# What is the area below the curve f(x)=x^3 -4x^2 +2x+ 8 and bounded by the y-axis, the x axis and x=3 expressed as an integral?

Jan 8, 2017

$S = \frac{69}{4}$
${\int}_{0}^{3} \left({x}^{3} - 4 {x}^{2} + 2 x + 8\right) \mathrm{dx} = {\left[{x}^{4} / 4 - 4 {x}^{3} / 3 + 2 {x}^{2} / 2 + 8 x\right]}_{0}^{3}$ =
$\frac{81}{4} - 36 + 9 + 24 = \frac{81}{4} - 3 = \frac{81 - 12}{4} = \frac{69}{4}$