Question

I have a calculus question about approximating areas, pic included

Answer 1

Since the equation of the curve is

`f(x)=2x^2-16x+36`

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

`A=\int_a^bf(x)dx`

a = 0, b = 2

`A=\int_0^2(2x^2-16x+36)dx`

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`

We will simplify the bracket

`A=[(2x^3)/(3)-8x^2+32x]_0^2`

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)]`

Simplify

`\begin{gathered} A=[(16)/(3)-32+64]-0 \\ \\ A=(16)/(3)-(96)/(3)+(192)/(3) \\ \\ A=(112)/(3)=37(1)/(3) \end{gathered}`

The area under the curve is 37 1/3 square units (37.3333)