Question

Evaluate :
${\int }_1^3(x^2+3x+e^x)dx$,
as the limit of the sum.

Answer 1

Let $I={\int }_1^3(x^2+3x+e^x)dx$

Here, $f\left(x\right)=x^2+3x+e^x,a=1,b=3$ and $nh=3-1=2$.

Now, ${\int }_1^3(x^2+3x+e^x)dx$

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

= $\underset{h\to 0}{lim}\left[\begin{array}{c}(1+h{)}^2+3(1+h)+e^{(1+h)}+(1+2h{)}^2)\\ +3(1+2h)+e^{(1+2h)+...+(1+nh{)}^2}\\ +3(1+nh)+e^{(1+nh)}\end{array}\right]$

= $\underset{h\to 0}{lim}\left[\begin{array}{c}1+h^2+2h+3+3h+e.e^h+1+2^2{h}^2\\ +4h+3+6h+e.e^{2h}+...+1+n^2{h}^2\\ +2nh+3+3nh+e.e^{nh}\end{array}\right]$

=$\underset{h\to 0}{lim}\left[\begin{array}{c}n+h^2(1+2^2+...+n^2)\\ +2h(1+2+...+n)+3n\\ +3h(1+2+...+n)\\ +e.(e^h+e^{2h}+...+e^{nh})\end{array}\right]$

=$\underset{h\to 0}{lim}\left[\begin{array}{c}n+h^2\frac{n(n+1)(2n+1)}{6}+2h\frac{n(n+1)}{2}\\ +3n+3h\frac{n(n+1)}{2}+e.e^h\left(\frac{e^{nh}-1}{e^h-1}\right)\end{array}\right]$

= $\underset{h\to 0}{lim}\left[\begin{array}{c}nh+\frac{nh(nh+h)(2nh+h)}{6}+nh(nh+h)\\ +3nh+\frac{3}{2}nh(nh+h)+h.e.e^h\left(\frac{e^{nh}-1}{e^h-1}\right)\end{array}\right]$
$=2+\frac{2\times 2\times 4}{6}+2\times 2+3\times 2+\frac{3}{2}\times 2\times 2+e\cdot \stackrel{lim}{h\to 0}{e}^h\left(\frac{e^2-1}{\frac{e^h-1}{h}}\right)$

= $2+\frac{16}{6}+4+6+6+e.e^0\left(\frac{e^2-1}{1}\right)$

= $\frac{62}{3}+e(e^2-1)$

=$\frac{62}{3}+e^3-e$