Question

Evaluate ${\int }_{-1}^2(e^{3x}+7x-5)dx$ as a limit of sums.

Answer 1

${\int }_{-1}^2(e^{3x}+7x-5)dx$

Here, $a=-1,b=2,nh=2-(-1)=3$

$=\underset{h\to 0}{lim}h\left[f\right(a)+f(a+h)+f(a+2h)+...+f(a+\left(\overline{n-1}\right)\left)h\right]$

$f\left(a\right)=f(-1)=e^{-3}-7-5$

$f(a+h)=f(h-1)=e^{3(h-1)}+7(h-1)-5$

$f(a+2h)=f(2h-1)=e^{3(2h-1)}+7(2h-1)-5$

$f(a+\overline{n-1}h)=f(\overline{n-1}h-1)$

$=e^{3(\overline{n-1}h-1)}+7(\overline{n-1}h-1)-5$

$=\underset{h\to 0}{lim}h\left[\right(e^{-3}-7-5)+(e^{3h}.e^{-3}+7h-7-5)+(e^{6h}.e^{-3}+14h-7-5)+...+e^{3(n-1)h}.e^{-3}+7(n-1)h-7-5]$

$=\underset{h\to 0}{lim}h\left[e^{-3}\right(1+e^{3h}+e^{2\left(3h\right)+...+e^{3(n-1)h}})-7n-5n+7h+2(7h)+...+(n-1\left)7h\right]$

$=\underset{h\to 0}{lim}h[e^{-3}\left(\frac{(e^{3h}{)}^n-1}{e^{3h}-1}\right)-7n-5n+7h\frac{\left(\right(n-1)\times n)}{2}]$

$=\underset{h\to 0}{lim}[e^{-3}.\frac{e^{3nh}-1}{3\times \frac{e^{3h}-1}{3h}}-7nh-5nh+\frac{7}{2}(nh-h\left)\right(nh\left)\right]$

$=[\frac{1}{3e^3}\times \frac{e^9-1}{1}-7(3)-5(3)+\frac{7}{2}(3-0\left)\right(3\left)\right]$

$=\frac{e^9-1}{3e^3}-21-15+\frac{7}{2}\times 9$

$=\frac{e^9-1}{3e^3}+\frac{63}{2}-36$

$=\frac{e^9-1}{3e^3}+\frac{63-72}{2}$

$=\frac{e^9-1}{3e^3}-\frac{9}{2}$