Question

Obtain as the limit of sum $\stackrel{lo{g}_e^7}{\underset{lo{g}_e^3}{\int }}{e}^{x}dx$.

Answer 1

Here $a=log3$ and b$=$log $7$ is given and $f\left(x\right)=e^x$
Function $f\left(x\right)$ is continuous in interval $[log3,log5]$ of set R. We devide interval $[log3,log5]$ in n sub intervals with length of each intervals $h=\frac{b-a}{n}$
$\therefore h=\frac{log7-log3}{n}$
$\therefore nh=log\left(\frac{7}{3}\right)$
Ans As $n\to \infty$ then $h\to 0$
Here $f\left(x\right)=e^x$
$\therefore f(a+ih)=e^a+ih$
$=e^a\cdot eih$
$=e^{lo{g}_e^3}\cdot e^{ih}$
$=3e^{ih}$
Now $\sum _{i=1}^{n}f(a+ih)$
$=\sum _{i=1}^{n}3e^{ih}$
$=3[e^h+e^{2h}+e^{3h}+...+e^{nh}]$
Here first term of geometric series a$=e^h$
And common ratio $r=e^h>1$
$\therefore$ Required sum $=a\left(\frac{r^h-1}{r-1}\right)$
$=e^h\left(\frac{e^{nh}-1}{e^h-1}\right)$
$\therefore \sum _{i=1}^{n}f(a+ih)=3\left[\frac{e^h(e^{nh}-1)}{e^h-1}\right]$
$\therefore$ Limit of sum $=\underset{h\to 0}{lim}h\sum _{r=1}^{n}f(a+ih)$
$=\underset{h\to 0}{lim}3h\left[\frac{e^h(e^{nh}-1)}{e^h-1}\right]$
$=3\left[\underset{h\to 0}{lim}\frac{e^h(e^{log\left(\frac{7}{3}\right)}-1)}{\left(\frac{e^h-1}{h}\right)}\right]$
$=3\frac{\left(e^0\right)(7_{/3}-1)}{lo{g}_{e}e}$
$=3\left(\frac{7-3}{3}\right)$ $(\because loge=1)$
$=\frac{3\left(4\right)}{3}$
$=4$
$\therefore$ Required limit if sum $=4$.