Question

The area bounded by the curve $y=x^2+x+4$, the x-axis and the ordinates $x=1$ and $x=3$ is

Answer 1

The required area A is entirely above the x-axis and so we can simply evaluate the integral between the required limits.

$A={\int }_1^{3}ydx$
$A={\int }_1^3(x^2+x+4)dx$
$A={[\frac{x^3}{3}+\frac{x^2}{2}+4x]}_1^3$
$A=[\frac{27}{3}+\frac{9}{2}+12]-[\frac{1}{3}+\frac{1}{2}+4]$
$A=12+8.66=20.66$