NCERT Solutions for Exercise 7.7 Class 12 Maths Chapter 7 - Integrals

# NCERT Solutions for Exercise 7.7 Class 12 Maths Chapter 7 - Integrals

Edited By Ramraj Saini | Updated on Dec 03, 2023 10:04 PM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.7

NCERT Solutions for Exercise 7.7 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 7.7 is the seventh exercise in the sequence of chapter 7 Integrals. This NCERT syllabus exercise is interesting from a learning point of view as it caters to special types of functions for integration which are quadratic functions under square root. A special approach is followed to solve such questions which can be learnt from solving a few questions given in this Class 12 Maths NCERT textbook exercise. Students should mug up the basic approach to solve the questions of this type in order to save time in the examination.

Exercise 7.7 Class 12 Maths hence cannot be avoided by the serious students. Students should religiously practice NCERT solutions for Class 12 Maths chapter 7 exercise 7.7 provided below. 12th class Maths exercise 7.7 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 7.7

### Question:1 Integrate the functions in Exercises 1 to 9.

Given function $\sqrt{4 - x^2}$ ,

So, let us consider the function to be;

$I = \int \sqrt{4-x^2}dx$

$= \int \sqrt{(2)^2-x^2}dx$

Then it is known that, $= \int \sqrt{a^2-x^2}dx =\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

Therefore, $I = \frac{x}{2}\sqrt{4-x^2} +\frac{4}{2}\sin^{-1}{\frac{x}{2}}+C$

$= \frac{x}{2}\sqrt{4-x^2} +2\sin^{-1}{\frac{x}{2}}+C$

### Question:2 Integrate the functions in Exercises 1 to 9.

Given function to integrate $\sqrt{1 - 4x^2}$

Now we can rewrite as

$= \int \sqrt{1 - (2x)^2}dx$

As we know the integration of this form is $\left [ \because \int \sqrt{a^2-x^2}dx =\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\sin^{-1}\frac{x}{a} \right ]$

$= \frac{(\frac{2x}{2})\sqrt{1^2-(2x)^2}+\frac{1^2}{2}\sin^{-1}\frac{2x}{1}}{2\rightarrow Coefficient\ of\ x\ in\ 2x} +C$

$= \frac{1}{2}\left [ x\sqrt{1-4x^2}+\frac{1}{2}\sin^{-1}2x \right ]+C$

$= \frac{x}{2}\sqrt{1-4x^2}+\frac{1}{4}\sin^{-1}2x+C$

### Question:3 Integrate the functions in Exercises 1 to 9.

Given function $\sqrt{x^2 + 4x + 6}$ ,

So, let us consider the function to be;

$I = \int\sqrt{x^2 + 4x + 6}dx$

$= \int\sqrt{(x^2 + 4x + 4)+2}dx = \int\sqrt{(x + 2)^2 +(\sqrt2)^2}dx$

And we know that, $\int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| +C$

$\Rightarrow I = \frac{x+2}{2}\sqrt{x^2+4x+6}+\frac{2}{2}\log\left | (x+2)+\sqrt{x^2+4x+6} \right |+C$

$= \frac{x+2}{2}\sqrt{x^2+4x+6}+\log\left | (x+2)+\sqrt{x^2+4x+6} \right |+C$

### Question:4 Integrate the functions in Exercises 1 to 9.

Given function $\sqrt{x^2 + 4x +1}$ ,

So, let us consider the function to be;

$I = \int\sqrt{x^2 + 4x + 1}dx$

$= \int\sqrt{(x^2 + 4x + 4)-3}dx = \int\sqrt{(x + 2)^2 -(\sqrt3)^2}dx$

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| +C$

$\therefore I = \frac{x+2}{2}\sqrt{x^2+4x+1}-\frac{3}{2}\log\left | (x+2)+\sqrt{x^2+4x+1} \right |+C$

### Question:5 Integrate the functions in Exercises 1 to 9.

Given function $\sqrt{1-4x-x^2}$ ,

So, let us consider the function to be;

$I = \int\sqrt{1-4x-x^2}dx$

$= \int\sqrt{1-(x^2+4x+4-4)}dx = \int\sqrt{1+4 -(x+2)^2}dx$

$= \int\sqrt{(\sqrt5)^2 -(x+2)^2}dx$

And we know that, $\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

$\therefore I = \frac{x+2}{2}\sqrt{1-4x-x^2}+\frac{5}{2}\sin^{-1}\left ( \frac{x+2}{\sqrt5} \right )+C$

### Question:6 Integrate the functions in Exercises 1 to 9.

Given function $\sqrt{x^2 + 4x - 5}$ ,

So, let us consider the function to be;

$I = \int\sqrt{x^2+4x-5}dx$

a $= \int\sqrt{(x^2+4x+4)-9}dx = \int\sqrt{(x+2)^2 -(3)^2}dx$

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}|+C$

$\therefore I = \frac{x+2}{2}\sqrt{x^2+4x-5}-\frac{9}{2}\log\left | (x+2)+ \sqrt{x^2+4x-5} \right |+C$

### Question:7 Integrate the functions in Exercises 1 to 9.

Given function $\sqrt{1 + 3x - x^2}$ ,

So, let us consider the function to be;

$I = \int\sqrt{1+3x-x^2}dx$

$= \int\sqrt{(1-\left ( x^2-3x+\frac{9}{4}-\frac{9}{4} \right )}dx = \int \sqrt{\left ( 1+\frac{9}{4} \right )-\left ( x-\frac{3}{2} \right )^2}dx$ $= \int \sqrt{\left ( \frac{\sqrt{13}}{2} \right )^2-\left ( x-\frac{3}{2} \right )^2}dx$

