#### Need solution for RD sharma maths class 12 chapter 19 Definite Integrals exercise 19.5 question 11

$\frac{4}{3}$

Hint:

To solve the given statement, we have to use the formula of addition limits.

Given:

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

Solution:

We have,

$\int_{a}^{b}\left ( fx \right )dx=\lim_{h\rightarrow 0}h\left [ f(a)+f(a+h)+f(a+2h)+.....f(a+(n-1)h) \right ]$

Where, $h=\frac{b-a}{n}$

Here,

$a=1,b=2,f(x)=x^{2}-1\\ h=\frac{2-1}{n}=\frac{1}{n}$

Thus, we have

\begin{aligned} I &=\int_{1}^{2}\left(x^{2}-1\right) d x \\ I &=\lim _{h \rightarrow 0} h[f(1)+f(1+h)+f(1+2 h)+\ldots f(1+(n-1)) h] \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} h\left[\left(1^{2}-1\right)\left((1+h)^{2}-1\right)+\left((1+2 h)^{2}-1\right)+\ldots+\left(1+((n-1) h)^{2}-1\right)\right] \\ &=\lim _{h \rightarrow 0} h\left[0+2 h(1+2+3+\ldots(n-1))+h^{2}\left(1^{2}+2^{2}+3^{2}+\ldots\right)\right] \\ \end{aligned}

\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{2}{n}\left(\frac{n(n-1)}{2}\right)+\frac{1}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)\right] \quad\left[\begin{array}{l} \mathrm{Q} h=\frac{1}{n} \text { if } \\ h \rightarrow 0, n \rightarrow \infty \end{array}\right] \\ \end{aligned}

\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{1}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right)+\frac{1}{6 n^{3}} \cdot n^{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right) \\ \end{aligned}

\begin{aligned} &=1+\frac{2}{6} \\ &=\frac{6+2}{6}=\frac{8}{6} \\ &=\frac{4}{3} \end{aligned}