NCERT Solutions Class 12 Maths Chapter-7 (Integrals)Exercise 7.8

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-7 (Integrals)Exercise 7.8 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Exercise 7.8 

Q1. Evaluate the following definite integral as limit of sums : ${\int }_a^{b}xdx$

Answer. $\begin{array}{c}\text{It is known that,}\\ {\int }_a^{b}f\left(x\right)dx=(b-a){lim}_{n\to \infty }\frac{1}{n}\left[f\right(a)+f(a+h)+\dots +f(a+(n-1)h\left)\right],\text{where}h=\frac{b-a}{n}\\ \text{Here,}a=a,b=b,\text{and}f\left(x\right)=x\end{array}$ $\therefore {\int }_a^{b}xdx=(b-a){lim}_{n\to \infty }\frac{1}{n}[a+(a+h)\dots (a+2h),a+(n-1\left)h\right]$ $\begin{array}{c}=(b-a){lim}_{n\to \infty }\frac{1}{n}\left[\right(a+a+a+\dots +a)+(h+2h+3h+\dots +(n-1)h\left)\right]\\ =(b-a){lim}_{n\to \infty }\frac{1}{n}[na+h(1+2+3+\dots +(n-1)\left)\right]\end{array}$ $\begin{array}{c}=(b-a){lim}_{n\to \infty }\frac{1}{n}[na+h\left\{\frac{(n-1)\left(n\right)}{2}\right\}]\\ =(b-a){lim}_{n\to \infty }\frac{1}{n}[na+\frac{n(n-1)h}{2}]\\ =(b-a){lim}_{n\to \infty }\frac{n}{n}[a+\frac{(n-1)h}{2}]\end{array}$ $\begin{array}{c}=(b-a){lim}_{n\to \infty }[a+\frac{(n-1)(b-a)}{2n}]\\ =(b-a){lim}_{n\to \infty }[a+\frac{(1-\frac{1}{n})(b-a)}{2}]\\ =(b-a)[a+\frac{(b-a)}{2}]\end{array}$ $\begin{array}{cc}& =(b-a)\left[\frac{2a+b-a}{2}\right]\\ & =\frac{(b-a)(b+a)}{2}\\ & =\frac{1}{2}(b^2-a^2)\end{array}$

Q2. Evaluate the following definite integral as limit of sums : ${\int }_0^5(x+1)dx$

Answer. $\begin{array}{c}\text{Let}I={\int }_0^5(x+1)dx\\ \text{It is known that,}\\ {\int }_a^{b}f\left(x\right)dx=(b-a){lim}_{n\to \infty }\frac{1}{n}\left[f\right(a)+f(a+h)\dots f(a+(n-1)h\left)\right],\text{where}h=\frac{b-a}{n}\end{array}$ $\begin{array}{c}\text{Here,}a=0,b=5,\text{and}f\left(x\right)=(x+1)\\ \Rightarrow h=\frac{5-0}{n}=\frac{5}{n}\\ \therefore {\int }_0^5(x+1)dx=(5-0){lim}_{n\to \infty }\frac{1}{n}\left[f\right(0)+f\left(\frac{5}{n}\right)+\dots +f\left(\right(n-1\left)\frac{5}{n}\right)]\end{array}$ $\begin{array}{cc}& =5\underset{n\to \infty }{lim}\frac{1}{n}[1+(\frac{5}{n}+1)+\dots \{1+\left(\frac{5(n-1)}{n}\right)\}]\\ & =5\underset{n\to \infty }{lim}\frac{1}{n}\left[\right(1+1+1\dots 1)+[\frac{5}{n}+2\cdot \frac{5}{n}+3\cdot \frac{5}{n}+\dots (n-1\left)\frac{5}{n}\right]]\\ & =5\underset{n\to \infty }{lim}\frac{1}{n}[n+\frac{5}{n}\{1+2+3\dots (n-1)\left\}\right]\end{array}$ $\begin{array}{cc}& =5\underset{n\to \infty }{lim}\frac{1}{n}[n+\frac{5}{n}\cdot \frac{(n-1)n}{2}]\\ & =5\underset{n\to \infty }{lim}\frac{1}{n}[n+\frac{5(n-1)}{2}]\end{array}$ $\begin{array}{cc}& =5\underset{n\to \infty }{lim}[1+\frac{5}{2}(1-\frac{1}{n})]\\ & =5[1+\frac{5}{2}]\\ & =5\left[\frac{7}{2}\right]\\ & =\frac{35}{2}\end{array}$

