Since the equation of the curve is
[tex]f(x)=2x^2-16x+36[/tex]
To find the area under the curve we will use the integration
[tex]A=\int_a^bf(x)dx[/tex]
a = 0, b = 2
[tex]A=\int_0^2(2x^2-16x+36)dx[/tex]
In integration, we add the power of x by 1 and divide the term by the new power
[tex]A=[\frac{2x^{2+1}}{2+1}-\frac{16x^{1+1}}{1+1}+32x]_0^2[/tex]
We will simplify the bracket
[tex]A=[\frac{2x^3}{3}-8x^2+32x]_0^2[/tex]
Now we will substitute x by 2 and 0, then subtract the values of them
[tex]A=[\frac{2(2^3)}{3}-8(2^2)+32(2)]-[\frac{2(0)}{3}-8(0)+32(0)][/tex]
Simplify
[tex]\begin{gathered} A=[\frac{16}{3}-32+64]-0 \\ \\ A=\frac{16}{3}-\frac{96}{3}+\frac{192}{3} \\ \\ A=\frac{112}{3}=37\frac{1}{3} \end{gathered}[/tex]
The area under the curve is 37 1/3 square units (37.3333)
