Question

${\int }_0^{\pi }sinx=\underset{n\to \infty }{lim}{\Sigma }_{i=1}^{n}sin\left(\frac{\pi i}{n}\right)\frac{\pi }{n}$
State whether the above statement is True or False?

• A. True
• B. False

Answer 1

The correct option is A True

We have to state whether ${\int }_0^{\pi }sinxdx$ can be represented as ${lim}_{n\to \infty }{\sum }_{i=1}^{n}sin\left(\frac{\pi i}{n}\right)\frac{\pi }{n}$ is true or false.
We know that ${\int }_a^{b}f\left(x\right)dx={lim}_{n\to \infty }\left[\frac{b-a}{n}{\sum }_{i=1}^{n}f(a+i\frac{b-a}{n})\right]$
Here we have $f\left(x\right)=sinx,a=0,b=\pi$
Therefore ${\int }_0^{\pi }sinxdx={lim}_{n\to \infty }\left[\frac{\pi -0}{n}{\sum }_{i=1}^{n}f(0+i\frac{\pi -0}{n})\right]$
$={lim}_{n\to \infty }\left[\frac{\pi }{n}{\sum }_{i=1}^{n}f\left(\frac{\pi i}{n}\right)\right]$
$={lim}_{n\to \infty }\left[\frac{\pi }{n}{\sum }_{i=1}^{n}sin\left(\frac{\pi i}{n}\right)\right]$
${\int }_0^{\pi }sinxdx={lim}_{n\to \infty }{\sum }_{i=1}^{n}sin\left(\frac{\pi i}{n}\right)\frac{\pi }{n}$
Hence the answer is TRUE.