Q3. Evaluate the following definite integral as limit of sums : ${\int }_2^3{x}^{2}dx$

Answer. $\begin{array}{c}{\int }_a^{b}f\left(x\right)dx=(b-a){lim}_{n\to \infty }\frac{1}{n}\left[f\right(a)+f(a+h)+f(a+2h)\dots f\{a+(n-1)h\left\}\right],\text{where}h=\frac{b-a}{n}\\ \text{Here,}a=2,b=3,\text{and}f\left(x\right)=x^2\end{array}$ $\begin{array}{c}\Rightarrow h=\frac{3-2}{n}=\frac{1}{n}\\ \therefore {\int }_2^3{x}^{2}dx=(3-2){lim}_{n\to \infty }\frac{1}{n}\left[f\right(2)+f(2+\frac{1}{n})+f(2+\frac{2}{n})\dots f\{2+(n-1\left)\frac{1}{n}\right\}]\end{array}$ $\begin{array}{c}={lim}_{n\to \infty }\frac{1}{n}\left[\right(2{)}^2+{(2+\frac{1}{n})}^2+{(2+\frac{2}{n})}^2+\dots {(2+\frac{(n-1)}{n})}^2]\\ ={lim}_{n\to \infty }\frac{1}{n}[2^2+\{2^2+{\left(\frac{1}{n}\right)}^2+2\cdot 2\cdot \frac{1}{n}\}+\dots +\left\{\right(2{)}^2+\frac{(n-1{)}^2}{n^2}+2\cdot 2\cdot \frac{(n-1)}{n}\}]\end{array}$ $\begin{array}{cc}& =\underset{n\to \infty }{lim}\frac{1}{n}[(2^2+\dots +2^2)+\{{\left(\frac{1}{n}\right)}^2+{\left(\frac{2}{n}\right)}^2+\dots +{\left(\frac{n-1}{n}\right)}^2\}+2\cdot 2\cdot \{\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+\dots +\frac{(n-1)}{n}\}]\\ & =\underset{n\to \infty }{lim}\frac{1}{n}[4n+\frac{1}{n^2}\{1^2+2^2+3^2\dots +(n-1{)}^2\}+\frac{4}{n}\{1+2+\dots +(n-1)\left\}\right]\end{array}$ $\begin{array}{cc}& =\underset{n\to \infty }{lim}\frac{1}{n}[4n+\frac{1}{n^2}\left\{\frac{n(n-1)(2n-1)}{6}\right\}+\frac{4}{n}\left\{\frac{n(n-1)}{2}\right\}]\\ & =\underset{n\to \infty }{lim}\frac{1}{n}[4n+\frac{n(1-\frac{1}{n})(2-\frac{1}{n})}{6}+\frac{4n-4}{2}]\end{array}$ $\begin{array}{cc}& =\underset{n\to \infty }{lim}[4+\frac{1}{6}(1-\frac{1}{n})(2-\frac{1}{n})+2-\frac{2}{n}]\\ & =4+\frac{2}{6}+2\\ & =\frac{19}{3}\end{array}$

Q4. Evaluate the following definite integral as limit of sums : ${\int }_1^4(x^2-x)dx$

