Question

Evaluate the following definite integrals as limit of sums.
${\int }_{-1}^1{e}^{x}dx$.

• A. $e^2-1$
• B. $\frac{e^2-1}{e}$
  
• C. $\frac{e^2-1}{2}$
  
• D. $\frac{e^2-1}{e^2}$

Answer 1

The correct option is B $\frac{e^2-1}{e}$

Here, the function $f\left(x\right)=e^x$ is continuous in $[-1,1]$. Since dividing the n length of $[-1,1]$ in the same length, the length of each subser is h. i.e., $h=\frac{b-a}{n}=\frac{1-(-1)}{n}=\frac{2}{n}\Rightarrow nh=2$

here $a=-1$ & $b=1$

Now, $f(a+ih)=f(-1+ih)=e^{-1+ih}=e^{-1}\cdot e^{ih}=\frac{e^{ih}}{e}$

By definition, ${\int }_{-1}^1{e}^{x}dx=\underset{h\to 0}{lim}h\sum _{i=1}^{n}f(a+ih)$

So, ${\int }_{-1}^1{e}^{x}dx=\underset{h\to 0}{lim}h\sum _{i=1}^n\left(\frac{e^{ih}}{e}\right)$

$=\underset{h\to 0}{lim}\frac{h}{e}\times [e^h+e^{2h}+e^{3h}+......+e^{nh}]$

Here we can see a geometric progression where $a=e^h$ & $r=\frac{e^{2h}}{e^h}=e^h$, $S_n=\frac{a(r^n-1)}{r-1}$

So, $=\underset{h\to 0}{lim}\frac{h}{e}\times \frac{e^h(e^{nh}-1)}{(e^h-1)}$ (here $nh=2$)

$=\frac{e^2-1}{e}\underset{h\to 0}{lim}\frac{e^h}{\left(\frac{e^h-1}{h}\right)}$

$=\frac{e^2-1}{e}\times \frac{e}{1}$

${\int }_{-1}^1{e}^{x}dx=\frac{e^2-1}{e}$.