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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?

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    $\begingroup$ The integral represents the area under the curve. What's the area under f(x) between x = -4 and x = 6? $\endgroup$ – Kaynex Jan 20 '18 at 5:38

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.



$g(x)$ is the signed area bounded by $f$ and the $x$-axis.

Hence if $x < -4$, $g(x)=0$.

$g(x)$ then increases linearly from $-4$ to $1$ and then it decreases from $-1$ to $5$.

After which, the value of $g$ doesn't change as $f$ is equal to $0$ for $x > 5$.

Hence $g(6)=g(5)$.

Try to evaluate $g(5)$, that is find the signed area from $-4$ to $5$. You might like to sketch out the graph. You just need to compute area of two rectangles and subtract them.

You are right that the maximum value occur at $x=-1$. Try to evaluate $2(-1-(-4))$.


Note that: $$g(6)=\int_{-4}^{6}f(t)\, dt$$ $$ = \int_{-4}^{-1} 2\, dt +\int_{-1}^{5}(-4)\, dt +\int_{5}^{6} 0\, dt$$ $$=2\times 3 +(-4)\times 6 = - 18$$

And also note that $g(x) $ increases linearly from $-4$ to $-1$ and decreases from then on. So, $$\max_{g(x)} =\int_{-4}^{-1} 2\, dt =6$$


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