Answer. $\begin{array}{cc}\text{Let}I& ={\int }_1^4(x^2-x)dx\\ & ={\int }_1^4{x}^{2}dx-{\int }_1^{4}xdx\end{array}$ $\begin{array}{c}\text{Let}I=I_1-I_2,\text{where}{I}_1=\int x^{2}dx\text{and}{I}_2=\int xdx\\ \text{It is known that,}\\ {\int }_a^{b}f\left(x\right)dx=(b-a){lim}_{n\to \infty }\frac{1}{n}\left[f\right(a)+f(a+h)+f(a+(n-1)h\left)\right],\text{where}h=\frac{b-a}{n}\end{array}$ $\begin{array}{c}\text{For}{I}_1={\int }_1^4{x}^{2}dx\\ a=1,b=4,\text{and}f\left(x\right)=x^2\\ \therefore h=\frac{4-1}{n}=\frac{3}{n}\end{array}$ $\begin{array}{cc}{I}_1& ={\int }_1^4{x}^{2}dx=(4-1)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(1)+f(1+h)+\dots +f(1+(n-1)h\left)\right]\\ & =3\underset{n\to \infty }{lim}\frac{1}{n}[1^2+{(1+\frac{3}{n})}^2+{(1+2\cdot \frac{3}{n})}^2+\dots {(1+\frac{(n-1)3}{n})}^2]\\ & =3\underset{n\to \infty }{lim}\frac{1}{n}[1^2+\{1^2+{\left(\frac{3}{n}\right)}^2+2\cdot \frac{3}{n}\}+\dots +\{1^2+{\left(\frac{(n-1)3}{n}\right)}^2+\frac{2\cdot (n-1)\cdot 3}{n}\}]\end{array}$ $=3{lim}_{n\to \infty }\frac{1}{n}[(1^2+\dots +1^2)+{\left(\frac{3}{n}\right)}^2\{1^2+2^2+\dots +(n-1{)}^2\}+2\cdot \frac{3}{n}\{1+2+\dots +(n-1)\left\}\right]$ $\begin{array}{cc}& =3\underset{n\to \infty }{lim}\frac{1}{n}[n+\frac{9}{n^2}\left\{\frac{(n-1)\left(n\right)(2n-1)}{6}\right\}+\frac{6}{n}\left\{\frac{(n-1)\left(n\right)}{2}\right\}]\\ & =3\underset{n\to \infty }{lim}\frac{1}{n}[n+\frac{9n}{6}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{6n-6}{2}]\\ & =3\underset{n\to \infty }{lim}[1+\frac{9}{6}(1-\frac{1}{n})(2-\frac{1}{n})+3-\frac{3}{n}]\end{array}$ $\begin{array}{c}=3[1+3+3]\\ =3\left[7\right]\\ I_1=\mathrm{21...}\left(2\right)\\ \text{For}{I}_2=\int xdx\\ a=1,b=4,\text{and}f\left(x\right)=x\end{array}$ $\begin{array}{cc}\Rightarrow h& =\frac{4-1}{n}=\frac{3}{n}\\ \therefore I_2& =(4-1)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(1)+f(1+h)+\dots f(a+(n-1)h\left)\right]\\ & =3\underset{n\to 1}{lim}\frac{1}{n}[1+(1+h)+\dots +(1+(n-1)h\left)\right]\end{array}$ $\begin{array}{c}=3{lim}_{n\to \infty }\frac{1}{n}[1+(1+\frac{3}{n})+\dots +\{1+(n-1\left)\frac{3}{n}\right\}]\\ =3{lim}_{n\to \infty }\frac{1}{n}\left[\right(1+1+\dots +1)+\frac{3}{n}(1+2+\dots +(n-1)\left)\right]\\ =3{lim}_{n\to \infty }\frac{1}{n}[n+\frac{3}{n}\left\{\frac{(n-1)n}{2}\right\}]\end{array}$ $\begin{array}{cc}& =3\underset{n\to \infty }{lim}\frac{1}{n}[1+\frac{3}{2}(1-\frac{1}{n})]\\ & =3[1+\frac{3}{2}]\\ & =3\left[\frac{5}{2}\right]\end{array}$ $\begin{array}{c}{I}_2=\frac{15}{2}\\ \text{From equations}\left(2\right)\text{and}\left(3\right),\text{we obtain}\end{array}$ $I=I_1+I_2=21-\frac{15}{2}=\frac{27}{2}$

Q5. Evaluate the following definite integral as limit of sums : ${\int }_{-1}^1{e}^{x}dx$

