Why Does Changing Variables in an Integral Preserve Equality?

[FlorenceC]

Homework Statement

prove that the ∫₀¹ f(x) = ∫₀1f(1-x)

Homework Equations

The Attempt at a Solution

I got all the way to ∫₀¹-f(u) du where u = 1-x but I don;t know how to prove it.

 

[ehild]

Homework Statement

prove that the ∫₀¹ f(x) = ∫₀1f(1-x)

Homework Equations

The Attempt at a Solution

I got all the way to ∫₀¹-f(u) du where u = 1-x but I don;t know how to prove it.

When you substitute the integration variable the limits also change.

 

[FlorenceC]

When you substitute the integration variable the limits also change.

Okay, so -∫₁ f(u) du = ∫₀¹f(u) du. but how do i relate this to ∫f(x)?

 

[ehild]

It does not matter how do you name the integration variable. It can be x instead of u. The function is the same, and so are the limits.

 

[Ray Vickson]

Okay, so -∫₁ f(u) du = ∫₀¹f(u) du. but how do i relate this to ∫f(x)?

Don't you see that $\int_0^1 f(x) \, dx = \int_0^1 f(u) \, du = \int_0^1 f(\text{anything}) \, d\text{anything}$?

 

[Mark44]

Okay, so -∫₁ f(u) du = ∫₀¹f(u) du. but how do i relate this to ∫f(x)?

Don't you see that $\int_0^1 f(x) \, dx = \int_0^1 f(u) \, du = \int_0^1 f(\text{anything}) \, d\text{anything}$?

To elaborate on what Ray said, I have added variables in the limits of integration to emphasize that in each integral we have a different dummy variable.
$$ \int_{x = 0}^1 f(x) \, dx = \int_{u = 0}^1 f(u) \, du = \int_{\text{anything} = 0}^1 f(\text{anything}) \, d\text{anything}$$

When you use substution to change the variable of integration, you need to either change the limits of integration (which was not done in the above) or undo the substitution.