# Simple Integration by Substitution

## Main Question or Discussion Point

The integrand is f(x) = x* sqr(x-1)
The interval [1,2]

Please draw it out in a gif file and send it to me via email.

-much appreciated.

Math Is Hard
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If I tell you how to do the substitution will you try to work it out from there? Or do you want someone to just do your homework for you?

Gokul43201
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The policy here is that we'll help you if you show what you've tried and where you appear to be stuck.

It seems to me that you may have a problem with algebra (in this case, at least), as the most obvious substitution should get you an easily integrable form.

Start with $$u=\sqrt{x-1}$$

Math Is Hard
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I wouldn't recommend that.
(but I am interested to see how you worked it out with that sub!)

Math Is Hard said:
I wouldn't recommend that.
(but I am interested to see how you worked it out with that sub!)
and why not?
::
u=sqrt(x-1)
x=u^2+1
dx=2udu

The integrand becomes (u^2+1)u*2udu
A simple polynomial which has to be integrated from 0 to 1 if i am not mistaken with calculations that are going in my head.
::

-- AI

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Gokul43201
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Tenali, I wish you hadn't done that in the thread. This leaves adiles with no work to do.

Math Is Hard
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I would use a slightly different (and simpler in my mind) substitution method, but I'll withhold, just to leave a little mystery and hopefully some "pleasure of discovery" for adiles.

since some one already solved it for him, the simplest sub i saw was...

f(x) = x* sqr(x-1)

let
u = x-1

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Hurkyl
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That's why you make the one example something different (but similar enough to demonstrate the point... maybe $\int \sqrt{x-1} \, dx$) The problem is that all the example in the world usually don't help unless the student actually does a few himself.

cronxeh
Gold Member
why the hell was my solution erased from this thread?

Cronxeh im guessing

Gokul43201 said:
I wish you hadn't done that in the thread. This leaves adiles with no work to do.