integration by substitution

**demode** · May 3rd 2010, 10:55 PM

(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?

**drumist** · May 3rd 2010, 11:07 PM

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 $-1<x<0$ , we should get $x=-\sqrt[4]{u}$ because x is always negative in this region. Using the same logic, for $0<x<1$ , 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:

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

**CalculusCrazed** · May 3rd 2010, 11:12 PM

I'm pretty sure that you need to set the limits also. For each of your integrals you approach a value that is undefined.

Thread: integration by substitution

LinkBack

Thread Tools

Display