NCERT Solutions for Exercise 7.11 Class 12 Maths Chapter 7

NCERT Solutions for Exercise 7.11 Class 12 Maths Chapter 7 - Integrals

Edited By Ramraj Saini | Updated on Dec 03, 2023 08:58 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.11

NCERT Solutions for Exercise 7.11 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. In NCERT solutions for Class 12 Maths chapter 7 exercise 7.11 is last in the series of exercises apart from miscellaneous exercise. Solutions to exercise 7.11 Class 12 Maths mainly deals with some of the complex problems. Definite integral evaluation is done in this exercise. NCERT solutions for Class 12 Maths chapter 7 exercise 7.11 with 19 subjective questions and 2 Multiple choice questions provides a bulky source to practice questions. The level of these NCERT book questions is at par with that of JEE Main and NEET.

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12th class Maths exercise 7.11 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: 7.11

Question:1 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

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

Answer:

We have _{.............................................................} (i)

By using

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get :-

$I\ =\ \int_0^\frac{\pi}{2}\sin^2 x dx$ ................................................................ (ii)

Adding both (i) and (ii), we get :-

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

or $\int_0^\frac{\pi}{2}\ (cos^2 x\ +\ sin^2 x) dx\ =\ 2I$

or $\int_0^\frac{\pi}{2}1. dx\ =\ 2I$

or $2I\ =\ \left [ x \right ] ^\frac{\Pi }{2}_0\ =\ \frac{\Pi }{2}$

or $I\ =\ \frac{\Pi }{4}$

Question:2 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

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

Answer:

We have _{..........................................................................} (i)

By using ,

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

or $I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+ \sqrt{\sin x}}dx$ .......................................................(ii)

Adding (i) and (ii), we get,

$2I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}\ +\ \sqrt{\cos x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx$

or $2I\ =\ \int_0^\frac{\pi}{2}1.dx$

or $2I\ =\ \left [ x \right ]^\frac{\Pi }{2}_0\ =\ \frac{\Pi }{2}$

Thus $I\ =\ \frac{\Pi }{4}$

Question:âââââââ3 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}$

Answer:

We have $I\ =\ \int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}$ ..................................................................(i)

By using :

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}(\frac{\pi}{2}-x)dx}{\sin^\frac{3}{2}(\frac{\pi}{2}-x) + \cos^{\frac{3}{2}}(\frac{\pi}{2}-x)}$

or $I\ =\ \int^{\frac{\pi}{2}}_0\frac{\cos^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}$ . ............................................................(ii)

Adding (i) and (ii), we get :

$2I\ =\ \int^{\frac{\pi}{2}}_0\frac{\ (sin^{\frac{3}{2}}x+cos^{\frac{3}{2}}x)dx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}$

or $2I\ = \int_{0}^{{\frac{\pi}{2}}}1.dx$

or $2I\ = \left [ x \right ]^{\frac{\pi}{2}}_ 0\ =\ {\frac{\pi}{2}}$

Thus $I\ =\ {\frac{\pi}{4}}$

Question:4 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

. $\int_0^\frac{\pi}{2} \frac{\cos^5 xdx}{\sin^5x + \cos^5x}$

Answer:

We have $I\ =\ \int_0^\frac{\pi}{2} \frac{\cos^5 xdx}{\sin^5x + \cos^5x}$ ..................................................................(i)

By using :

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\frac{\pi}{2} \frac{\cos^5 (\frac{\pi}{2}-x)dx}{\sin^5(\frac{\pi}{2}-x) + \cos^5(\frac{\pi}{2}-x)}$

or $I\ =\ \int_0^\frac{\pi}{2} \frac{\sin^5 xdx}{\sin^5x + \cos^5x}$ . ............................................................(ii)

Adding (i) and (ii), we get :

$2I\ =\ \int_0^\frac{\pi}{2} \frac{\ ( sin^5x \ +\ cos^5 x)dx}{\sin^5x + \cos^5x}$

or $2I\ = \int_{0}^{{\frac{\pi}{2}}}1.dx$

or $2I\ = \left [ x \right ]^{\frac{\pi}{2}}_ 0\ =\ {\frac{\pi}{2}}$

Thus $I\ =\ {\frac{\pi}{4}}$

Question:5 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

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

Answer:

We have, $I\ =\ \int_{-5}^5|x+2|dx$

For opening the modulas we need to define the bracket :

If (x + 2) < 0 then x belongs to (-5, -2). And if (x + 2) > 0 then x belongs to (-2, 5).

