When the denominator has a sqrt(x), to integrate this, I would "flip" it with the numerator to make it a (x)^-1/2. And then if I take the integral of this function I get "2x^(1/2)+C".... When adding 1 to "-(1\2)" you get just (1/2), so how does the the 2 end up top (numerator) as a coefficient of x^(1/2)?

Question

anonymous · Answer

Can you explain why 3 divided by 1/2 is 6?

anonymous · Answer

Because you flip the fraction on the bottom and multiply it to the numerator... right?

anonymous · Answer

Exactly. So same thing here $$\frac{x^{1/2}}{1/2} $$ means you flip the 1/2 in the bottom and what is 1/2 flipped over?

anonymous · Answer

I'm just confused because you "bring up" the sqrt(x) and it becomes a negative exponent, but after the integral happens it is now positive... if it's now a positive exponent shouldn't it stay at the top? in which case why would you flip the fraction taken from the integral process?

anonymous · Answer

$$\int\limits_{}^{}1/\sqrt{x}dx = \int\limits_{}^{}x^{-1/2}dx = (1/2)x ^{1/2} +C$$

anonymous · Answer

So you are talking about 2 separate processes: rewriting (this is just for appearances - no 'math' goes on) and integration (math happens here). The process of integration says to do 2 steps: add 1 to the exponent, then divide by the new exponent.

anonymous · Answer

so, it's a simplification step to then multiply the flipped denominator? (sorry, I have poor math vocabulary)