Consider deleting eq. 6.140 and 6.141

[Chechgm]

Describe the mistake Above eq. 6.140, it is written "Note that the integral on the right-hand side is with respect to $x$, but we need an integral with respect to $y$ because we are differentiating with respect to $y$". After performing the substitution, you claim using the fundamental theorem of calculus to arrive to eq. 6.142. I couldn't figure out how to get to eq. 6.142 using this, so I double checked the fundamental theorem of calculus (theorem 6.20 and 6.21 in Rudin's principles, but the wikipedia site has the same), and you don't need the integral to be with respect to $y$, but the limits of the integral to be with respect to $y$. This is already the case in eq. 6.139 . In summary: one can go from 6.139 to 6.142, in a less confusing way.

Location Please provide the

1. version (bottom of page): Print version, although I downloaded the book yesterday and it has the same content.
2. Chapter: 6
3. Page: 195
4. line number/equation number: 6.140, 6.141

Proposed solution Delete the text above eq. 6.140, invoking the fundamental theorem of calculus on 6.139 and jumping to 6.142 without using 6.140 and 6.141 .

Additional context If I am incorrect in my reasoning, I think having a slightly more detailed explanation of the jump from 6.141 to 6.142 would be useful.

[pclucas14]

I think You couldn't make it to 6.142 because there is a typo in the bounds of the integral : When applying the change of variables formula from 6.139 to 6.140 the bounds should go from [a, U^{-1}(y)] to [U(a), U(U^{-1}(y))] = [U(a), y]

Regarding @Chechgm's point, I had the same question : Why can't we just integrate f(x) in 6.139 to get d/dy F(U^{-1}(y)) and just apply the chain rule to get 6.142 ?

Great job on the book :)