So the integral becomes :-

$I\ =\ \int_{-5}^{-2} -(x+2)dx\ +\ \int_{-2}^{5} (x+2)dx$

or $I\ =\ -\left [ \frac{x^2}{2}\ +\ 2x \right ]^{-2} _{-5}\ +\ \left [ \frac{x^2}{2}\ +\ 2x \right ]^{5} _{-2}$

This gives $I\ =\ 29$

Question:6âââââââ By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_2^8|x-5|dx$

Answer:

We have, $I\ =\ \int_{2}^8|x-5|dx$

For opening the modulas we need to define the bracket :

If (x - 5) < 0 then x belongs to (2, 5). And if (x - 5) > 0 then x belongs to (5, 8).

So the integral becomes:-

$I\ =\ \int_{2}^{5} -(x-5)dx\ +\ \int_{5}^{8} (x-5)dx$

or $I\ =\ -\left [ \frac{x^2}{2}\ -\ 5x \right ]^{5} _{2}\ +\ \left [ \frac{x^2}{2}\ -\ 5x \right ]^{8} _{5}$

This gives $I\ =\ 9$

Question:7 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int^1_0x(1-x)^ndx$

Answer:

We have $I\ =\ \int^1_0x(1-x)^ndx$

U sing the property : -

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get : -

$I\ =\ \int^1_0x(1-x)^ndx\ =\ \int^1_0(1-x)(1-(1-x))^ndx$

or $I\ =\ \int^1_0(1-x)x^n\ dx$

or $I\ =\ \int^1_0(x^n\ -\ x^{n+1}) \ dx$

or $=\ \left [ \frac{x^{n+1}}{n+1}\ -\ \frac{x^{n+2}}{n+2} \right ]^1_0$

or $=\ \left [ \frac{1}{n+1}\ -\ \frac{1}{n+2} \right ]$

or $I\ =\ \frac{1}{(n+1)(n+2)}$

Question:8 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\frac{\pi}{4}\log(1+\tan x)dx$

Answer:

We have $I\ =\ \int_0^\frac{\pi}{4}\log(1+\tan x)dx$

By using the identity

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\frac{\pi}{4}\log(1+\tan x)dx\ =\ \int_0^\frac{\pi}{4}\log(1+\tan (\frac{\pi}{4}-x))dx$

or $I\ =\ \int_0^\frac{\pi}{4}\log(1+\frac{1-\tan x}{1+\tan x})dx$

or $I\ =\ \int_0^\frac{\pi}{4}\log(\frac{2}{1+\tan x})dx$

or $I\ =\ \int_0^\frac{\pi}{4}\log{2}dx\ -\ \int_0^\frac{\pi}{4}\log(1+ \tan x)dx$

or $I\ =\ \int_0^\frac{\pi}{4}\log{2}dx\ -\ I$

or $2I\ =\ \left [ x\log2 \right ]^{\frac{\Pi }{4}}_0$

or $I\ =\ \frac{\Pi }{8}\log2$

Question:9 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^2x\sqrt{2-x}dx$

Answer:

We have $I\ =\ \int_0^2x\sqrt{2-x}dx$

By using the identity

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get :

$I\ =\ \int_0^2x\sqrt{2-x}dx\ =\ \int_0^2(2-x)\sqrt{2-(2-x)}dx$

or $I\ =\ \int_0^2(2-x)\sqrt{x}dx$

or $I\ =\ \int_0^2(2\sqrt{x}\ -\ x^\frac{3}{2} dx$

or $=\ \left [ \frac{4}{3}x^\frac{3}{2}\ -\ \frac{2}{5}x^\frac{5}{2} \right ]^2_0$

or $=\ \frac{4}{3}(2)^\frac{3}{2}\ -\ \frac{2}{5}(2)^\frac{5}{2}$

or $I\ =\ \frac{16\sqrt{2}}{15}$

Question:10 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\frac{\pi}{2} (2\log\sin x- \log\sin 2x)dx$

