Integral Bounds: Explaining Inequalities & f(u) to f(x)

[Speedking96]

Homework Statement

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)?

 

[SteamKing]

Homework Statement

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

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.

 

[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.

 

[Mark44]

Homework Statement

This is the problem with the solution:

View attachment 79564

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

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.

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

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.

 

[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'.

 

[SteamKing]

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'.

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?

 

[Speedking96]

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

Wow, it was that simple! Hehe.

 

[Mark44]

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?

The substitution is u = a - x.

 

[Ray Vickson]

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.

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?