Answer. $\begin{array}{c}\text{Let}I={\int }_1{e}^{x}dx\dots \left(1\right)\\ \text{It is known that,}\\ {\int }_a^{b}f\left(x\right)dx=(b-a){lim}_{n\to \infty }\frac{1}{n}\left[f\right(a)+f(a+h)\dots f(a+(n-1)h\left)\right],\text{where}h=\frac{b-a}{n}\end{array}$ $\begin{array}{c}\text{Here,}a=-1,b=1,\text{and}f\left(x\right)=e^x\\ \therefore h=\frac{1+1}{n}=\frac{2}{n}\end{array}$ $\therefore I=(1+1){lim}_{n\to \infty }\frac{1}{n}\left[f\right(-1)+f(-1+\frac{2}{n})+f(-1+2\cdot \frac{2}{n})+\dots +f(-1+\frac{(n-1)2}{n})]$ $\therefore I=(1+1){lim}_{n\to \infty }\frac{1}{n}\left[f\right(-1)+f(-1+\frac{2}{n})+f(-1+2\cdot \frac{2}{n})+\dots +f(-1+\frac{(n-1)2}{n})]$ $=2{lim}_{n\to \infty }\frac{1}{n}[e^{-1}+e^{(-1\frac{2}{n})}+e^{(-1+2\frac{2}{n})}+\dots e^{(-1+(n-1{)}^{\frac{2}{n}})}]$ $=2{lim}_{n\to \infty }\frac{1}{n}\left[e^{-1}\{1+e^{\frac{2}{n}}+e^{\frac{4}{n}}+e^{\frac{6}{n}}+e^{(n-1{)}^{\frac{2}{n}}}\}\right]$ $\begin{array}{c}=2{lim}_{n\to \infty }\frac{e^{-1}}{n}\left[\frac{e^{\frac{2n}{n}-1}}{e^{\frac{2}{n}-1}}\right]\\ =e^{-1}\times 2{lim}_{n\to \infty }\frac{1}{n}\left[\frac{e^2-1}{e^{\frac{2}{n}-1}}\right]\end{array}$ $=\frac{e^{-1}\times 2(e^2-1)}{{lim}_{\frac{2}{n}\to 0}\left(\frac{e^{\frac{2}{n}}-1}{n}\right)\times 2}$ $\begin{array}{ccc}& =e^{-1}\left[\frac{2(e^2-1)}{2}\right]& [\underset{h\to 0}{lim}\left(\frac{e^A-1}{h}\right)=1]\\ & =\frac{e^2-1}{c}& \\ & =(e-\frac{1}{e})\end{array}$

Q6. Evaluate the following definite integral as limit of sums : ${\int }_0^4(x+e^{2x})dx$

Answer. $\begin{array}{c}\text{It is known that,}\\ {\int }_a^{b}f\left(x\right)dx=(b-a){lim}_{n\to \infty }\frac{1}{n}\left[f\right(a)+f(a+h)+\dots +f(a+(n-1)h\left)\right],\text{where}h=\frac{b-a}{n}\\ \text{Here,}a=0,b=4,\text{and}f\left(x\right)=x+e^{2x}\\ \therefore h=\frac{4-0}{n}=\frac{4}{n}\end{array}$ $\Rightarrow {\int }_0^4(x+e^{2x})dx=(4-0){lim}_{n\to \infty }\frac{1}{n}\left[f\right(0)+f(h)+f(2h)+\dots +f((n-1)h\left)\right]$ $\begin{array}{c}=4{lim}_{n\to \infty }\frac{1}{n}[(0+e^0)+(h+e^{2h})+(2h+e^{22h})+\dots +\left\{\right(n-1)h+e^{2(n-1)h}\}]\\ =4{lim}_{n\to \infty }\frac{1}{n}[1+(h+e^{2h})+(2h+e^{4h})+\dots +\left\{\right(n-1)h+e^{2(n-1)k}\}]\end{array}$ $=4{lim}_{n\to \infty }\frac{1}{n}\left[\right\{h+2h+3h+\dots +(n-1)h\}+(1+e^{2h}+e^{4h}+\dots +e^{2(n-1)h})]$ $=4{lim}_{n\to \infty }\frac{1}{n}[\frac{\left(h\right(n-1\left)n\right)}{2}+\left(\frac{e^{2hin}-1}{e^{2h}-1}\right)]$ $=4{lim}_{n\to \infty }\frac{1}{n}[\frac{4}{n}\cdot \frac{(n-1)n}{2}+\left(\frac{e^8-1}{e^{\frac{8}{n}}-1}\right)]$ $=4\left(2\right)+4{lim}_{n\to \infty }\frac{(e^8-1)}{\frac{e^{\frac{8}{n}}-1}{\frac{8}{n}})8}$ $\begin{array}{c}=8+\frac{4\cdot (e^8-1)}{8}({lim}_{x\to 0}\frac{e^x-1}{x}=1)\\ =8+\frac{e^8-1}{2}\\ =\frac{15+e^8}{2}\end{array}$

Chapter-7 (integrals)

• Exercise 7.1
• Exercise 7.2
• Exercise 7.3
• Exercise 7.4
• Exercise 7.5
• Exercise 7.6
• Exercise 7.7
• Exercise 7.9
• Exercise 7.10
• Exercise 7.11