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NCERT Solutions for Exercise 7.8 Class 12 Maths Chapter 7 - Integrals

NCERT Solutions for Exercise 7.8 Class 12 Maths Chapter 7 - Integrals

#CBSE Class 12th
Updated on 03 Dec 2023, 10:08 PM IST

NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.8

NCERT Solutions for Exercise 7.8 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 7.8 is another exercise in the series of 12 exercises. Solutions to exercise 7.8 Class 12 Maths basically deals with the concept of limit of sums. There are 6 questions based on the same concept. Even practicing 2-3 questions can suffice for understanding this concept in detail. NCERT solutions for Class 12 Maths chapter 7 exercise 7.8 provided here are in a manner that one can understand the concept without even reading the theory.

12th class Maths exercise 7.8 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 7.8

Question:1 Evaluate the following definite integrals as a limit of sums.

$\int_a^b x dx$

Answer:

We know that,
$\int_{a}^{b}f(x)dx = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+...+f(a+(n-1)h)]$
$\therefore \int_{a}^{b}xdx = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[a+(a+h)...(a+2h)..a+(n-1)h]$
$\\ = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[(a+a...a+a)_{n}+(h+2h+3h....(n-1)h)]\\ = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[na+h(1+2+3..+n-1)]\\ = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[na+h(\frac{n(n-1)}{2})]\\ = (b-a)\lim_{x\rightarrow \infty }[a+\frac{n-1}{2}h]\\ = (b-a)\lim_{x\rightarrow \infty }[a+\frac{(n-1)(b-a)}{2n}]\\ = (b-a)\lim_{x\rightarrow \infty }[a+\frac{(1-\frac{1}{n})(b-a)}{2}]\\ = (b-a)[a+\frac{(b-a)}{2}]\\ =(b-a)(b+a)/2\\ =\frac{(b^2-a^2)}{2}$

This is how the integral is evaluated using limit of a sum

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Question:2 Evaluate the following definite integrals as limit of sums.

$\int_0^5 (x + 1)dx$

Answer:

We know that
let $I =\int_{0}^{5}(x+1)dx$
$\\\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]\\ h = \frac{b-a}{n}$
Here a = 0, b = 5 and $f(x)=(x+1)$
therefore $h=\frac{5}{n}$


$\int_{0}^{5}(x+1)dx=5\lim_{x\rightarrow \infty }\frac{1}{n}[f(0)+f(5/n)+.....+f((n-1)5/n)]$

$=5\lim_{x\rightarrow \infty }\frac{1}{n}[1+(5/n+1)+....+(1+\frac{5(n-1)}{n})]\\ =5\lim_{x\rightarrow \infty }\frac{1}{n}[(1+1..+1)_{n}+\frac{5}{n}(1+2+3+...+n-1)]\\ =5\lim_{x\rightarrow \infty }\frac{1}{n}[n+\frac{5}{n}\frac{n(n-1)}{2}]\\ =5\lim_{x\rightarrow \infty }\frac{1}{n}[n+\frac{5(n-1)}{2}]\\ =5\lim_{x\rightarrow \infty }[1+\frac{5(1-\frac{1}{n})}{2}]\\ =5[1+\frac{5}{2}]\\ =\frac{35}{2}$


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Question:3 Evaluate the following definite integrals as limit of sums.

$\int_2^3 x^2dx$

Answer:

We know that

$\\\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]\\ h = \frac{b-a}{n}$
here a = 2 and b = 3 , so h = 1/n


$\int_{2}^{3}x^2dx=(3-2)\lim_{x\rightarrow \infty }\frac{1}{n}[f(2)+f(2+\frac{1}{n})+f(2+\frac{2}{n})+....+f(2+\frac{(n-1)}{n})]$

