#### please solve RD sharma class 12 chapter 19 Definite Integrals exercise 19.5 question 1 maths textbook solution

$\frac{33}{2}$

Hint:

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

Given:

$\int_{0}^{3}\left ( x+4 \right )dx$

Solution:

We have,

$\int_{0}^{3}\left ( x+4 \right )dx$

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

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

Here,

$a=0,b=3,f(x)=(x+4)\\ h=\frac{3}{n}\Rightarrow nh=3$

Thus, we have

\begin{aligned} I &=\int_{0}^{3}(x+4) 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[4+(h+4)+(2 h+4)+\ldots((n-1) h+4)] \\ &=\lim _{h \rightarrow 0} h[4 n+h(1+2+3+\ldots(n-1))] \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} h\left[4 n+h\left(\frac{n(n-1)}{2}\right)\right] \\ &=\lim _{n \rightarrow \infty} \frac{3}{n}\left[4 n+\frac{3}{n} \frac{\left(n^{2}-1\right)}{2}\right] \\ \end{aligned}

\begin{aligned} &=\lim _{n \rightarrow \infty} 12+\frac{9}{2}\left(1-\frac{1}{n}\right) \\ &=12+\frac{9}{2}=\frac{33}{2} \end{aligned}