How do you find the area between f(x)=2-1/2x and g(x)=2-sqrtx?

Nov 12, 2016

The area is $\frac{4}{3}$ square units.

Explanation:

Start by finding the intersection points.

$\left\{\begin{matrix}y = 2 - \frac{1}{2} x \\ y = 2 - \sqrt{x}\end{matrix}\right.$

$2 - \sqrt{x} = 2 - \frac{1}{2} x$

$\frac{1}{2} x = \sqrt{x}$

${\left(\frac{1}{2} x\right)}^{2} = {\left(\sqrt{x}\right)}^{2}$

$\frac{1}{4} {x}^{2} = x$

$\frac{1}{4} {x}^{2} - x = 0$

$x \left(\frac{1}{4} x - 1\right) = 0$

$x = 0 \mathmr{and} 4$

$\therefore y = 2 - \sqrt{x}$

$y = 2 - \sqrt{0} \mathmr{and} y = 2 - \sqrt{4}$

$y = 2 \mathmr{and} 0$

The intersection points are hence $\left\{0 , 2\right\}$ and $\left\{4 , 0\right\}$.

Now, we do a rudimentary sketch of the graphs on one grid (only if you don't have access to a graphing calculator, of course).

Our process here is to subtract the area of the lower graph ($y = 2 - \sqrt{x}$), from the area of the upper graph ($y = - \frac{1}{2} x + 2$). This will all be in the interval 0 ≤ x ≤ 4.

We start by finding the area of the upper graph.

${\int}_{0}^{4} \left(- \frac{1}{2} x + 2\right) \mathrm{dx}$

$\implies \left(- \frac{1}{4} {x}^{2} + 2 x\right) {|}_{0}^{4}$

$\implies - \frac{1}{4} \left({4}^{2}\right) + 2 \left(4\right) - \left(- \frac{1}{4} {\left(0\right)}^{2} + 2 \left(0\right)\right)$

$\implies - 4 + 8 - 0$

$\implies 4$

Now for the lower graph.

${\int}_{0}^{4} \left(2 - \sqrt{x}\right) \mathrm{dx}$

$\implies \left(2 x - \frac{2}{3} {x}^{\frac{3}{2}}\right) {|}_{0}^{4}$

$\implies 2 \left(4\right) - \frac{2}{3} {\left(4\right)}^{\frac{3}{2}} - \left(2 \left(0\right) - \frac{2}{3} {\left(0\right)}^{\frac{3}{2}}\right)$

$\implies 8 - \frac{16}{3}$

$\implies \frac{8}{3}$

We now subtract the first area from the second:

$4 - \frac{8}{3} = \frac{4}{3}$

Hence, the area of the region between $f \left(x\right) = 2 - \frac{1}{2} x$ and $g \left(x\right) = 2 - \sqrt{x}$ is $\frac{4}{3} {\text{ units}}^{2}$.