So I have "guessed" all but d and the first part of e correctly but I must admit I had no idea what I was doing other than using the rule of the fundamental theorem of calculus (the integral simply being $f(t)$) that I thought would apply here. But it doesn't seem to be working for the ones I got wrong.

I drew a sketch of $f(x)$ and I assumed I would then sketch $g(x)$. But I am not entirely sure how I would do that because you can't exactly sketch an integral precisely simply through its derivative.

Any help?

## Best Answer

For part (d) $$g(6) =2(3)-4(6)=-18 $$

For part (e) the value is $g(-1) =2(3)=6$

Note that the integral of a constant function on an interval is the length of the interval times the constant.