# How do you find the area between y=1/2x^3+2, y=x+1, x=0, x=2?

Nov 9, 2016

graph{(1/2x^3+2-y)(x+1-y)(x-2)x=0 [-6.67, 9.14, -0.72, 7.18]}

$A = 2$

#### Explanation:

First of all
$\frac{1}{2} {x}^{3} + 2 > x + 1 , \forall 0 \le x \le 2$
To explain this call $f \left(x\right) = \frac{1}{2} {x}^{3} - x + 1$

Then ${f}^{'} \left(x\right) = \frac{3}{2} {x}^{2} - 1$ so ${f}^{'} \left(x\right) = 0$ only when $x = \pm \sqrt{\frac{2}{3}}$

and ${f}^{' '} \left(x\right) = 3 x > 0 , \forall 0 < x \le 2$

So f has a local minimum in $x = \sqrt{\frac{2}{3}}$ and this minimum is
$f \left(\sqrt{\frac{2}{3}}\right) = 1 - \frac{2}{3} \sqrt{\frac{2}{3}} > 0$

Then the area is given by

$A = {\int}_{0}^{2} \left(\frac{1}{2} {x}^{3} - x + 1\right) \mathrm{dx} = {\left[{x}^{4} / 8 - {x}^{2} / 2 + x\right]}_{0}^{2} = \frac{16}{8} - \frac{4}{2} + 2 = 2$