$\\=(1)\lim_{x\rightarrow \infty }\frac{1}{n}[2^2+(2+\frac{1}{n})^2+......+(2+\frac{(n-1)}{n})^2]\\ =\lim_{x\rightarrow \infty }\frac{1}{n}[(2^2+2^2+...2^2)_{n}+(\frac{1}{n})^2+(\frac{2}{n})^2+....(\frac{n-1}{n})^2+4(\frac{1}{n}+\frac{2}{n}+.....+\frac{n-1}{n})\\ =\lim_{x\rightarrow \infty }\frac{1}{n}[4n+\frac{n(n-1)(2n-1)}{6n^2}+\frac{4}{n}.\frac{n(n-1)}{2}]\\ =\lim_{x\rightarrow \infty }\frac{1}{n}[4n+(1-\frac{(1-\frac{1}{n})(2n-1)}{6})+\frac{4(n-1)}{2}]$
$\\=\lim_{x\rightarrow \infty }\frac{1}{n}[4n+(1-\frac{n(1-\frac{1}{n})(2-\frac{1}{n})}{6})+\frac{4(n-1)}{2}]\\ =\lim_{x\rightarrow \infty }\frac{1}{n}.n[4+(1-\frac{(1-\frac{1}{n})(2-\frac{1}{n})}{6})+2-\frac{2}{n}]\\ =4+\frac{2}{6}+2 =\frac{19}{3}$



Question:4 Evaluate the following definite integrals as limit of sums.

$\int_{1}^4(x^2-x)dx$

Answer:

Let
$\\I = \int_{1}^{4}(x^2-x)dx =\int_{1}^{4}x^2dx-\int_{1}^{4}xdx\\ I = I_1-I_2$

$\int_{1}^{4}x^2dx=(4-1)\lim_{x\rightarrow \infty }\frac{1}{n}[f(1)+f(1+h)+f(1+2h)+.....+f(1+(n-1)h)]$

$=(4-1)\lim_{x\rightarrow \infty }\frac{1}{n}[f(1)+f(1+h)+f(1+2h)+.....+f(1+(n-1)h)]\\ =3\lim_{x\rightarrow \infty }\frac{1}{n}[1^2+(1+\frac{3}{n})^2+(1+2.\frac{3}{n})^2+......+(1+(n-1).\frac{3}{n})^2]\\ =3\lim_{x\rightarrow \infty }\frac{1}{n}[(1^2+..1^2)_{n}+(\frac{3}{n})^2(1^2+2^2+3^2+....+(n-1)^2)+2.\frac{3}{n}(1+2+3..+n-1)]\\ =3\lim_{x\rightarrow \infty }\frac{1}{n}[n+\frac{9}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{6}{n}(\frac{n(n-1)}{2})]$

$=3\lim_{x\rightarrow \infty }\frac{1}{n}[n+\frac{9}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{6}{n}(\frac{n(n-1)}{2})]\\ =3\lim_{x\rightarrow \infty }[1+\frac{9}{6}(1-\frac{1}{n})(2-\frac{1}{n})+3(1-\frac{1}{n})]\\ =3[1+\frac{9}{6}.2+3]\\ = 21$

for the second part, we already know the general solution of $\int_{a}^{b}xdx = \frac{(b^2-a^2)}{2}$
So, here a = 1 and b = 4
therefore $\int_{1}^{4}xdx = \frac{(4^2-1^2)}{2}=\frac{15}{2}$

So, $I = 21-\frac{15}{2} = \frac{27}{2}$



Question:5 Evaluate the following definite integrals as limit of sums.

. $\int_{-1}^1 e^xdx$

Answer:

let $I = \int_{-1}^{1}e^xdx$
We know that
$\\\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]\\ h = \frac{b-a}{n}$
Here a =-1, b = 1 and $f(x) = e^x$
therefore h = 2/n
$I = 2.\lim_{x\rightarrow \infty }\frac{1}{n}[f(-1)+f(-1+\frac{2}{n})+.....+f(-1+(n-1).\frac{2}{n})]$
$\\ =2.\lim_{x\rightarrow \infty }\frac{1}{n}[e^{-1}+e^{-1+\frac{2}{n}}+e^{-1+2.\frac{2}{n}}+...+e^{-1+(n-1).\frac{2}{n}}]\\ = 2.\lim_{x\rightarrow \infty }\frac{1}{n}[e^{-1}(1+e^{2/n}+e^{4/n}+...+e^{(n-1).\frac{2}{n}})]\\ =$
By using sum of n terms of GP $S =\frac{a(r^n-1)}{r-1}$ ....where a = 1st term and r = ratio

