NCERT Solutions for Exercise 7.10 Class 12 Maths Chapter 7

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

Edited By Ramraj Saini | Updated on Dec 03, 2023 10:15 PM IST | #CBSE Class 12th

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NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.10

NCERT Solutions for Exercise 7.10 Class 12 Maths Chapter 7 Integrals are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for Class 12 Maths chapter 7 exercise 7.10 is more or less similar to the exercise 7.9. Both these exercises cater to questions in which the value of the integrals is found out. Solutions to exercise 7.10 provided below are in detail. Some of the questions are multiple choice based which are also discussed in detail. NCERT solutions for Class 12 Maths chapter 7 exercise 7.10 discussed here are in detail. Students can choose to practice at least a few of the questions before examination. Overall for integral chapters, NCERT book exercise questions are enough to score well. Also students can refer NCERT chapter Integrals for practice.

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NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.10
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Integrals Class 12 Chapter 7 Exercise: 7.10
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Also Read| Integrals Class 12 Notes
Benefits of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10
Key Features Of NCERT Solutions for Exercise 7.10 Class 12 Maths Chapter 7
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12th class Maths exercise 7.10 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Integrals Class 12 Chapter 7 Exercise: 7.10

Question:1 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^1\frac{x}{x^2 +1}dx$

Answer:

$\int_0^1\frac{x}{x^2 +1}dx$
let $x^2+1 = t \Rightarrow xdx =dt/2$
when x = 0 then t = 1 and when x =1 then t = 2
$\therefore \int_{o}^{1}\frac{x}{x^2+1}dx=\frac{1}{2}\int_{1}^{2}\frac{dt}{t}$
$\\=\frac{1}{2}[\log\left | t \right |]_{1}^{2}\\ =\frac{1}{2}\log 2$

Question:2 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int^\frac{\pi}{2}_0\sqrt{\sin\phi}\cos^5\phi d\phi$

Answer:

$\int^\frac{\pi}{2}_0\sqrt{\sin\phi}\cos^5\phi d\phi$
let $\sin \phi = t \Rightarrow \cos \phi d\phi = dt$
when $\phi =0,t\rightarrow 0$ and $\phi =\pi/2,t\rightarrow 1$

using the above substitution we can evaluate the integral as

$\\\therefore \int_{0}^{1}\sqrt{t}(1-t^2)dt\\ =\int_{0}^{1} t^\frac{1}{2}(1+t^4-2t^2)dt\\ =\int_{0}^{1}t^\frac{1}{2}dt+\int_{0}^{1}t^{9/2}dt-2\int_{0}^{1}t^{5/2}dt\\ =[2t^{3/2}/3+2t^{11/2}/11+4t^{7/2}/7]^1_0\\ =\frac{64}{231}$

Question:3 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx$

Answer:

$\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx$
let $x = \tan\theta\Rightarrow dx =\sec^2\theta d\theta$
when x = 0 then $\theta= 0$ and when x = 1 then $\theta= \pi/4$

$\\=\int_{0}^{\pi/4}\sin^{-1}(\frac{2\tan\theta}{1+\tan\theta})\sec^2\theta d\theta\\ =\int_{0}^{\pi/4}\sin^{-1}(\sin 2\theta)\sec^2\theta d\theta\\ =\int_{0}^{\pi/4}2\theta \sec^2\theta d\theta\\$
Taking $\theta$ as a first function and $\sec^2\theta$ as a second function, by using by parts method

$\\=2[\theta\int \sec^2\theta d\theta-\int(\frac{d}{d\theta}\theta.\int \sec^2\theta\ d\theta)d\theta]^{\pi/4}_0\\ =2[\theta\tan\theta-\int \tan\theta\ d\theta]^{\pi/4}_0\\ =2[\theta\tan\theta+\log\left | \cos\theta \right |]^{\pi/4}_0\\ =2[\pi/4+\log(1/\sqrt{2})]\\ =\pi/4-\log 2$

