Zenarthra
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\displaystyle\int^2_0\sqrt[x]{2x+5}\ dx

If i was integrating using substitution of u^2 = 2x+5 , would i need to change the values of 2 and 0 or not?
Since i replace u at the end anyway?

Thanks!
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usycool1
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#2
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(Original post by Zenarthra)
\displaystyle\int^2_0\sqrt[x]{2x+5}\ dx

If i was integrating using substitution of u^2 = 2x+5 , would i need to change the values of 2 and 0 or not?
Since i replace u at the end anyway?

Thanks!
If you replace it, no, but probably best to make it clear that x = 2 or x = 0 (as opposed to u) in the limits.
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davros
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(Original post by Zenarthra)
\displaystyle\int^2_0\sqrt[x]{2x+5}\ dx

If i was integrating using substitution of u^2 = 2x+5 , would i need to change the values of 2 and 0 or not?
Since i replace u at the end anyway?

Thanks!
Firstly, I don't think you wrote what you meant to - I presume your integral should be of a square root, not the x-th root as you've written it.

Secondly, when you make a substitution you have to keep everything consistent - either write your final result as a function of u and change your limits to u limits, or convert your final result back into a function of x and use the original x limits.
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TenOfThem
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(Original post by Zenarthra)
\displaystyle\int^2_0\sqrt[x]{2x+5}\ dx

If i was integrating using substitution of u^2 = 2x+5 , would i need to change the values of 2 and 0 or not?
Since i replace u at the end anyway?

Thanks!
I would agree with Davros (I have been doing that a lot lately) but would go further and take usycool's point

The minute your function includes any u your limits should read x=

My practice is to
Write limits as values
Introduce u and change limits to x=
(if sticking with u)Change limits to u=
Then continue with values
(if planning n a return to x)Continue with x=
Only return to numbers after the re-substitution
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Zenarthra
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(Original post by TenOfThem)
I would agree with Davros (I have been doing that a lot lately) but would go further and take usycool's point

The minute your function includes any u your limits should read x=

My practice is to
Write limits as values
Introduce u and change limits to x=
(if sticking with u)Change limits to u=
Then continue with values
(if planning n a return to x)Continue with x=
Only return to numbers after the re-substitution
(Original post by davros)
Firstly, I don't think you wrote what you meant to - I presume your integral should be of a square root, not the x-th root as you've written it.

Secondly, when you make a substitution you have to keep everything consistent - either write your final result as a function of u and change your limits to u limits, or convert your final result back into a function of x and use the original x limits.
Yes you're right davros, i didnt mean to write it out like that. It's my first time using latex so forgive me.
And also TenOfThem thanks, but would be easier when you have limits in x, change them to u and just substitute the limits into u when integrated?
Would I get the same answer?

ThankS!
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davros
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(Original post by Zenarthra)
Yes you're right davros, i didnt mean to write it out like that. It's my first time using latex so forgive me.
And also TenOfThem thanks, but would be easier when you have limits in x, change them to u and just substitute the limits into u when integrated, so it would say replacing u?
I said get the same answer right?

ThankS!
Personally, I would go even further and write u explicitly as a function of x like this:

"By making the substitution u = \sqrt{2x + 5} we have ... "

I see a lot of books trying to be "clever" by saying things like "Let u^2 = f(x) ...", but this hides the subtlety that you have a choice of which square root to take when you change the limits.

You can write u = \sqrt{f(x)} or u = -\sqrt{f(x)} and get the correct answer either way, but it makes it explicit what you're doing when you convert the limits!
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Zenarthra
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(Original post by usycool1)
If you replace it, no, but probably best to make it clear that x = 2 or x = 0 (as opposed to u) in the limits.
Thanks!
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Zenarthra
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(Original post by davros)
Personally, I would go even further and write u explicitly as a function of x like this:

"By making the substitution u = \sqrt{2x + 5} we have ... "

I see a lot of books trying to be "clever" by saying things like "Let u^2 = f(x) ...", but this hides the subtlety that you have a choice of which square root to take when you change the limits.

You can write u = \sqrt{f(x)} or u = -\sqrt{f(x)} and get the correct answer either way, but it makes it explicit what you're doing when you convert the limits!
Sorry im a little confused here's what i mean:
Image

Would i replace the u's or just get the u's from the x values and find the area that way, would it give me the same answer?

ThankS!
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user2020user
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(Original post by Zenarthra)
...
When you make a change of variable, you must change the limits to your new variable. When you revert back to your original variable, revert your limits back to normal.

\displaystyle \begin{aligned} \int_{x_i}^{x_j} F'(x) \text{ d}x & \ \overset{\text{sub}}= \int_{u_i}^{u_j} G'(u) \text{ d}u \\ & = G(u) \Big|_{u_i}^{u_j} \\ & = F(x) \Big|_{x_i}^{x_j} \end{aligned}
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JerzyDudek
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Also how would you go about solving the intagral in the OP? Is it possible to do with a-level (including fm) knowledge? Is it even possible to find the closed form of the integral?
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davros
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(Original post by JerzyDudek)
Also how would you go about solving the intagral in the OP? Is it possible to do with a-level (including fm) knowledge? Is it even possible to find the closed form of the integral?
If you mean the one that appeared to have an x-th root in it, then no, I don't think there's a closed form for it!

Actually, looking at the OP's last post I think I'd assumed that x wasn't supposed to be there at all whereas in fact I think the OP intended it to be outside the root and multiplying it, but I don't think that changes any of the arguments above - either the root on its own or the root multiplied by x can be integrated.

OP: if you let u = \sqrt{2x + 5} then when you change to an integral in u, the limits must be changed too:
x = 0 => u = \sqrt{5}
x = 2 => u = 3
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cuckoo99
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(Original post by JerzyDudek)
Also how would you go about solving the intagral in the OP? Is it possible to do with a-level (including fm) knowledge? Is it even possible to find the closed form of the integral?
it would look horrible as would have to divide (u^(1/x))/2 (assuming you let u = 2x+5) by 1/x+1 xD

edit: impossible to do as its impossible to get it all in terms of 1 variable with A-level knowledge(i think) therefore limits cant be applied
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Zenarthra
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(Original post by davros)
If you mean the one that appeared to have an x-th root in it, then no, I don't think there's a closed form for it!

Actually, looking at the OP's last post I think I'd assumed that x wasn't supposed to be there at all whereas in fact I think the OP intended it to be outside the root and multiplying it, but I don't think that changes any of the arguments above - either the root on its own or the root multiplied by x can be integrated.

OP: if you let u = \sqrt{2x + 5} then when you change to an integral in u, the limits must be changed too:
x = 0 => u = \sqrt{5}
x = 2 => u = 3
I understand, thank you!
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Zenarthra
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(Original post by Khallil)
When you make a change of variable, you must change the limits to your new variable. When you revert back to your original variable, revert your limits back to normal.

\displaystyle \begin{aligned} \int_{x_i}^{x_j} F'(x) \text{ d}x & \ \overset{\text{sub}}= \int_{u_i}^{u_j} G'(u) \text{ d}u \\ & = G(u) \Big|_{u_i}^{u_j} \\ & = F(x) \Big|_{x_i}^{x_j} \end{aligned}
Thanks!
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Dilzo999
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(Original post by Zenarthra)
Sorry im a little confused here's what i mean:
Image

Would i replace the u's or just get the u's from the x values and find the area that way, would it give me the same answer?

ThankS!
Dem paint skills
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Zenarthra
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(Original post by dilzo999)
dem paint skills
dats how i roll!
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