please solve RD sharma class 12 chapter 19 Definite Integrals exercise 19.5 question 3 maths textbook solution

8

Hint:

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

Given:

$\int_{1}^{3}\left ( 3x-2 \right )dx$

Solution:

We have,

$\int_{1}^{3}\left ( 3x-2 \right )dx$

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

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

Here,

$a=1,b=3,f(x)=(3x-2)\\ h=\frac{2}{n}$

Thus, we have

\begin{aligned} I &=\int_{1}^{3}(3 x-2) 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[1+(3(1+h)-2)+\{3(1+2 h)-2\}+\ldots\{3(1+(n-1) h)-2\}] \\ &=\lim _{h \rightarrow 0} h[n+3 h(1+2+3+\ldots(n-1))] \\ \end{aligned}\begin{aligned} &=\lim _{h \rightarrow 0} h\left[n+3 h\left(\frac{n(n-1)}{2}\right)\right] \\ &=\lim _{n \rightarrow \infty} \frac{2}{n}\left[n+\frac{6}{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} 2+\frac{6}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right) \\ &=2+6=8 \end{aligned}