# Question #31a2b

Jun 28, 2016

Use the reverse power rule to integrate $4 x - {x}^{2}$ from $0$ to $4$, to end up with an area of $\frac{32}{3}$ units.

#### Explanation:

Integration is used to find the area between a curve and the $x$- or $y$-axis, and the shaded region here is exactly that area (between the curve and the $x$-axis, specifically). So all we have to do is integrate $4 x - {x}^{2}$.

We also need to figure out the bounds of integration. From your diagram, I see that the bounds are the zeros of the function $4 x - {x}^{2}$; however, we have to find out numerical values for these zeros, which we can accomplish by factoring $4 x - {x}^{2}$ and setting it equal to zero:
$4 x - {x}^{2} = 0$
$x \left(4 - x\right) = 0$
$x = 0$$\textcolor{w h i t e}{X X} \mathmr{and} \textcolor{w h i t e}{X X}$$x = 4$

We will therefore integrate $4 x - {x}^{2}$ from $0$ to $4$:
${\int}_{0}^{4} 4 x - {x}^{2} \mathrm{dx}$
$= {\left[2 {x}^{2} - {x}^{3} / 3\right]}_{0}^{4} \to$ using reverse power rule ($\int {x}^{n} \mathrm{dx} = \frac{{x}^{n + 1}}{n + 1}$)
$= \left(\left(2 {\left(4\right)}^{2} - {\left(4\right)}^{3} / 3\right) - \left(2 {\left(0\right)}^{2} - {\left(0\right)}^{3} / 3\right)\right)$
$= \left(\left(32 - \frac{64}{3}\right) - \left(0\right)\right)$
$= \frac{32}{3} \approx 10.67$