#### please solve RD sharma class 12 chapter 19 Definite Integrals exercise 19.5 question 4 maths textbook solution

6

Hint:

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

Given:

$\int_{-1}^{1}(x+3) d x$

Solution:

We have,

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

Here,

\begin{aligned} &a=-1, b=1, f(x)=x+3 \\ &h=\frac{b-a}{n}=\frac{1-(-1)}{n}=\frac{2}{n} \end{aligned}

Thus, we have

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

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

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

\begin{aligned} &=\lim _{n \rightarrow 0} 4+2\left(1-\frac{1}{\infty}\right) \\ &=4+2=6 \end{aligned}