Homework Help: Integration with respect to a higher power

1. Jun 23, 2007

steven10137

1. The problem statement, all variables and given/known data
$$\int {x^2 ,d(x^4)}$$

2. Relevant equations
as a starter; the previous problem was:
$$\int {x^2 ,d(x^2)}$$

and I managed to solve this by letting u=x^2 then integrating u:

$$\int {u}$$

=$$\frac{x^4}{2}+C$$

3. The attempt at a solution
can someone please explain the thoery behind this?
my textbook gives no explanation and i dont really know what im looking for...
I tried following the working from the previous problem through, but got (x^6)/3 when the answer was supposed to be 2(x^6)/3

Steven

2. Jun 23, 2007

Moridin

In the previous problem, you made the differential to du. That is an interesting tactic for this problem too..

3. Jun 23, 2007

steven10137

so I am on the right track i take it?
If so;
$$\int {x^2 d(x^4)} = \int {u^2 d(u)}$$
am I correct?

4. Jun 23, 2007

Moridin

Not quite. If

$$u = x^4$$

then

$$x^2 = ?$$

5. Jun 23, 2007

MathematicalPhysicist

dx^4/dx=4x^3
dx^4=4x^3dx

6. Jun 23, 2007

HallsofIvy

In general,
$$\int f(x)dg(x)= \int f(x) \frac{dg}{dx} dx$$

7. Jun 24, 2007

steven10137

thankyou all for your help, I understand it now :)
$$dx^4 = 4x^3 dx$$
therefore
$$\int {x^2 dx^4}$$
$$= \int {x^2 4x^3 dx}$$
$$= 4 \int {x^2 x^3 dx}$$
$$= 4 \int {x^5 dx}$$
$$= 4 \frac {x^6}{6} + C$$
$$= \frac {2x^6}{3} + C$$

cheers
Steven