#### Explain solution rd sharma class 12 chapter 19 Definite Integrals exercise 19.5, question 14

$\frac{7}{2}$

Hint:

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

Given:

$\int_{1}^{0}\left ( 3x^{2}-5x \right )dx$

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,

$a=0,b=1,f(x)=3x^{2}+5x\\ h=\frac{1-0}{n}=\frac{1}{n}$

Thus, we have

\begin{aligned} I &=\int_{0}^{1}\left(3 x^{2}+5 x\right) d x \\ I &=\lim _{h \rightarrow 0} h[f(0)+f(h)+f(2 h)+\ldots f(0+(n-1)) h] \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} h\left[0+\left(3 h^{2}+5 h\right)+\left(3(2 h)^{2}+5(2 h)\right)+\ldots\right] \\ &=\lim _{h \rightarrow 0} h\left[3 h^{2}\left(1^{2}+2^{2}+3^{2}+\ldots(n-1)^{2}\right)+5 h(1+2+3+\ldots(n-1))\right] \\ \end{aligned}

\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{3}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)+\frac{5}{n}\left(\frac{n(n-1)}{2}\right)\right] \\ &=\lim _{n \rightarrow \infty} \frac{3}{n^{3}} \cdot \frac{n^{3}}{6}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)+\frac{5}{2 n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right) \\ \end{aligned}

\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{1}{2}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)+\frac{5}{2}\left(1-\frac{1}{n}\right) \\ \end{aligned}

\begin{aligned} &=\frac{1}{2}(1-0)(2-0)+\frac{5}{2}(1-0) \\ &=1+\frac{5}{2} \\ &=\frac{2+5}{2} \\ &=\frac{7}{2} \end{aligned}