# Definite integral

1. Feb 3, 2007

### dnt

1. The problem statement, all variables and given/known data

if the integral from 0 to 4 of f(x) = -1

then what is the integral from -2 to 0 of x[f(x^2)]?

2. Relevant equations

n/a

3. The attempt at a solution

my first instinct is that this is an even/odd definition of an integral problem. the x squared in the function makes it even, which means its symmetrical over the y axis and thus you can double the area.

however, the x which is being multiplied by f(x^2) is confusing me as is the fact that 0 to 4 isnt the same as -2 to 2 (the symmetrical part).

can someone help me out? thanks.

2. Feb 3, 2007

### Dick

I would think very seriously about a variable change u=x^2.

3. Feb 4, 2007

### dnt

thanks for the help but im still quite confused. i dont understand how to relate the two integrals especially since the two end points are different.

4. Feb 5, 2007

### dnt

ok well i did the u=x^2 substitution and got this:

u = x^2

du = 2x dx

therefore i have:

the integral from 4 to 0 of f(u) du times (1/2)

now, according to the original problem, the integral from 0 to 4 of f(x) is -1.

how do i relate my new integral to the original? i know if i switch the two end points i can simply change the value of the integral by multiplying by -1 but my new integral is with u...the original is with x.

5. Feb 5, 2007

### dextercioby

For definite integrals, the integration variable is a dummy variable and it van be anything you want. So it doesn't matter the notation, the number the integral is equal to is independent of the way you denote the integration variable.

6. Feb 5, 2007

### HallsofIvy

Staff Emeritus
So you have
$$\int_{-2}^0 xf(x^2)dx= \frac{1}{2} \int_0^4 f(u)du$$
and you know what that second integral is!

7. Feb 5, 2007

### Staff: Mentor

Halls, you have a sign error in the above. (Limits of integration are switched.)

8. Feb 5, 2007

### dnt

(-1/2) x (-1) = 1/2

correct?

9. Feb 5, 2007

### Dick

Quite correct.

10. Feb 5, 2007

### dnt

thank you.

11. Feb 5, 2007

### dnt

even though i got the answer, im still interested in learning more about this concept. is there a link that would explain it in more detail (the idea that for definite integrals the integraion variable doesnt really matter).

thanks.

12. Feb 6, 2007

### HallsofIvy

Staff Emeritus
Any text book should explain that- it's simply a matter of the fact that the definite integral of a function is a number and so doesn't depend on any variable.
$$\int_0^1 x^2 dx= \frac{1}{3}$$
$$\int_0^1 y^2 dy= \frac{1}{3}$$
$$\int_0^1 t^2 dt= \frac{1}{3}$$
$$\int_0^1 a^2 da= \frac{1}{3}$$

Are you surprized?

13. Feb 6, 2007

### dnt

not surprised but it still doesnt quite make 100% sense to me yet. i need to keep reading on it until i get it.