Question

Express the following integral as limits of sum and hence evalute it. ${\int }_0^4(x-x^2)dx$

Answer 1

We have,

$I={\int }_0^4(x-x^2)dx$

$I={\int }_0^{4}xdx-{\int }_0^4{x}^{2}dx$

$I={{\left[\frac{x^2}{2}\right]}_0}^4-{{\left[\frac{x^3}{3}\right]}_0}^4$

$I=\left[\frac{4^2-0^2}{2}\right]-\left[\frac{4^3-0^3}{3}\right]$

$I=\frac{16}{2}-\frac{64}{3}$

$=\frac{48-128}{6}$

$=-\frac{80}{6}$

$=-\frac{40}{3}$

Hence, the is the answer.