#### Provide solution for RD sharma maths class 12 chapter19 Definite Integrals exercise 19 point 5 question 8

$\frac{14}{3}$

Hint:

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

Given:

$\int_{0}^{2}\left(x^{2}+1\right) d x$

Solution:

We have,

$\int_{a}^{b}(f x) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots f(a+(n-1) h)]$

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

Here,

\begin{aligned} &a=0, b=2, f(x)=x^{2}+1 \\ &h=\frac{2}{n} \end{aligned}

Thus, we have

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

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

$=\lim _{n \rightarrow \infty} \frac{2}{n}\left[n+\frac{4}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)\right] \quad\left[\begin{array}{l} \mathrm{Q} h=-\mathrm{if} \\ n \\ h \rightarrow 0, n \rightarrow \infty \end{array}\right]$

\begin{aligned} &=\lim _{n \rightarrow \infty} 2+\frac{4}{3 n^{2}} \cdot n^{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right) \\ &=2+\frac{4}{3} \times 2 \times 1 \\ &=\frac{14}{3} \end{aligned}