# Integral Bounds

1. Feb 24, 2015

### Speedking96

1. The problem statement, all variables and given/known data
This is the problem with the solution:

Can someone please explain how the new bounds were computed, I don't quite understand what's going on with the inequalities?

Also, in the final two steps, how can the f(u) change to f(x)?

2. Feb 24, 2015

### SteamKing

Staff Emeritus
The inequalities are just another mathematical expression which describes the bounds of integration.

0 ≤ x ≤ a is just math talk for "all x between 0 and a, including both 0 and a."

After making the substitution u = x - a, the bounds of the u-integral are obtained by substituting the bounds of the x-integral (0 ≤ x ≤ a) into the expression for u.

Also, in the final two steps, how can the f(u) change to f(x)?[/QUOTE]

Once you have the integrand in the same form after u-substitution as the original form of the integrand for the x-integral, you can change u to x or whatever. These are essentially 'dummy' variables at this point.

3. Feb 24, 2015

### Speedking96

Okay, when you say: After making the substitution u = x - a, the bounds of the u-integral are obtained by substituting the bounds of the x-integral (0 ≤ x ≤ a) into the expression for u.

How would you substitute an inequality (0 ≤ x ≤ a) into an equation (u = x - a) ?

Thanks for the help.

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4. Feb 24, 2015

### Staff: Mentor

I think they have the second set of inequalities backwards; i.e., $a \le u \le 0$.
The substitution u = a - x is equivalent to u + x = a. Both u and a range between 0 and a, but as x increases from 0 to a (assumed to positive, or at least nonnegative), u decreases from a to 0.
The x and u in the two integrals are dummy variables. With regard to integration, $\int f(u) du$ is exactly the same as $\int f(x) dx$. It's only when you undo the substitution that you get something different.

5. Feb 24, 2015

### Speedking96

Like, when the bounds are numbers, I have no problem changing the bounds when doing the substitution, but here I can't seem to wrap my head around it since the bounds are not 'numbers'.

6. Feb 24, 2015

### SteamKing

Staff Emeritus
You're overthinking this. You don't substitute the inequality into the expression for u, you substitute the values at the ends of the interval into the expression.

If u = x - a, what's the value of u when x = 0? When x = a?

7. Feb 24, 2015

### Speedking96

When x = 0, the value would be -a, and when x = a, the value would be 0.

Wow, it was that simple! Hehe.

8. Feb 24, 2015

### Staff: Mentor

The substitution is u = a - x.

9. Feb 24, 2015

### Ray Vickson

Look at u = a-x for 0 <= x <= a. When x = 0, what is the value of u? When x = a, what is the value of u?