Question:4 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^2x\sqrt{x+2}$ . (Put ${x+2} = t^2$ )

Answer:

Let $x+2 = t^2\Rightarrow dx =2tdt$
when x = 0 then t = $\sqrt{2}$ and when x=2 then t = 2

$I=\int_{0}^{2}x\sqrt{x+2}dx$

$\\=2\int_{\sqrt{2}}^{2}(t^2-2)t^2dt\\ =2\int_{\sqrt{2}}^{2}(t^4-2t^2)dt\\ =2[t^5/5-\frac{2}{3}t^3]^2_{\sqrt{2}}\\ =2[\frac{32}{5}-\frac{16}{3}-\frac{4\sqrt{2}}{5}+\frac{4\sqrt{2}}{3}]\\ =\frac{16\sqrt{2}(\sqrt{2}+1)}{15}$

Question:5 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1 + \cos^2 x}dx$

Answer:

$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1 + \cos^2 x}dx =I$
let $\cos x =t\Rightarrow -\sin x dx = dt$
when x=0 then t = 1 and when x= $\pi/2$ then t = 0

$\\I=\int_{1}^{0}\frac{dt}{1+t^2}\\ =[\tan ^{-1}t]^0_1\\ =\pi/4$

Question:6 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^2\frac{dx}{x + 4 - x^2}$

Answer:

By adjusting, the denominator can also be written as $(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2 =x+4-x^2$
Now,
$\Rightarrow \int_{0}^{2}\frac{dx}{(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2}$
let $x-1/2 = t\Rightarrow dx=dt$
when x= 0 then t =-1/2 and when x =2 then t = 3/2

$\\\Rightarrow\int_{-1/2}^{3/2}\frac{dt}{(\frac{\sqrt{17}}{2})^2-t^2}\\ =\frac{1}{2.\frac{\sqrt{17}}{2}}\log\frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}\\ =\frac{1}{\sqrt{17}}[\log\frac{\sqrt{17}/2+3/2}{\sqrt{17}/2-3/2}-\log\frac{\sqrt{17}/2-1/2}{\sqrt{17}/2+1/2}]\\ =\frac{1}{\sqrt{17}}[\log\frac{\sqrt{17}+3}{\sqrt{17}-3/}.\frac{\sqrt{17}+1}{\sqrt{17}+1}]$
$\\ =\frac{1}{\sqrt{17}}[\log (\frac{17+3+4\sqrt{17}}{17+3-4\sqrt{17}})]\\ =\frac{1}{\sqrt{17}}[\log (\frac{5+\sqrt{17}}{5-\sqrt{17}})]$
On rationalisation, we get

$=\frac{1}{\sqrt{17}}\log \frac{21+5\sqrt{17}}{4}$

Question:7 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_{-1}^1\frac{dx}{x^2 +2x + 5}$

Answer:

$\int_{-1}^1\frac{dx}{x^2 +2x + 5}$
the Dr can be written as $x^2+2x+5 = (x+1)^2+2^2$
and put x+1 = t then dx =dt

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

$\\\Rightarrow \int_{0}^{2}\frac{dt}{t^2+2^2}\\ =\frac{1}{2}[\tan^{-1}\frac{t}{2}]^2_0\\ =\frac{1}{2}( \pi/4)\\ =\frac{\pi}{8}$

Question:8 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx$

Answer:

$\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx$
let $2x =t \Rightarrow 2dx =dt$
when x = 1 then t = 2 and when x = 2 then t= 4

$\\=\frac{1}{2}\int_{2}^{4}(\frac{2}{t}-\frac{2}{t^2})e^tdt\\$
let
$\frac{1}{t} = f(t)\Rightarrow f'(t)=-\frac{1}{t^2}$
$\Rightarrow \int_{2}^{4}(\frac{1}{t}-\frac{1}{t^2})e^tdt =\int_{2}^{}4e^t[f(t)+f'(t)]dt$
$\\=[e^tf(t)]^4_2\\ =[e^t.\frac{1}{t}]^4_2\\ =\frac{e^4}{4}-\frac{e^2}{2}\\ =\frac{e^2(e^2-2)}{4}$

Question:9 Choose the correct answer in Exercises 9 and 10.

