Need solution for RD sharma maths class 12 chapter 19 Definite Integrals exercise 19.5 question 10

$\frac{43}{3}$

Hint:

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

Given:

$\int_{2}^{3}\left(2 x^{2}+1\right) d x$

Solution:

We have,

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

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

Here,

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

Thus, we have

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

\begin{aligned} &=\lim _{h \rightarrow 0} h\left[\left(2 \times 2^{2}+1\right)\left(2(2+h)^{2}+1\right)+\left(2(2+2 h)^{2}+1\right)+\ldots+\left(2(2+(n-1) h)^{2}+1\right)\right] \\ &=\lim _{h \rightarrow 0} h\left[9 n+8 h(1+2+3+\ldots)+2 h^{2}\left(1^{2}+2^{2}+3^{2}+\ldots\right)\right] \\ \end{aligned}

\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{1}{n}\left[9 n+\frac{8}{n}\left(\frac{n(n-1)}{2}\right)+\frac{2}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)\right] \end{aligned}

\begin{aligned} &=\lim _{n \rightarrow \infty} 9+\frac{4}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right)+\frac{1}{3 n^{3}} \cdot n^{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right) \\ \end{aligned}

\begin{aligned} &=9+4-0+\frac{2}{3} \\ &=\frac{29+12+2}{3}=\frac{43}{3} \end{aligned}