# Finding where an integral is positive or negative given the graph of the function.

This problem on my math test was slightly confusing, as I was not sure how to solve it:

I am guessing that since the integral is positive or negative, we can find where the derivative is increasing and decreasing, and apply the same intervals for the question.
Thanks.

• Can you see that $g(0)=0?$ And that $g>0$ on $(0,4]?$ – zhw. Nov 19 '16 at 20:43
• $g(x)$ is area under the graph bounded by lines $t=0$ and $t=x$. – user160738 Nov 19 '16 at 20:44
• @zhw, yes that is quite obvious. However, I am unsure whether the problem is simply asking for this: g(x) > 0 on [-4,-3] U [0,4], g(x) < 0 on [-3,0] U [4,6]. Is this the case? – Shreyas B. Nov 19 '16 at 20:46
• No, that is far from the case. – zhw. Nov 19 '16 at 21:35

## 1 Answer

In the comments you agreed $g(0)= 0,$ $g>0$ on $(0,4].$ What about $g(5)?$ Now we encounter area that is to be subtracted, namely the area of that triangle. But the area of that triangle is clearly less than the area that gives $g(4).$ So we have $g>0$ on $(0,5].$ Now think about $g$ on $[5,6].$

For $x<0,$ recall that $\int_0^x f = - \int_x^0 f$ by definition. Because $f<0$ on $(-3,0)$ we will therefore have $g>0$ on $[-3,0).$ Keeping that in mind, visually inspect the graph to see what happens for $x<-3.$