Changing bounds (integration by substitution) (1 Viewer)

ooo · Nov 26, 2014

Suppose my substitution was x = u^2 -1
When changing bounds, do I take u = +- sqrt(x+1) ? Explain why.

FrankXie · Nov 26, 2014

ooo said:
Suppose my substitution was x = u^2 -1
When changing bounds, do I take u = +- sqrt(x+1) ? Explain why.

You can take either one, but not both. Because the substitution needs to be one-to-one mapping.

ooo · Nov 27, 2014

No, you can't take either one. You can try it out with a random definite integral; only one will give the correct answer.

FrankXie · Nov 27, 2014

ooo said:
No, you can't take either one. You can try it out with a random definite integral; only one will give the correct answer.

are you serious?

$$ Lets randomly do $ \int_0^1xdx, $ which is easily evaluated to be $ \frac12$

$$ Now lets take substitution $ x=u^2-1, $ and first take $ u=\sqrt{x+1}. $ So $ dx=2udu, $ and $ x=0\rightarrow u=1, x=1\rightarrow u=\sqrt2, $ thus the integral $=\int_1^{\sqrt2}(u^2-1)2udu=\left[\frac{u^4}{2}-u^2\right]_1^{\sqrt2}=(\frac42-2)-(\frac12-1)=\frac12$

$$ Secondly, same substitution but take $ u=-\sqrt{x+1}. $ This time, $ x=0\rightarrow u=-1, x=1\rightarrow u=-\sqrt2, $ thus the integral $=\int_{-1}^{-\sqrt2}(u^2-1)2udu=\left[\frac{u^4}{2}-u^2\right]_{-1}^{-\sqrt2}=(\frac42-2)-(\frac12-1)=\frac12$

Therefore, either one gives both correct answer!

SilentWaters · Nov 27, 2014

Still works. We end up with even powers throughout the integrands, making it a case of evaluating symmetric areas under an even function in u.

SilentWaters · Nov 27, 2014

braintic said:
Not true.

When you take the limits -1 and -sqrt2, they are in REVERSE order. So you get MINUS the other answer.
Try it.

You forget that we reverse the limits due to the substitution being $u = -\sqrt{x+1}$ in the second case. This means that in our integrand the $\sqrt{x+1}$ becomes $-u$ .

FrankXie · Nov 27, 2014

anyone wanna a bet? lol

SilentWaters · Nov 28, 2014

$x = u^2 -1, 2u\,du = dx$ for $\int_0^1 {x\sqrt{x+1}} \,dx$
Case 1: $u = +\sqrt{x+1}.$ Integral is $\int\limits_{1}^{\sqrt{2}} {\left(u^2 -1\right)u\cdot2u\,du}$
Case 2: $u = -\sqrt{x+1}.$ Integral is $\int\limits_{-1}^{-\sqrt{2}} {\left(u^2 -1\right)(-u)\cdot2u\,du} = \int\limits_{-\sqrt{2}}^{-1} {\left(u^2 -1\right)u\cdot2u\,du}$

Perfectly equivalent, as the common 'u' integrand is even.

SilentWaters · Nov 28, 2014

Sorry, which substitution has been changed? I stuck with whatever was given in the thread

You always consider both the substitution and the limits for any changes to a definite integral anyway.

FrankXie · Nov 28, 2014

braintic said:
Hang on a moment .... You changed the actual substitution.

In Frank Xie's example, he used the SAME algebraic substitution in each case, using the negative version ONLY to get the new limits.

Yes, if you use the minus formula for BOTH the algebraic substitution AND the new limits, you will always get the same answer, but that is NOT what he did, and I was debating the issue based on that premise.

lol. I only change limits but keep other parts unchanged because in my example I did NOT have the term $\sqrt{x+1}$ , so I did not need to change the integrand.

I totally agree with SilentWaters. -- He did NOT change the substitution at all

Changing bounds (integration by substitution) (1 Viewer)

ooo

New Member

FrankXie

Active Member

ooo

New Member

FrankXie

Active Member

SilentWaters

Member

SilentWaters

Member

FrankXie

Active Member

SilentWaters

Member

SilentWaters

Member

FrankXie

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)