Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.8

Last Updated : 23 Jul, 2025

NCERT solutions Class 12 math's Chapter 7 Exercise 7.8 consists of 6 questions that impart a clear understanding of the definite integrals, their intervals, and limits. Learn about the concept used and the solution to Chapter 7– Integrals Exercise 7.8 in this article.

Question 1: \int_{a}^{b} x \,dx

Solution:

We known that,

\int_{a}^{b} f(x) \,dx =(b-a) \lim_{n\to\infty}\frac{1}{n} [f(a)+f(a+h)+.....f(a+(n-1)h)]where \,, h = \frac{b-a}{n}

Here, a = a, b = b and f(x) = x

\int_{a}^{b} x \,dx =(b-a) \lim_{n\to\infty}\frac{1}{n} [(a+a+...+a\,(n\,times))+(h+2h+..+(n-1)h)]

=(b-a) \lim_{n\to\infty}\frac{1}{n} [na+h(1+2+3..(n-1))]

=(b-a) \lim_{n\to\infty}\frac{1}{n} [na+h{\frac{(n-1)(n)}{2}}]

=(b-a) \lim_{n\to\infty}\frac{n}{n} [a+{\frac{(n-1)(h)}{2}}]

=(b-a) \lim_{n\to\infty} [a+{\frac{(n-1)(h)}{2}}]

=(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}}]

=(b-a) [{\frac{2a+b-a}{2}}]

={\frac{(b-a)(b+a)}{2}}

={\frac{(b^2-a^2)}{2}}

Question 2:\int_{0}^{5} (x+1) \,dx

Solution:

Let I = \int_{0}^{b} (x+1) \,dx

\int_{a}^{b} f(x) \,dx =(b-a) \lim_{n\to\infty}\frac{1}{n} [f(a)+f(a+h)+.....f(a+(n-1)h)] where \,, h = \frac{b-a}{n}

Here, a = 0, b = b and f(x) = (x+1)

=> h = \frac{5-0}{n} = \frac{5}{n}

\int_{0}^{5} (x+1) \,dx = (5-0) \lim_{n\to\infty}\frac{1}{n} [f(0)+f(\frac{5}{n})+f(\frac{(n-1)5}{n})]

=5 \lim_{n\to\infty}\frac{1}{n} [1+(\frac{5}{n}+1)+...+[1+(\frac{(n-1)5}{n})]

=5 \lim_{n\to\infty}\frac{1}{n} [(1+1..+1(n\, times)\,)+[\frac{5}{n}+2.\frac{5}{n}+3.\frac{5}{n}+...+(\frac{(n-1)5}{n})]

=5 \lim_{n\to\infty}\frac{1}{n} [n+{\frac{5(n-1).n}{n.2}}]

=5 \lim_{n\to\infty}\frac{1}{n} [n+{\frac{5(n-1)}{2}}]

=5 \lim_{n\to\infty} [1+{\frac{5(1-\frac{1}{n})}{2}}]

= 5[1+\frac{5}{2}]

= 5[\frac{7}{2}]

= [\frac{35}{2}]

Question 3:\int_{2}^{3} x^2 \,dx

Solution:

We known that,

\int_{a}^{b} f(x) \,dx =(b-a) \lim_{n\to\infty}\frac{1}{n} [f(a)+f(a+h)+.....f(a+(n-1)h)] where \,, h = \frac{b-a}{n}

Here, a = 2, b = 3 and f(x) = x2 => h = (3-2)/n = 1/n

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

=\lim_{n\to\infty}\frac{1}{n} [(2)^2+(2+\frac{1}{n}))^2+...+[(2+(\frac{(n-1)}{n}))^2]

=\lim_{n\to\infty}\frac{1}{n} [(2)^2+[2^2+[\frac{1}{n}]^2+2.2.\frac{1}{n}]+...+[2^2+(\frac{(n-1)}{n})^2+2.2.\frac{n-1}{n}]\,]

