# Math Help - question about area/perimeter/volume using integrals

1. ## question about area/perimeter/volume using integrals

AP CALC AB problem (high school)

1)
f(x) = e^(-x/4) + sin (x^2), g(x) = (1/3)x^2 - x, and x-axis supposedly create a region where I am to find the area/perimeter/volume. When I graph this, there is no region such that satisfies the condition of all 3. Am I graphing this incorrectly? or is this AP calculus AB question just poorly/incorrectly written?

2)
g(x) = (5 e^(-x/6) ) (sin ((x^2) / 6)) and the x-axis enclose a region R. Again, I'm supposed to find area/perimeter/volume.

I am pretty sure the region R enclosed by g(x) and the x-axis is infinite as it goes on and on and on...which would set up an infinite amounts of integral [x1,x2] + integral [x2,x3] + etc as the upper and lower bounds keep alternating due to the sin wave of the function.
plot &#40;5e&#94;&#40;-x&#47;6&#41;&#41;&#40;sin&#40;&#40;x&#94;2&#41;&#4 7;6&#41;&#41; - Wolfram|Alpha

no domain was specified. I doubt the teacher meant to trick high school students with this. Am I looking at these two questions wrong? Thanks!

2. ## Re: question about area/perimeter/volume using integrals

Dude, you'll make more people interested of this post if you use latex to display your functions/formulas etc. :-)

3. ## Re: question about area/perimeter/volume using integrals

I see two separate regions. One section begins where $(1/3)x^2 - x= e^{-x/4}+ sin(x^2)$ and ends where $e^{-x/4}+ sin(x^2)= 0$. The second region begins where that function is 0 again, then ends where $(1/3)x^2 - x= e^{-x/4}+ sin(x^2)$.

4. ## Re: question about area/perimeter/volume using integrals

Originally Posted by liquidFuzz
Dude, you'll make more people interested of this post if you use latex to display your functions/formulas etc. :-)
I had already typed the formula out so I copy/pasted without even considering it. I probably should have went the extra effort :P

Originally Posted by HallsofIvy
I see two separate regions. One section begins where $(1/3)x^2 - x= e^{-x/4}+ sin(x^2)$ and ends where $e^{-x/4}+ sin(x^2)= 0$. The second region begins where that function is 0 again, then ends where $(1/3)x^2 - x= e^{-x/4}+ sin(x^2)$.
I see the two regions as one unless you are considering the area between $e^{-x/4}+ sin(x^2)$ and x axis. The two regions within the two domains you mention don't seem to cross on my graph, so I consider that as just 1 bound region. I do see 2 other regions bound by f(x) and g(x) but not x-axis.

After much zooming in and out, I did finally find a very small region where x is approximately 3.1 and 2 larger regions to the right of it where x is approximately 3.5. This entire setup though seems incredibly unintuitive.

However, I am more interested in the second setup with the infinite bound regions. I am pretty sure I am graphing it correctly. I just wouldn't know how to setup, for example, an area formula using integrals for an infinite number of regions without some complex sigma notation.