Show that
Question:

$\int_{2}^{3} x^{2} d x$

Solution:

It is known that,

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

Here, $a=2, b=3$, and $f(x)=x^{2}$

$\Rightarrow h=\frac{3-2}{n}=\frac{1}{n}$

$\therefore \int_{2}^{3} x^{2} d x=(3-2) \lim _{n \rightarrow \infty} \frac{1}{n}\left[f(2)+f\left(2+\frac{1}{n}\right)+f\left(2+\frac{2}{n}\right) \ldots f\left\{2+(n-1) \frac{1}{n}\right\}\right]$

$=\lim _{n \rightarrow \infty} \frac{1}{n}\left[(2)^{2}+\left(2+\frac{1}{n}\right)^{2}+\left(2+\frac{2}{n}\right)^{2}+\ldots\left(2+\frac{(n-1)}{n}\right)^{2}\right]$

$=\lim _{n \rightarrow \infty} \frac{1}{n}\left[2^{2}+\left\{2^{2}+\left(\frac{1}{n}\right)^{2}+2 \cdot 2 \cdot \frac{1}{n}\right\}+\ldots+\left\{(2)^{2}+\frac{(n-1)^{2}}{n^{2}}+2 \cdot 2 \cdot \frac{(n-1)}{n}\right\}\right]$

$=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\left(2^{2}+\ldots+2^{2}\right)+\left\{\left(\frac{1}{n}\right)^{2}+\left(\frac{2}{n}\right)^{2}+\ldots+\left(\frac{n-1}{n}\right)^{2}\right\}+2 \cdot 2 \cdot\left\{\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+\ldots+\frac{(n-1)}{n}\right\}\right]$

$=\lim _{n \rightarrow \infty} \frac{1}{n}\left[4 n+\frac{1}{n^{2}}\left\{1^{2}+2^{2}+3^{2} \ldots+(n-1)^{2}\right\}+\frac{4}{n}\{1+2+\ldots+(n-1)\}\right]$

$=\lim _{n \rightarrow \infty} \frac{1}{n}\left[4 n+\frac{1}{n^{2}}\left\{\frac{n(n-1)(2 n-1)}{6}\right\}+\frac{4}{n}\left\{\frac{n(n-1)}{2}\right\}\right]$

$=\lim _{n \rightarrow \infty} \frac{1}{n}\left[4 n+\frac{n\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)}{6}+\frac{4 n-4}{2}\right]$

$=\lim _{n \rightarrow \infty}\left[4+\frac{1}{6}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)+2-\frac{2}{n}\right]$

$=4+\frac{2}{6}+2$

$=\frac{19}{3}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.