=\lim_{n\to\infty}\frac{1}{n} [(2^2+2^2..+2^2(n\, times)\,)+[(\frac{1}{n})^2+(\frac{2}{n})^2+....+(\frac{n-1}{n})^2]+2.2\{\frac{1}{n}+\frac{2}{n}+....+\frac{(n-1)}{n}\}]
=\lim_{n\to\infty}\frac{1}{n} [ \,4n+\frac{1}{n^2}\{1^2+2^2+...+(n-1)^2\}+\frac{4}{n}\{1+2+...+(n-1)\}] \,

=\lim_{n\to\infty}\frac{1}{n} [ \,4n+\frac{1}{n^2}\{\frac{n(n-1)(2n-1)}{6}\}+\frac{4}{n}\{\frac{n(n-1)}{2}\}] \,

=\lim_{n\to\infty}\frac{1}{n} [ \,4n+\{\frac{n(1-\frac{1}{n})(2-\frac{1}{n})}{6}\}+\{\frac{4(n-1)}{2}\}] \,

=\lim_{n\to\infty} [ \,4+\frac{1}{6}(1-\frac{1}{n})(2-\frac{1}{n})+2-\frac{2}{n}] \,

= 4+\frac{2}{6}+2 = \frac{19}{3}

Question 4: \int_{1}^{4} (x^2-x) \,dx

Solution:

Let I =\int_{1}^{4} (x^2-x) \,dx

=\int_{1}^{4} x^2 \,dx -\int_{1}^{4} x \,dx

Let \, I = I_1-I_2,where I1=\int_{1}^{4} x^2 \,dx \, and \, I2=\int_{1}^{4} x \,dx ..........(1)

We known that,

\int_{a}^{b} f(x) \,dx =(b-a) \lim_{n\to\infty}\frac{1}{n} [f(a)+f(a+h)+.....f(a+(n-1)h)] where \,, h = \frac{b-a}{n}

For I_1=\int_{1}^{4} x^2 \,dx \,

a = 1, b = 4, and f(x) = x2

h = (4-1)/n = 3/n

I_1=\int_{1}^{4} x^2 \,dx \,=(4-1) \lim_{n\to\infty}\frac{1}{n} [f(1)+f(1+h)+...+f(1+(n-1)h)]

=3\lim_{n\to\infty}\frac{1}{n} [(1)^2+(1+\frac{3}{n}))^2+...+[(1+(\frac{(n-1)3}{n}))^2]

=3\lim_{n\to\infty}\frac{1}{n} [(1)^2+[1^2+[\frac{3}{n}]^2+2.3.\frac{1}{n}]+...+[1^2+(\frac{(n-1)3}{n})^2+2.3.\frac{n-1}{n}]\,]

=3\lim_{n\to\infty}\frac{1}{n} [1^2+1^2+..+1^2(n\, times)\,)+(\frac{3}{n})^2[ \,1^2+2^2+...+(n-1)^2] \,+2.\frac{3}{n}\{1+2++...+(n-1)\}] \,

=3\lim_{n\to\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_{n\to\infty}\frac{1}{n} [ \,n+\{\frac{9n(1-\frac{1}{n})(2-\frac{1}{n})}{6}\}+\{\frac{6(n-1)}{2}\}] \,

=3\lim_{n\to\infty} [ \,1+\frac{9}{6}(1-\frac{1}{n})(2-\frac{1}{n})+3-\frac{3}{n}] \,

= 3[1+3+3]

I1 = 3[7] = 21 .............(2)

For I2 =\int_{1}^{4} x \,dx

a = 1, b = 4 and f(x) = x

=> h = (4-1)/n = 3/n

I_2=(4-1) \lim_{n\to\infty}\frac{1}{n} [f(1)+f(1+h)+...+f(1+(n-1)h)]

=3\lim_{n\to\infty}\frac{1}{n} [(1)+(1+\frac{3}{n})+...+[(1+(\frac{(n-1)3}{n})]

=3\lim_{n\to\infty}\frac{1}{n} [1+1+...+1(n\, times)\,)+\frac{3}{n}[ \,1+2+..+(n-1)] \,

=3\lim_{n\to\infty}\frac{1}{n} [ \,n+\frac{3}{n}\{\frac{n(n-1)}{2}\}] \,

=3\lim_{n\to\infty} \frac{1}{n}[ \,1+\frac{3}{2}(1-\frac{1}{n})