And we know that, $\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

$\therefore I = \frac{x-\frac{3}{2}}{2}\sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1}\left ( \frac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}} \right )+C$

$= \frac{2x-3}{4}\sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1}\left ( \frac{2x-3}{\sqrt{13}} \right )+C$

### Question:8 Integrate the functions in Exercises 1 to 9.

Given function $\sqrt{x^2 + 3x}$ ,

So, let us consider the function to be;

$I = \int\sqrt{x^2+3x}dx$

$= \int\sqrt{x^2+3x+\frac{9}{4}-\frac{9}{4}}dx$

$= \int\sqrt{\left ( x+\frac{3}{2} \right )^2-\left ( \frac{3}{2} \right )^2 }dx$

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| +C$

$\therefore I = \frac{x+\frac{3}{2}}{2}\sqrt{x^2+3x}-\frac{\frac{9}{4}}{2}\log \left | \left ( x+\frac{3}{2} \right )+\sqrt{x^2+3x} \right |+C$

$= \frac{2x+3}{4}\sqrt{x^2+3x}-\frac{9}{8}\log\left | \left ( x+\frac{3}{2} \right )+\sqrt{x^2+3x} \right |+C$

### Question:9 Integrate the functions in Exercises 1 to 9.

Given function $\sqrt{1 + \frac{x^2}{9}}$ ,

So, let us consider the function to be;

$I = \int\sqrt{1+\frac{x^2}{9}}dx = \frac{1}{3}\int \sqrt{9+x^2}dx$

$= \frac{1}{3}\int \sqrt{3^2+x^2}dx$

And we know that, $\int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| +C$

$\therefore I = \frac{1}{3}\left [ \frac{x}{2}\sqrt{x^2+9} +\frac{9}{2}\log|x+\sqrt{x^2+9}| \right ]+C$

$= \frac{x}{6}\sqrt{x^2+9} +\frac{3}{2}\log\left | x+\sqrt{x^2+9} \right |+C$

### Question:10 Choose the correct answer in Exercises 10 to 11.

$\int \sqrt{1+x^2}dx$ is equal to

(A) $\frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\log\left |\left(x + \sqrt{1+x^2} \right )\right| +C$

(B) $\frac{2}{3}(1+x^2)^{\frac{3}{2}} + C$

(C) $\frac{2}{3}x(1+x^2)^{\frac{3}{2}} + C$

(D) $\frac{x^2}{2}\sqrt{1+x^2} + \frac{1}{2}x^2\log\left |x + \sqrt{1+x^2} \right| +C$

As we know that, $\int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| +C$

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

Therefore the correct answer is A.

### Question:11 Choose the correct answer in Exercises 10 to 11.

$\int \sqrt{x^2 - 8x+7}dx$ is equal to

(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) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} -3\sqrt2\log\left|x-4+\sqrt{x^2 -8x+7}\right| +C$

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

Given integral $\int \sqrt{x^2 - 8x+7}dx$

So, let us consider the function to be;

$I = \int\sqrt{x^2-8x+7}dx =\int\sqrt{(x^2-8x+16)-(9)}dx$

$=\int\sqrt{(x-4)^2-(3)^2}dx$

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| +C$

$I = \frac{(x-4)}{2}\sqrt{x^2-8x+7}-\frac{9}{2}\log|(x-4)+\sqrt{x^2-8x+7}| +C$

Therefore the correct answer is D.

## More About NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.7

NCERT solutions for Class 12 Maths chapter 7 exercise 7.7 consists of 11 questions in total and almost all questions are based on the same concept. While solving exercise 7.7 Class 12 Maths students can skip the questions repeated with the same concepts to save their time. NCERT solutions for Class 12 Maths chapter 7 exercise 7.7 cannot be avoided to learn the process of integration of quadratic equations integration.

Also Read| Integrals Class 12 Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.7

• The Class 12th Maths chapter 7 exercise provided below is in detail prepared in step by step manner.
• Practicing exercise 7.7 Class 12 Maths is highly recommended to learn special types of problems.
• These Class 12 Maths chapter 7 exercise 7.7 solutions can be asked directly in the Board exams as well as competitive exams.
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## Key Features Of NCERT Solutions for Exercise 7.7 Class 12 Maths Chapter 7

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 7.7 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 7.7, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 7.7 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 7.7 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 7.7 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 7.7 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
##### JEE Main Important Mathematics Formulas

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

## Subject Wise NCERT Exemplar Solutions

Happy learning!!

1. Define Integration ?

Integration is defined as the process of finding the antiderivative of a function which represents the area under the simple curve. Integration in higher forms also can be used to find centre of mass, centre of gravity etc.

2. Mention the areas in which Integration is used ?

Integration is used to find the area, volume, centre of mass, centre of gravity etc. of many functions.

3. Questions of how many marks are asked in the Board exam from this chapter ?

The Integral chapter covers around 10 to 15 marks questions in board exams.

4. Is this chapter useful in other subjects also ?

Yes, Integration has its applications in physics and physical chemistry.

5. List out few topics in Exercise 7.7 Class 12 Maths

Topics which emphasis on the integration of quadratic functions is mentioned in the Exercise 7.7 Class 12 Maths

6. How many total questions are there Exercise 7.7 Class 12 Maths

There are 11 questions in Exercise 7.7 Class 12 Maths

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

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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.

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.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

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 help you.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

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

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 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.

With increase of temperature, which of these changes?

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

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 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