Question

Obtain: ${\int }_a^b\sin xdx$ as limit of sum.

Answer 1

Here lower limit $=a$, and upper limit $=b,f\left(x\right)=\sin x$
Now, By definition
$f(A+ih)=f(a+ih)=\sin (a+ih)$
$f$ is continuous on $(a,b)$ and we divide $(a,b)$ into $n$ sub-intervals of equal length.
$h=\frac{b-a}{n}$
$\therefore nh=b-a$
Also as $n\to \infty ,h\to 0$
${\int }_a^b\sin xdx={lim}_{h\to 0}h{\sum }_{i=1}^{n}f(a+ih)$
${\sum }_{i=1}^{n}f(a+ih)={lim}_{h\to 0}h{\sum }_{i=1}^n\sin (a+ih)={lim}_{h\to 0}h\frac{[\cos (a+\frac{h}{2})-\cos (a+nh+\frac{h}{2})]}{2\sin \frac{h}{2}}$
$={lim}_{h\to 0}h\frac{[\cos (a+\frac{h}{2})-\cos (b+\frac{h}{2})]}{\frac{\sin \frac{h}{2}}{\frac{h}{2}}}$
$\therefore {\int }_a^b\sin xdx=\cos a-\cos b$ is required limit of sum