#### Explain solution rd sharma class 12 chapter 19 Definite Integrals exercise 19.5, question 13

$\frac{27}{2}$

Hint:

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

Given:

$\int_{4}^{1}\left ( x^{2}-x \right )dx$

Solution:

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

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

Here,

$a=1,b=4,f(x)=x^{2}-x\\ h=\frac{4-1}{n}=\frac{3}{n}$

Thus, we have

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

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

\begin{aligned}&=\lim _{n \rightarrow \infty} \frac{3}{n}\left[\frac{3}{n}\left(\frac{n(n-1)}{2}\right)+\frac{9}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)\right]\left[\begin{array}{c} \mathrm{Q} h=\frac{3}{n} \text { if } \\ h \rightarrow 0, n \rightarrow \infty \end{array}\right] \\ \end{aligned}

\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{9}{2 n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right)+\frac{9}{2 n^{3}} \cdot n^{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right) \\ &=\lim _{n \rightarrow \infty} \frac{9}{2}\left(1-\frac{1}{n}\right)+\frac{9}{2}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right) \end{aligned}

$=\frac{9}{2}\left ( 1-0 \right )+\frac{9}{2}\left ( 1-0 \right )(2-0)\\ =\frac{9}{2}+9\\ =\frac{9+18}{2}\\ =\frac{27}{2}$