Question

Evaluate ${\int }_0^3(2x^2+3x+5)dx$ as limit of a sum.

Answer 1

@page { margin: 2cm } p { margin-bottom: 0.25cm; line-height: 120% }

$\underset{0}{\overset{3}{\int }}\left(2x^2+3x+5\right)dx$

here ,

$f\left(x\right)=2x^2+3x+5a=0,b=3$

$nh=b-a=3-0=3$

$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{h\to 0}{lim}h\left[f\left(a\right)+f\left(a+h\right)+....+f\left(a+\left(n-1\right)h\right)\right]$

$\Rightarrow \underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{h\to 0}{lim}h\left[f\left(0\right)+f\left(h\right)+f\left(2h\right)+...+f\left(n-1\right)h\right]$

$=\underset{h\to 0}{lim}h\left[5+\left(2h^2+3h+5\right)+\left(8h^2+6h+5\right)+.....+\left\{2{\left(n-1\right)}^{2}h+3\left(n-1\right)h+5\right\}\right]$

$=\underset{h\to 0}{lim}h\left[5n+h^2\left(2+8+.......+2{\left(n-1\right)}^2\right)+h\left(3+6+.....3\left(n-1\right)\right)\right]$

$=\underset{h\to 0}{lim}h\left[5n+2h^2\left(1+4+.......+{\left(n-1\right)}^2\right)+3h\left(1+2+.....\left(n-1\right)\right)\right]$

$=\underset{h\to 0}{lim}h\left[5n+2h^2\times \frac{n\left(n-1\right)\left(2n-1\right)}{6}+3h\frac{n\left(n-1\right)}{2}\right]$

$=\underset{h\to 0}{lim}h\left[5nh+\frac{nh\left(nh-h\right)\left(2nh-h\right)}{3}+\frac{3nh\left(nh-h\right)}{2}\right]$

$=\underset{h\to 0}{lim}h\left[5\times 3+\frac{3\left(3-h\right)\left(6-h\right)}{3}+\frac{9\left(3-h\right)}{2}\right]$

$=15+3\times 6+\frac{9\times 3}{2}$

$=15+18+\frac{27}{2}$

$=33+\frac{27}{2}$

$=\frac{66+27}{2}$

$=\frac{93}{2}$