Question

Evaluate ${\int }_1^3(2x^2+5x)dx$ as a limit of a sum.

Answer 1

Let $f\left(x\right)=2x^2+5x$

Here $a=1,b=3$

Therefore, $h=\frac{b-a}{n}=\frac{3-1}{n}=\frac{2}{n}$

$\Rightarrow nh=2$

Also $n\to \infty \iff h\to 0$

We know that, ${\int }_b^{a}f\left(x\right)dx=\underset{h\to 0}{lim}h\left[f\right(a)+f(a+h)+...+f(a+(n-1)h\left)\right]$

${\int }_1^3(2x^2+5x)dx=\underset{h\to 0}{lim}h\left[f\right(1)+f(1+h)+...+f(1+(n-1)h\left)\right]$

$={lim}_{h\to 0}h\left[\right(2\times 1^2+5\times 1)+2(1+h{)}^2+5(1+h)+....+2(1-(n-1)h^2+5(1+(n-1)d\left)\right)]$

$={lim}_{h\to 0}h\left[\right(2+5)+(2+4h+2h^2+5+5h)+....+(2+4(n-1)h+2(n-1{)}^2{h}^2+5+5(n-1\left)h\right)]$

$={lim}_{h\to 0}h\left[\right(7)+(7+9h+2h^2)+....+(7+9(n-1)h+2(n-1{)}^2{h}^2]$

$={lim}_{h\to 0}h\left[\right(7n+9h(1+2+....+n-1)+2h^2(1^2+2^2+....(n-1{)}^2\left)\right)]$

$={lim}_{h\to 0}[7nh+9h^2\frac{n(n-1)}{2}+2h^3\frac{(n-1).n(2n-1)}{6}]$

$={lim}_{h\to 0}[7nh+\frac{9(nh{)}^2.(1-\frac{1}{n})}{2}+\frac{2(nh{)}^3.(1-\frac{1}{n}).(2-\frac{1}{n})}{6}]$

$={lim}_{n\to \infty }[14+\frac{36(1-\frac{1}{n})}{2}+\frac{16(1-\frac{1}{n}).(2-\frac{1}{n})}{6}]$ [$\therefore nh=2$]

$={lim}_{n\to \infty }[14+18(1-\frac{1}{n})+\frac{8}{3}(1-\frac{1}{n}).(2-\frac{1}{n})]$

$14+18+\frac{8}{3}\times 1\times 2$

$=32+\frac{16}{3}=\frac{96+16}{3}=\frac{112}{3}$.