The value of the integral $\int_{\frac{1}{3}}^1\frac{(x-x^3)^\frac{1}{3}}{x^4}dx$ is

(A) 6

(B) 0

(D) 4

Answer:

The value of integral is (A) = 6

$\int_{\frac{1}{3}}^1\frac{(x-x^3)^\frac{1}{3}}{x^4}dx$
$\int_{\frac{1}{3}}^1\frac{(\frac{1}{x^2}-1)^\frac{1}{3}}{x^3}dx\\$
let
$\frac{1}{x^2}-1 = t\Rightarrow \frac{dx}{x^3}=-dt/2$
now, when x = 1/3, t = 8 and when x = 1 , t = 0

therefore

$\\=-\frac{1}{2}\int_{8}^{0}t^{1/3}dt\\ =-\frac{1}{2}.\frac{3}{4}[t^4/3]^0_8\\ =-\frac{3}{8}[-2^4]\\ =6$

Question:10 Choose the correct answer in Exercises 9 and 10.

If $f(x) = \int_0^x t \sin t dt$ , then $f'(x)$ is

(A) $\cos x + x\sin x$

(B) $x\sin x$

(D) $\sin x + x\cos x$

Answer:

The correct answer is (B) = $x\sin x$

$f(x) = \int_0^x t \sin t dt$
by using by parts method,
$\\=t\int \sin t dt - \int (\frac{d}{dt}t\int \sin t dt)dt\\ =[t(-\cos t )+\sin t]^x_0$

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

Benefits of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10

The Class 12th Maths chapter 7 exercise provided here is of high quality which can be referred directly by the student .
NCERT syllabus Exercise 7.10 Class 12 Maths provides some good questions which can be asked in the examination.
Class 12 Maths chapter 7 exercise 7.10 solutions can be found similar to previous exercises. Hence students can skip some of the similar questions.

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Key Features Of NCERT Solutions for Exercise 7.10 Class 12 Maths Chapter 7

Comprehensive Coverage: The solutions encompass all the topics covered in ex 7.10 class 12, ensuring a thorough understanding of the concepts.
Step-by-Step Solutions: In this class 12 maths ex 7.10, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
Accuracy and Clarity: Solutions for class 12 ex 7.10 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
Conceptual Clarity: In this 12th class maths exercise 7.10 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
Inclusive Approach: Solutions for ex 7.10 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
Relevance to Curriculum: The solutions for class 12 maths ex 7.10 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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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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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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Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
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Seek Support:
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- Join Study Groups: Collaborating with peers can create a supportive learning environment.

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,

Qualifications:
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Hope you find this useful!

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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

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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

NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.10

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Integrals Class 12 Chapter 7 Exercise: 7.10

Question:1 Evaluate the integrals in Exercises 1 to 8 using substitution.

Answer:

Question:2 Evaluate the integrals in Exercises 1 to 8 using substitution.

Answer:

Question:3 Evaluate the integrals in Exercises 1 to 8 using substitution.

Answer:

Question:4 Evaluate the integrals in Exercises 1 to 8 using substitution.

Answer:

Question:5 Evaluate the integrals in Exercises 1 to 8 using substitution.

Answer:

Question:6 Evaluate the integrals in Exercises 1 to 8 using substitution.

Answer:

Question:7 Evaluate the integrals in Exercises 1 to 8 using substitution.

Answer:

Question:8 Evaluate the integrals in Exercises 1 to 8 using substitution.

Answer:

Question:9 Choose the correct answer in Exercises 9 and 10.

Answer:

Question:10 Choose the correct answer in Exercises 9 and 10.

Answer:

More About NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10

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