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Integration

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
how.JPG

How did WolframAlpha get from step 1 to 2?

I don't get where that extra "two" came from (outside of the integral), and I don't get how there is an extra u factored out.

My initial thought was maybe they mutlipled the outside by two, and inside by 1/2 so that equals 1, but it still doesn't account for the missing 1/sqrt(x) to complete "du".
 

agentmulder

Active member
Feb 9, 2012
33
From

$ du = \frac{1}{2 \sqrt{x}}dx $

We get $ 2 \sqrt{x} \ du = dx $

[tex] 2u \ du = dx [/tex]

Now substitute and pull out the constant 2 ...

:D
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
How did I miss that the first time :( The ramifications of doing calculus up all night for fun.
Thanks!
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
A simpler approach would be to expand .
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715