2⁡θdθ
Taking θ as a first function and sec2⁡θ as a second function, by using by parts method

=2[θ∫sec2⁡θdθ−∫(ddθθ⋅∫sec2⁡θdθ)dθ]0π/4=2[θtan⁡θ−∫tan⁡θdθ]0π/4=2[θtan⁡θ+log⁡|cos⁡θ|]0π/4=2[π4+log⁡(12)]=π2−log⁡2


Question 4: Evaluate the integral using substitution.

∫02xx+2 . (Put x+2=t2 )

Answer:

Let x+2=t2⇒dx=2tdt
when x = 0 then t = 2 and when x=2 then t = 2

I=∫02xx+2dx

=2∫22(t2−2)t2dt

=2∫22(t4−2t2)dt

=2[t55−23t3]22

=2[325−163−425+423]

=162(2+1)15


Question 5: Evaluate the integral using substitution.

∫0π2sin⁡x1+cos2⁡xdx

Answer:

∫0π2sin⁡x1+cos2⁡xdx=I
let cos⁡x=t⇒−sin⁡xdx=dt
when x=0 then t = 1 and when x= π/2 then t = 0

I=∫10dt1+t2=[tan−1⁡t]10=π/4


Question 6: Evaluate the integral using substitution.

∫02dxx+4−x2

Answer:

By adjusting, the denominator can also be written as (172)2−(x−12)2=x+4−x2
Now,
⇒∫02dx(172)2−(x−12)2
let x−1/2=t⇒dx=dt
when x= 0 then t =-1/2 and when x =2 then t = 3/2

⇒∫−1/23/2dt(172)2−t2
=12⋅172log⁡172+t172−t
=117[log⁡172+32172−32−log⁡172−12172+12]
=117[log⁡(17+317−3⋅17+117−1)]
=117[log⁡(17+3+41717+3−417)]
=117log⁡(5+175−17)
On rationalisation, we get

=117log⁡21+5174


Question 7: Evaluate the integral using substitution.

∫−11dxx2+2x+5

Answer:

∫−11dxx2+2x+5
it can be written as x2+2x+5=(x+1)2+22
and put x+1 = t then dx =dt

when x= -1 then t = 0 and when x = 1 then t = 2

⇒∫02dtt2+22=12[tan−1⁡t2]02=12(π/4)=π8


Question 8: Evaluate the integral using substitution.

∫12(1x−12x2)e2xdx

Answer:

∫12(1x−12x2)e2xdx
let 2x=t⇒2dx=dt
when x = 1 then t = 2 and when x = 2 then t= 4

=12∫24(2t−2t2)etdt
let
1t=f(t)⇒f′(t)=−1t2
⇒∫24(1t−1t2)etdt=∫24et[f(t)+f′(t)]dt
=[etf(t)]24=[et.1t]24=e44−e22=e2(e2−2)4


Question 9: Choose the correct answer

The value of the integral ∫131(x−x3)13x4dx is

(A) 6

(B) 0

(C) 3

(D) 4

Answer:

∫131(x−x3)13x4dx
∫131(1x2−1)13x3dx
let
1x2−1=t⇒dxx3=−dt/2
now, when x = 1/3, t = 8 and when x = 1 , t = 0

=−12∫80t1/3dt=−12.34[t4/3]80=−38[−24]=6

Hence, The value of integral is 6


Question 10: Choose the correct answer

If f(x)=∫0xtsin⁡tdt , then f′(x) is

(A) cos⁡x+xsin⁡x

(B) xsin⁡x

(C) xcos⁡x

(D) sin⁡x+xcos⁡x

Answer:

f(x)=∫0xtsin⁡tdt
by using by parts method,
=t∫sin⁡tdt−∫(ddtt∫sin⁡tdt)dt=[t(−cos⁡t)+sin⁡t]0x

f(x)=−xcosx+sinx

So, f′(x)=−cosx+xsinx+cosx=xsinx

Hence, The correct answer is xsin⁡x


Also Read,

Topics covered in Chapter 7, Integrals: Exercise 7.9

The main topics covered in class 12 maths chapter 7 of Integrals, exercise 7.9 are:

  • Evaluation of definite integrals by substitution: We substitute a part of the integral with a new variable and change the limits accordingly. In this way, we can solve complex and tricky integral problems easily.
  • Application of the fundamental theorem of calculus: The fundamental theorem of calculus is widely used in the evaluation of definite integrals. It is expressed as follows:
    ∫abf(x)dx=F(b)−F(a), where F(x) is the antiderivative of f(x).
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NCERT Solutions Subject Wise

Below are some useful links for subject-wise NCERT solutions for class 12.

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As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

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NCERT Exemplar Solutions Subject Wise

Here are some links to subject-wise solutions for the NCERT exemplar class 12.

Frequently Asked Questions (FAQs)

1. What is the weightage of Exercise 7.10 Class 12 Maths ?

Alteat 1 questions will be there in the board examination. Hence students can expect 5 marks minimum. It can go upto 10 marks also. 

2. Integrals without upper and lower limits are known as ………... ?

Integrals without upper and lower limits are known as Indefinite integrals.

3. Give the number of questions discussed in Exercise 7.10 Class 12 Maths ?

There are 10 questions discussed in Exercise 7.10 Class 12 Maths

4. How many multiple choice questions are there in Exercise 7.10 Class 12 Maths ?

There are 2 Multiple choice questions in Exercise 7.10 Class 12 Maths 

5. Are questions repeated in Board examination from this chapter?

Exact questions are rare to observe but questions are repeated based on the same concepts. 

6. How important is Exercise 7.10 Class 12 Maths for NEET and JEE?

In JEE mains and NEET, questions are asked on similar lines of Exercise 7.10 Class 12 Maths .

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Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

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Manasvini Gupta 30 January,2025
quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

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Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







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Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

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Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

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Ashvadha 12 July,2024
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