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

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. Students can refer to NCERT chapter Integrals other exercises given below:

  • Integrals Exercise 7.1

  • Integrals Exercise 7.2

  • Integrals Exercise 7.3

  • Integrals Exercise 7.4

  • Integrals Exercise 7.5

  • Integrals Exercise 7.6

  • Integrals Exercise 7.7

  • Integrals Exercise 7.9

  • Integrals Exercise 7.10

  • Integrals Exercise 7.11

  • Integrals Miscellaneous Exercise

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


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}



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.

Also see-

  • NCERT Exemplar Solutions Class 12 Maths Chapter 7

  • NCERT Solutions for Class 12 Maths Chapter 7

NCERT Solutions Subject Wise

  • NCERT Solutions Class 12 Chemistry

  • NCERT Solutions for Class 12 Physics

  • NCERT Solutions for Class 12 Biology

  • NCERT Solutions for Class 12 Mathematics

Subject Wise NCERT Exemplar Solutions

  • NCERT Exemplar Class 12 Maths

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  • NCERT Exemplar Class 12 Chemistry

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