Unraveling the Mystery: How WolframAlpha Got from Step 1 to 2

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The discussion focuses on the mathematical transition in WolframAlpha's calculation from step 1 to step 2, specifically addressing the factor of 2 and the manipulation of the differential "du." The user clarifies that the expression $ du = \frac{1}{2 \sqrt{x}}dx $ leads to the transformation $ 2 \sqrt{x} \ du = dx $, which explains the presence of the constant 2 in the equation. The conversation highlights the importance of understanding these calculus fundamentals and the impact of late-night study sessions on comprehension.

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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".
 

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

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

2u \ du = dx

Now substitute and pull out the constant 2 ...

:D
 
How did I miss that the first time :( The ramifications of doing calculus up all night for fun.
Thanks!
 
A simpler approach would be to expand .
 
ZaidAlyafey said:
A simpler approach would be to expand .

Very true, but I wanted to practice without it :) Thanks guys!
 

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