Find the area of the region between the graphs of f(x)=3x+8 and

Question

g(x)=x^2 + 2x+2 over [0,2].

I got 34/3.
Calculus - Steve
∫[0,2] (x^2+2x+2) dx
= 1/3 x^3 + x^2 + 2x [0,2]
= 8/3 + 4 + 4
= 32/3

Why are you taking the antiderivative of x^2 +2x+2 when we are trying to find the area between the two graphs? Am I supposed to to take the area of the top graph and subtract the area if the bottom graph to find the area between two curves at [0,2]?

Here is my work:

∫[0,2] 3x+8-(x^2+2x+2)dx=

∫[0,2] -x^2 +x+6 dx=

[1/3 x^3 +1/2 x^2+6x] from 0 to 2= 34/3

Answer 1

Oops. My bad.
I missed f(x).

You are correct.

Answer 2

correct