# Thread: Trig Sub Integral: What are the bounds?

1. ## Trig Sub Integral: What are the bounds?

I'm trying to integrate f(x) = (2/3)(4-x^2)^(3/2) with respect to x from -2 to 2. So far I've used trig sub with x/2 = cos u to get f(u) = (-32/3)(sin(u)) ^ 4. I think this is correct but I'm not sure what bounds to use when I integrate f(u). When I tried going from -pi to 0 I get - 4pi rather than 4pi which I know to be the correct answer. Can someone explain to me why I can't go from -pi to 0? Or if I've made a mistake elseware can you explain what I did? Thanks a lot!

2. Originally Posted by eay444
I'm trying to integrate f(x) = (2/3)(4-x^2)^(3/2) with respect to x from -2 to 2. So far I've used trig sub with x/2 = cos u to get f(u) = (-32/3)(sin(u)) ^ 4. I think this is correct but I'm not sure what bounds to use when I integrate f(u). When I tried going from -pi to 0 I get - 4pi rather than 4pi which I know to be the correct answer. Can someone explain to me why I can't go from -pi to 0? Or if I've made a mistake elseware can you explain what I did? Thanks a lot!
This presentation is messy. I assumed that you meant to calculate:

$\displaystyle\int^{2}_{-2}\frac{2}{3}(4-x^2)^{\frac{3}{2}}dx$

I assume then that you've taken the following steps:

= $\displaystyle\int^{2}_{-2}\frac{2}{3}\times4^{\frac{3}{2}}(1-\frac{x^2}{4})^{\frac{3}{2}}dx$

$=\displaystyle\frac{16}{3}\int^{2}_{-2}(1-\frac{x^2}{4})^{\frac{3}{2}}dx$

Then made the substitution of $\frac{x}{2}=cos(u)$ etc...? Or have I misinterpreted?

As your bounds with respect to $x$ are $2$ and $-2$, and $\frac{x}{2}=cos(u)$

The upper bound becomes where: $\frac{2}{2}=cos(u)$
$Cos(u)=1$
$u=0$

The lower bound becomes where $Cos(u)=-1$
Which is where $u=\pi$

If you switch the order of the bounds, and have $u=\pi$ as your upper bound, then your answer becomes negative. I don't see why you've used $u=-\pi$.

3. I'm sorry about the presentation but yes that is exactly what I was trying to calculate. I chose u = -pi as my lower bound as that satisfied the condition cos(u) = -1 and I choose 0 as my upper bound as that satisfies cos(u) = 1. These were the first solutions to the cos(u) = x equation that I thought of. I know there are an infinite number of solutions to both cos(u) = 1 and cos(u) = -1 so how exactly do you pick out two that work? I see now that it definitely matters what you choose but I'm not sure why?

Again, sorry if this isn't clear enough. Let me know if you don't understand what I mean.

4. Oh, I understand now. Importantly, as it's an area, you are looking for the magnitude of your result, so the sign isn't usually very important. Obviously, if you take solutions which are far apart from each other, then you will receive the wrong answer. Taking a look at the graph of y=Sin^4(x) might clarify; see
Sin^4(x)

I have to be honest, I'm struggling to provide a strong answer. Hopefully someone else of greater ability can provide more sound reasoning, but here is my attempt:

If you integrate between 0 and -pi, or between 0 and pi, you will receive the same answer - the area is the same, obviously. However, the direction you measure the area from, and to, changes. This is what the negative sign is indicating. If you measure from 0 to +pi, the area increases as x increases, If you measure from 0 to -pi, the area increases as x decreases, as reflected by the change on sign. This is not a sufficient explanation, however .

http://www.wolframalpha.com/input/?i=Integrate+sin^4+x+from+0+to+pi
http://www.wolframalpha.com/input/?i=Integrate+sin^4+x+from+0+to+-pi