=3[ \,1+\frac{3}{2}] \,

I_2 = \frac{15}{2} \,\,\,..........(3)

From eq (2) and (3) we obtain,

I = I1 - I2 = 21 - \frac{15}{2} = \frac{27}{2}

Question 5:\int_{-1}^{1} {\rm e^x} \,dx

Solution:

Let I = \int_{-1}^{1} {\rm e^x} \,dx

We known that,

\int_{a}^{b} f(x) \,dx =(b-a) \lim_{n\to\infty}\frac{1}{n} [f(a)+f(a+h)+.....f(a+(n-1)h)] where \,, h = \frac{b-a}{n}

Here a = -1, b = 1 and f(x) = ex

=> h = (1+1)/n = 2/n

=(1+1)\lim_{n\to\infty}\frac{1}{n} [f(-1)+f(-1+\frac{2}{n})+..+f(-1+\frac{2(n-1)}{2})]

=2\lim_{n\to\infty}\frac{1}{n} [ e^{(-1)}+e^{(-1+\frac{2}{n})}+..+e^{(-1+\frac{2(n-1)}{n})}]

=2\lim_{n\to\infty}\frac{1}{n} [ e^{-1}\{1+e^{\frac{2}{n}}+e^{\frac{4}{n}}+e^{\frac{6}{n}}+...+e^{\frac{2(n-1)}{n}}\}]

=2\lim_{n\to\infty}\frac{e^{-1}}{n} [ \,\frac{e^{\frac{2n}{n}-1}}{e^{\frac{2}{n}-1}}] \,

=e^{-1}*2\lim_{n\to\infty}\frac{1}{n} [ \,\frac{e^{2-1}}{e^{\frac{2}{n}-1}}] \,

= \frac{e^{-1}*2(e^{2}-1)}{\lim_{\frac{2}{n}\to0}( \,\frac{e^{\frac{2}{n}}-1}{\frac{2}{n}}) \,*2 }

= e^{-1}[ \,\frac{2(e^{2}-1)}{2}] \,

=\frac{e^{2}-1}{e}

=\{e-\frac{1}{e}\}

Question 6:\int_{0}^{4} (x+{\rm e^{2x}}) \,dx

Solution:

Let I = \int_{0}^{4} (x+{\rm e^{2x}}) \,dx

We known that,

\int_{a}^{b} f(x) \,dx =(b-a) \lim_{n\to\infty}\frac{1}{n} [f(a)+f(a+h)+.....f(a+(n-1)h)] where h = \frac{b-a}{n}

Here a = 0, b = 4, and f(x) = x+e2x

h = \frac{4-0}{n} = \frac{4}{n}

=> \int_{0}^{4} (x+{\rm e^{2x}}) \,dx =(4-0)\lim_{n\to\infty}\frac{1}{n} [(f(0)+f(h)+f(2h)+....+f((n-1)h)]

=4\lim_{n\to\infty}\frac{1}{n} [(0+{\rm e}^0)+(h+{\rm e}^{2h})+(2h+e^{2.2h})+..+\{(n-1)h+e^{2(n-1)h}\}]

=4\lim_{n\to\infty}\frac{1}{n} [1+(h+{\rm e}^{2h})+(2h+e^{4h})+..+\{(n-1)h+e^{2(n-1)h}\}]

=4\lim_{n\to\infty}\frac{1}{n} [\{h+2h+3h+...+(n-1)h\}+(1+e^{2h}+e^{4h}+..+e^{2(n-1)h}]

=4\lim_{n\to\infty}\frac{1}{n} [\{h\{1+2+..(n-1)\}+ (\frac{e^{2hn}-1}{e^{2h}-1})]

=4\lim_{n\to\infty}\frac{1}{n} [\frac{4}{n}.\frac{n(n-1)}{2}+\{\frac{e^8-1}{e\frac{8}{n}-1}\}

=4(2)+4\lim_{n\to\infty} \frac{e^8-1}{(\frac{e\frac{8}{n}-1}{\frac{8}{n}})8}

=8+\frac{4(e^8-1)}{8}

=8+\frac{e^8-1}{2} = \frac{15+e^8}{2}

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