#### please solve RD sharma class 12 chapter 19 Definite Integrals exercise 19.5 question 2 maths textbook solution

8

Hint:

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

Given:

$\int_{0}^{2}\left ( x+3 \right )dx$

Solution:

We have,

$\int_{0}^{2}\left ( x+3 \right )dx$

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

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

Here,

$a=0,b=2,f(x)=(x+3)\\ h=\frac{2}{n}$

Thus, we have

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

\begin{aligned} &=\lim _{h \rightarrow 0} h\left[3 n+h\left(\frac{n(n-1)}{2}\right)\right] \\ &=\lim _{n \rightarrow \infty} \frac{2}{n}\left[3 n+\frac{2}{n} \frac{n(n-1)}{2}\right] \end{aligned}                           $\left[\begin{array}{l} \mathrm{Q} h \rightarrow 0 \& h=\frac{2}{n} \\ n \rightarrow \infty \end{array}\right]$

\begin{aligned} &=\lim _{n \rightarrow \infty}\left[6+\frac{2}{n} \cdot n^{2}\left(1-\frac{1}{n}\right)\right] \\ &=6+2=8\end{aligned}