1. ## Calc. Problem

Calculate the area under the graph of f(x) = 3x^2 and above the x-axis on the interval from x= -2 to x= 2.

I entered this graph into my calculator and there is no area under the graph, so I assumed it was 0. Then, I did fnInt(3x^2,x,-2,2) and got 16 as my solution. Just was wondering if I could get a second opinion to see if I did this right.

2. Originally Posted by redsoxfan0825
Calculate the area under the graph of f(x) = 3x2 and above the x-axis on the interval from x= -2 to x= 2.

I entered this graph into my calculator and there is no area under the graph, so I assumed it was 0. Then, I did fnInt(3x2,x,-2,2) and got 16 as my solution. Just was wondering if I could get a second opinion to see if I did this right.
The graph of $3x^2$ is symmetrical about $x=0$ so we can consider one side of x=0 and multiply by 2

$\int^2_{-2}3x^2\,.dx = 2\, \int^2_0 3x^2\,.dx$

That integral should be simple enough and equal 8 so when we multiply that by 2 we get 16

Not sure how you got 0 either because integrating the original expression yields $8-(-8)=16$

edit: attached is the graph of $3x^2$ and it clearly shows an area