Answer:

We have $I\ =\ \int_0^\frac{\pi}{2} (2\log\sin x- \log\sin 2x)dx$

or $I\ =\ \int_0^\frac{\pi}{2} (2\log\sin x- \log(2\sin x\cos x))dx$

or $I\ =\ \int_0^\frac{\pi}{2} (\log\sin x- \log\cos x\ -\ \log2)dx$ ..............................................................(i)

By using the identity :

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get :

$I\ =\ \int_0^\frac{\pi}{2} (\log\sin (\frac{\pi}{2}-x)- \log\cos (\frac{\pi}{2}-x)\ -\ \log2)dx$

or $I\ =\ \int_0^\frac{\pi}{2} (\log\cos x- \log\sin x\ -\ \log2)dx$ ....................................................................(ii)

Adding (i) and (ii) we get :-

$2I\ =\ \int_0^\frac{\pi}{2} (- \log 2 -\ \log 2)dx$

or $I\ =\ -\log 2\left [ \frac{\Pi }{2} \right ]$

or $I\ =\ \frac{\Pi }{2}\log\frac{1}{2}$

Question:11 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^2 x dx$

Answer:

We have $I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^2 x dx$

We know that sin ² x is an even function. i.e., sin ² (-x) = (-sinx) ² = sin ² x.

Also,

$I\ =\ \int_{-a}^af(x) dx\ =\ 2\int_{0}^af(x) dx$

So,

$I\ =\ 2\int_0^\frac{\pi}{2}\sin^2 x dx\ =\ 2\int_0^\frac{\pi}{2}\frac{(1-\cos2x)}{2} dx$

or $=\ \left [ x\ -\ \frac{\sin2x}{2} \right ]^{\frac{\Pi }{2}}_0$

or $I\ =\ \frac{\Pi }{2}$

Question:12 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\pi\frac{xdx}{1+\sin x}$

Answer:

We have $I\ =\ \int_0^\pi\frac{xdx}{1+\sin x}$ ..........................................................................(i)

By using the identity :-

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\pi\frac{xdx}{1+\sin x}\ =\ \int_0^\pi\frac{(\Pi -x)dx}{1+\sin (\Pi -x)}$

or $I\ =\ \int_0^\pi\frac{(\Pi -x)dx}{1+\sin x}$ ............................................................................(ii)

Adding both (i) and (ii) we get,

$2I\ =\ \int_0^\pi\frac{\Pi}{1+\sin x} dx$

or $2I\ =\ \Pi \int_0^\pi\frac{1-\sin x}{(1+\sin x)(1-\sin x)} dx\ =\ \Pi \int_0^\pi\frac{1-\sin x}{\cos^2 x} dx$

or $2I\ =\ \Pi \int_0^\pi (\sec^2\ -\ \tan x \sec x) x dx$

or $I\ =\ \Pi$

Question:13 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx$

Answer:

We have $I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx$

We know that $\sin^7x$ is an odd function.

So the following property holds here:-

$\int_{-a}^{a}f(x)dx\ =\ 0$

Hence

$I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx\ =\ 0$

Question:14 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^{2\pi}\cos^5xdx$

Answer:

We have $I\ =\ \int_0^{2\pi}\cos^5xdx$

I t is known that :-

$\int_0^{2a}f(x)dx\ =\ 2\int_0^{a}f(x)dx$ If f (2a - x) = f(x)

$=\ 0$ If f (2a - x) = - f(x)

Now, using the above property

$\cos^5(\Pi - x)\ =\ - \cos^5x$

Therefore, $I\ =\ 0$

Question:15 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int^\frac{\pi}{2} _0\frac{\sin x - \cos x }{1+\sin x\cos x}dx$

Answer:

We have $I\ =\ \int^\frac{\pi}{2} _0\frac{\sin x - \cos x }{1+\sin x\cos x}dx$ ................................................................(i)

