# Integration by Substitution

• November 15th 2011, 01:37 PM
GrigOrig99
Integration by Substitution
Hi, I'm new to integration. Can anyone help me figure out where I went wrong?
Many thanks.

Q. Evaluate the following:

• November 15th 2011, 01:44 PM
tom@ballooncalculus
Re: Integration by Substitution
Notice your u version of the integrand has no traces of the x in the numerator of the original. To bring that part with you, you can re-express it (x, that is) in terms of u, or else use integration by parts.
• November 15th 2011, 01:47 PM
e^(i*pi)
Re: Integration by Substitution
Quote:

Originally Posted by GrigOrig99
Hi, I'm new to integration. Can anyone help me figure out where I went wrong?
Many thanks.

Q. Evaluate the following:

Your numerator disappeared when changing the variable. Express x in terms of u.

edit: drat, beaten to it (Doh)
• November 15th 2011, 01:47 PM
Quacky
Re: Integration by Substitution
You haven't taken into account the $x$ on the numerator.

We have:

$\int^4_0{\frac{x}{\sqrt{2x+1}}}dx$

I'm going to let $u^2=2x+1$ so that $2u~du=2~dx$ and $dx=u~du$

When $x=4$, $u=3$ and when $x=0$, $u=1$

I have:

$\int^4_0{\frac{x}{\sqrt{2x+1}}}dx$

= $\int^3_1{\frac{u^2-1}{2u}}\cdot{u}~du$

Take it!
• November 15th 2011, 02:07 PM
tom@ballooncalculus
Re: Integration by Substitution
The parts version...

http://www.ballooncalculus.org/draw/parts/thirty.png

... where (key in spoiler) ...

Spoiler:
http://www.ballooncalculus.org/asy/prod.png

... is the product rule, straight continuous lines differentiating downwards with respect to x.

http://www.ballooncalculus.org/asy/maps/parts.png

... is lazy integration by parts, doing without u and v.

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• November 15th 2011, 02:39 PM
GrigOrig99
Re: Integration by Substitution
Thanks guys.