One question Calculus U substitution?

[Wamaster]

integral (2x-1)^3 from 0 to 1

Why is this wrong:

Work:
U= 2x-1
du= 2dx

New limit 2x-1 for 0 & 1.... 1 & -1

1/8[(2x-4)^4] plug in 1 & -1

I get -10.... the correct answer is 0
It seems the only way to get 0 is to use the old limit... why is this right. Like why wouldn't I plug in the new limit?
I checked with wolfram the answer is 0

... Like when do I use new and old limit to solve
Thanks in advance

 

[DrPhil]

integral (2x-1)^3 from 0 to 1

Why is this wrong:

Work:
U= 2x-1
du= 2dx

New limit 2x-1 for 0 & 1.... 1 & -1..x=0 --> u=-1,...x=1 --> u=1

\(\displaystyle \displaystyle \dfrac 12\ \int_{-1}^{1}u^3\ du \implies \left. \dfrac 18 u^4\right|_{u=-1}^{u=1}\)

1/8[(2x-4)^4] plug in 1 & -1

I get -10.... the correct answer is 0
It seems the only way to get 0 is to use the old limit... why is this right. Like why wouldn't I plug in the new limit?
I checked with wolfram the answer is 0

... Like when do I use new and old limit to solve
Thanks in advance

When you express the integral as a function of u, then you use u-limits.

If you backsubstitute to a function of x, you use x-limits. Generally I DON'T backsubstitute - it is likely to be easier to evaluate in terms of u.

EDIT: that is for a definite integral. For indefinite, "always" backsubstitute.

 

[HallsofIvy]

integral (2x-1)^3 from 0 to 1

Why is this wrong:

Work:
U= 2x-1
du= 2dx

New limit 2x-1 for 0 & 1.... 1 & -1

Yes, when x= 0, u= -1 and when x= 1, u= 1.

1/8[(2x-4)^4] plug in 1 & -1

No, after integrating with respect to u you get (1/8)u^4 and set u= -1 and 1. (1/8)(1)^4- (1/8)(-1)^4= 0

If you go back to the x variable, to get \(\displaystyle (1/8)(2x- 1)^4\) then you use the x values: x= 0 and x= 1.
(1/8)(2(1)- 1)^4- (1/8)(2(0)-1)^4 (1/8)(1)^4- (1/8)(-1)^4= 0.

I get -10.... the correct answer is 0.
It seems the only way to get 0 is to use the old limit... why is this right. Like why wouldn't I plug in the new limit?

Because you are not using the new variable.

I checked with wolfram the answer is 0

... Like when do I use new and old limit to solve
Thanks in advance

You use the old limits if you are using the old variable, the new limits if you are using the new variable!

 

[HallsofIvy]

Yes, when x= 0, u= -1 and when x= 1, u= 1.


No, after integrating with respect to u you get (1/8)u^4 and set u= -1 and 1. (1/8)(1)^4- (1/8)(-1)^4= 0

If you go back to the x variable, to get \(\displaystyle (1/8)(2x- 1)^4\) then you use the x values: x= 0 and x= 1.
(1/8)(2(1)- 1)^4- (1/8)(2(0)-1)^4 (1/8)(1)^4- (1/8)(-1)^4= 0.


Because you are not using the new variable.


You use the old limits if you are using the old variable, the new limits if you are using the new variable!

It might help you to actually write the variable with the limits of integration as a reminder:]
\(\displaystyle \int_{x= 0}^1 (2x-1)^3 dx= \int_{u= -1}^1 u^3 du\)