By using the property :-

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get ,

$I\ =\ \int^\frac{\pi}{2} _0\frac{\sin (\frac{\pi}{2}-x) - \cos (\frac{\pi}{2}-x) }{1+\sin (\frac{\pi}{2}-x)\cos (\frac{\pi}{2}-x)}dx$

or $I\ =\ \int^\frac{\pi}{2} _0\frac{\cos x - \sin x }{1+\sin x\cos x}dx$ ......................................................................(ii)

Adding both (i) and (ii), we get

$2I\ =\ \int^\frac{\pi}{2} _0\frac{0 }{1+\sin x\cos x}dx$

Thus I = 0

Question:16 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\pi\log(1 +\cos x)dx$

Answer:

We have $I\ =\ \int_0^\pi\log(1 +\tan x)dx$ .....................................................................................(i)

By using the property:-

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\pi\log(1 +\cos (\Pi -x))dx$

$I\ =\ \int_0^\pi\log(1 -\cos x)dx$ ....................................................................(ii)

Adding both (i) and (ii) we get,

$2I\ =\ \int_0^\pi\log(1 +\cos x)dx\ +\ \int_0^\pi\log(1 -\cos x)dx$

or $2I\ =\ \int_0^\pi\log(1 -\cos^2 x)dx\ =\ \int_0^\pi\log \sin^2 xdx$

or $2I\ =\ 2\int_0^\pi\log \sin xdx$

or $I\ =\ \int_0^\pi\log \sin xdx$ ........................................................................(iii)

or $I\ =\ 2\int_0^ \frac{\pi}{2} \log \sin xdx$ ........................................................................(iv)

or $I\ =\ 2\int_0^ \frac{\pi}{2} \log \cos xdx$ .....................................................................(v)

Adding (iv) and (v) we get,

$I\ =\ -\pi \log2$

Question:17 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx$

Answer:

We have $I\ =\ \int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx$ ................................................................................(i)

By using, we get

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx\ =\ \int_0^a \frac{\sqrt {(a-x)}}{\sqrt {(a-x)} + \sqrt{x}}dx$ .................................................................(ii)

Adding (i) and (ii) we get :

$2I\ =\ \int_0^a \frac{\sqrt x\ +\ \sqrt{a-x}}{\sqrt x + \sqrt{a-x}}dx$

or $2I\ =\ \left [ x \right ]^a_0 = a$

or $I\ =\ \frac{a}{2}$

Question:18 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^4 |x-1|dx$

Answer:

We have, $I\ =\ \int_{0}^4|x-1|dx$

For opening the modulas we need to define the bracket :

If (x - 1) < 0 then x belongs to (0, 1). And if (x - 1) > 0 then x belongs to (1, 4).

So the integral becomes:-

$I\ =\ \int_{0}^{1} -(x-1)dx\ +\ \int_{1}^{4} (x-1)dx$

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

This gives $I\ =\ 5$

Question:19 Show that $\int_0^a f(x)g(x)dx = 2\int_0^af(x)dx$ if $f$ and $g$ are defined as $f(x) = f(a-x)$ and $g(x) + g(a-x) = 4$

Answer:

Let $I\ =\ \int_0^a f(x)g(x)dx$ ........................................................(i)

This can also be written as :

$I\ =\ \int_0^a f(a-x)g(a-x)dx$

or $I\ =\ \int_0^a f(x)g(a-x)dx$ ................................................................(ii)

Adding (i) and (ii), we get,

$2I\ =\ \int_0^a f(x)g(a-x)dx +\ \int_0^a f(x)g(x)dx$

$2I\ =\ \int_0^a f(x)4dx$

or $I\ =\ 2\int_0^a f(x)dx$

Question:âââââââ20 Choose the correct answer in Exercises 20 and 21.