$\\=2\lim_{n\rightarrow \infty }\frac{e^{-1}}{n}[\frac{1.(e^{\frac{2}{n}.n}-1)}{e^\frac{2}{n}-1}]\\ =2\lim_{n\rightarrow \infty }\frac{e^{-1}}{n}(\frac{e^2-1}{e^{2/n}-1})\\ =\frac{e^{-1}(e^2-1)}{\lim_{\frac{2}{n}\rightarrow \infty }\frac{e^{2/n}-1}{2/n}}\\ =\frac{e^2-1}{e}$ .........using $[\lim_{x\rightarrow \infty }(\frac{e^x-1}{x})=1]$


Question:6 Evaluate the following definite integrals as limit of sums.

$\int_0^4(x + e^{2x})dx$

Answer:

It is known that,


$\int_{0}^{4}(x+e^{2x})dx = 4\lim_{x\rightarrow \infty }\frac{1}{n}[f(0)+f(h)+f(2h)+....+f(n-1)h]$
$\\=4\lim_{x\rightarrow \infty }\frac{1}{n}[(0+e^0)+(h+e^2h)+(2h+e^4h)+......+((n-1)h+e^{2(n-1)h})]\\ = 4\lim_{x\rightarrow \infty }\frac{1}{n}[h(1+2+3+.....+n-1)+(\frac{e^{2nh}-1}{e^{2h}-1})]\\ = 4\lim_{x\rightarrow \infty }\frac{1}{n}[\frac{4}{n}(\frac{n(n-1)}{2})+(\frac{e^8-1}{e^{8/n}-1})]$
$\\=4\lim_{x\rightarrow \infty }[4.\frac{1-\frac{1}{n}}{2}+\frac{\frac{e^8-1}{8}}{\frac{e^{8/n}-1}{\frac{8}{n}}}]\\ =4(2)+4[(\frac{e^8-1}{8})]\\ ==8+e^8/2-1/2\\ =\frac{15+e^8}{2}$ ..........................( $\lim_{x\rightarrow 0}\frac{e^x-1}{x}=1$ )

More About NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.8

The NCERT Class 12 Maths chapter Integrals which covers almost 6 questions in detail is prepared by experienced subject matter experts. Exercise 7.8 Class 12 Maths can be of great help to understand the concept of limit of sums in detail. NCERT Solutions for Class 12 Maths chapter 7 exercise 7.8 can be referred to without the help of any book.

Also Read| Integrals Class 12 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.8

  • The Class 12th Maths chapter 7 exercise provided here is prepared by subject Faculties in step by step manner.
  • Practicing exercise 7.8 Class 12 Maths can help even an average student to score well in the examination.
  • These Class 12 Maths chapter 7 exercise 7.8 solutions covers mainly one concept i.e limit of sums.
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Key Features Of NCERT Solutions for Exercise 7.8 Class 12 Maths Chapter 7

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 7.8 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 7.8, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 7.8 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 7.8 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 7.8 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 7.8 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions Subject Wise

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Happy learning!!!

Frequently Asked Questions (FAQs)

Q: Are there different types of Integrals ?
A:

Yes, There are mainly 2 types of integration, i.e Definite and indefinite Integrals. 

Q: Define Indefinite Integrals ?
A:

Indefinite integrals are those which are without upper and lower limits i.e its range is not defined. 

Q: Mention some of the applications of Integrals ?
A:

Application of Integrals includes finding the quantities of area, volume, displacement etc. 

Q: Is the concept of limit of sums difficult?
A:

No, It has a pattern which students need to understand. Same pattern can be used to solve almost every problem based on the same concept. 

Q: What are the topics mentioned in Exercise 7.8 Class 12 Maths?
A:

It mainly deals with the concept of evaluating the definite integrals as limit of sums. 

Q: With how many questions Exercise 7.8 Class 12 Maths deals with?
A:

Exercise 7.8 Class 12 Maths deals with a total of 6 questions.

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