Question

Evaluate ${\int }_0^2{e}^{x}dx$ as the limit of a sum.

Answer 1

${\int }_a^b=\underset{h\to 0}{lim}h\left[f\right(a)+f(a+n)+f(a+2n)+...+f(a+(n-1)n\left)\right]$
$f\left(x\right)=e^x$
$f\left(a\right)=e^a$
$f(a+n)=e^{a+n}$
$f(a+(n-1\left)h\right)=e^{a+(n-1)h}$
${lim}_{h\to 0}h[e^a+e^{a+n}+e^{a+2n}+...+e^{a+(n-1)h}]$
$={lim}_{h\to 0}h\left[e^a\right(1+e^h+....+e^{(n-1)h}]$
$={lim}_{h\to 0}h{e}^a\frac{1.(e^h{)}^n-1}{e^n-1}$
$=e^a\frac{({lim}_{h\to 0}{e}^{nh}-1)}{{lim}_{h\to 0}\frac{(e^h-1)}{h}}$
$=e^a{lim}_{h\to 0}(e^{nh}-1)$ $\therefore h=(b-a)$
${\int }_a^b{e}^{x}dx=e^a(e^{b-a}-1)=e^b-e^a$
${\int }_0^2{e}^{x}dx=e^2-e^0=e^2-1$