The value of is $\int_\frac{-\pi}{2}^\frac{\pi}{2}(x^3 + x\cos x + \tan^5 x + 1)dx$ is

(A) 0

(B) 2

(D) 1

Answer:

We have

$I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}(x^3 + x\cos x + \tan^5 x + 1)dx$

This can be written as :

$I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}x^3dx +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} x\cos x +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} \tan^5 x +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} 1dx$

Also if a function is even function then $\int_{-a}^{a}f(x)\ dx\ =\ 2\int_{0}^{a}f(x)\ dx$

And if the function is an odd function then : $\int_{-a}^{a}f(x)\ dx\ =\ 0$

Using the above property I become:-

$I\ =\ 0+0+0+ 2\int_{0}^{\frac{\Pi }{2}}1.dx$

or $I\ =\ 2\left [ x \right ]^{\frac{\Pi }{2}}_0$

or $I\ =\ \Pi$

Question:âââââââ21 Choose the correct answer in Exercises 20 and 21.

The value of $\int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx$ is

Answer:

We have

$I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx$ .................................................................................(i)

By using :

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin (\frac{\pi}{2}-x)}{4+3\cos (\frac{\pi}{2}-x)} \right )dx$

or $I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\cos x}{4+3\sin x} \right )dx$ .............................................................................(ii)

Adding (i) and (ii), we get:

$2I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx\ +\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\cos x}{4+3\sin x} \right )dx$

or $2I\ =\ \int_0^\frac{\pi}{2}\log1.dx$

Thus $I\ =\ 0$

Benefits of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.11

The Class 12th Maths chapter 7 exercise has a substantial number of questions to practice.
Exercise 7.11 Class 12 Maths can take a lot of time without knowledge of basic concepts. Hence one should go for previous exercises before doing this.
These Class 12 Maths chapter 7 exercise 7.11 solutions have similar questions to that of exercise 7.10.

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Key Features Of NCERT Solutions for Exercise 7.11 Class 12 Maths Chapter 7

Comprehensive Coverage: The solutions encompass all the topics covered in ex 7.11 class 12, ensuring a thorough understanding of the concepts.
Step-by-Step Solutions: In this class 12 maths ex 7.11, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
Accuracy and Clarity: Solutions for class 12 ex 7.11 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
Conceptual Clarity: In this 12th class maths exercise 7.11 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 7.11 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 7.11 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. Two types of integration are …………. And ………….. ?

Two types of integration are Definite and indefinite Integrals.

2. What can be the limit of an integral ?

It can be anything in which the given function is real.

3. Integrals are used in …………... ?

Integrals are used in finding area, volume, displacement etc.

4. Can one skip exercise 7.11 Class 12 Maths ?

No, one should do this exercise as 5 marks questions can be asked in the Board examination.

5. Which topics are dealt in Exercise 7.11 Class 12 Maths?

Some definite integrals of advance level are discussed in this exercise.

6. Can we take some assumptions in proof related questions ?

Yes, provided it fulfills the demand of the question.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

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!

Read Complete Answer

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.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
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:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- 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:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

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.

Read Complete Answer

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.

Read Complete Answer

kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

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 7.11 Class 12 Maths Chapter 7 - Integrals

NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.11

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Access NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.11

Integrals Class 12 Chapter 7 Exercise: 7.11

Question:1 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:2 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:âââââââ3 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:4 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:5 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:6âââââââ By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:7 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:8 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:9 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:10 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:11 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:12 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:13 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:14 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:15 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:16 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:17 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

Question:18 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

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Question:19 Show that if and are defined as and

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Question:âââââââ20 Choose the correct answer in Exercises 20 and 21.

Answer:

Question:âââââââ21 Choose the correct answer in Exercises 20 and 21.

Answer:

More About NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.11

Also Read| Integrals Class 12 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.11

Key Features Of NCERT Solutions for Exercise 7.11 Class 12 Maths Chapter 7

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Question:âââââââ3 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Question:6âââââââ By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Question:19 Show that $\int_0^a f(x)g(x)dx = 2\int_0^af(x)dx$ if $f$ and $g$ are defined as $f(x) = f(a-x)$ and $g(x) + g(a-x) = 4$

Question:âââââââ20 Choose the correct answer in Exercises 20 and 21.

Question:âââââââ21 Choose the correct answer in Exercises 20 and 21.