# integration by substitution

• May 3rd 2010, 11:55 PM
demode
integration by substitution
http://img413.imageshack.us/img413/6839/49887281.gif

(a) Let $u=x^4$ then $du=4x^3dx$ and therefore $\frac{du}{4x^3}=dx$. We get:

$\int^b_{a} f(x^4)dx= \int^{1^4}_{(-1)^4} 1+u \frac{du}{4x^3} =\int^{1}_{1} 1+u \frac{du}{4x^3}$

Both limits of integration in the new integral are 1, so the value of the definite integral is 0.

(b)
$\frac{du}{dx}=4x^3$

It is only zero when x=0. Jacobian become zero at (0,0) so it vanishes at this point.

(c) This is the part I'm really stuck on: from part (a) I know what to substitute for "dx" when I make the substitution $u=x^4$. So I think this is what the question is asking for:

$\int^0_{-1} 1+u \frac{du}{4x^3} + \int^1_0 1+u \frac{du}{4x^3}$

But expressions involving x should not appear in the new integrals. How do I get rid of the " $4x^3$"? Can anyone show me how to do this part?
• May 4th 2010, 12:07 AM
drumist
Since $u=x^4$ is the substitution being made, you can take a fourth root on both sides to get $x=\pm \sqrt[4]{u}$. The $\pm$ is important here, and is why the integral needs to be split into two regions.

In the region $-1, we should get $x=-\sqrt[4]{u}$ because x is always negative in this region. Using the same logic, for $0, we should use $x=\sqrt[4]{u}$. Now that you have an expression for x in both regions you can make the substitution.

By the way, make sure you use parentheses in your integral. It should be like this:

$\int^0_{-1} (1+u) \frac{du}{4x^3} + \int^1_0 (1+u) \frac{du}{4x^3}$
• May 4th 2010, 12:12 AM
CalculusCrazed
I'm pretty sure that you need to set the limits also. For each of your integrals you approach a value that is undefined.