Since the equation of the curve is

To find the area under the curve we will use the integration

a = 0, b = 2

In integration, we add the power of x by 1 and divide the term by the new power
![A=[(2x^(2+1))/(2+1)-(16x^(1+1))/(1+1)+32x]_0^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/g8uoih2exd834xcvf3k47wdgcxncxpyi48.png)
We will simplify the bracket
![A=[(2x^3)/(3)-8x^2+32x]_0^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/7f9vip8ugynj9ajtw0onmzydmd7f0c7key.png)
Now we will substitute x by 2 and 0, then subtract the values of them
![A=[(2(2^3))/(3)-8(2^2)+32(2)]-[(2(0))/(3)-8(0)+32(0)]](https://img.qammunity.org/2023/formulas/mathematics/high-school/8lux5huad5qme25hvfgz5amhpf2r0aefiq.png)
Simplify
![\begin{gathered} A=[(16)/(3)-32+64]-0 \\ \\ A=(16)/(3)-(96)/(3)+(192)/(3) \\ \\ A=(112)/(3)=37(1)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/2avjgt1tk583ryqftmhz473rsnoiwli6lc.png)
The area under the curve is 37 1/3 square units (37.3333)