0
$\begingroup$


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

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.

$\endgroup$
  • $\begingroup$ Can you see that $g(0)=0?$ And that $g>0$ on $(0,4]?$ $\endgroup$ – zhw. Nov 19 '16 at 20:43
  • $\begingroup$ $g(x)$ is area under the graph bounded by lines $t=0$ and $t=x$. $\endgroup$ – user160738 Nov 19 '16 at 20:44
  • $\begingroup$ @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? $\endgroup$ – Shreyas B. Nov 19 '16 at 20:46
  • $\begingroup$ No, that is far from the case. $\endgroup$ – zhw. Nov 19 '16 at 21:35
0
$\begingroup$

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 $x<-3.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.