Question

Find the anti-derivative of F(x)=1/sqrt(x)

Answer 1

\[
F(x) = x^{-1/2}
\]

Answer 2

how did you get that?

Answer 3

Now, remember that \[
(x^n) '= nx^{n-1}
\]Which means: \[
\left(\frac{x^n}{n}\right)' = x^{n-1}
\]

Answer 4

Well... \[
\frac{1}{\sqrt x} = \frac{1}{x^{1/2}} = x^{-1/2}
\]

Answer 5

my textbook says the answer is 2t^1/2+c

Answer 6

Now, remember that \[
(x^n) '= nx^{n-1}
\]Which means: \[
\left(\frac{x^n}{n}\right)' = x^{n-1}
\]If we let \(m=n-1\) then: \[
\left(\frac{x^{m+1}}{m+1}\right)' = x^{m}
\]

Answer 7

\[\int\limits_{ }^{ } x^n~dx~=\frac{n^{n+1}}{n+1}+C\]

Answer 8

I never found the anti derivative yet.

Answer 9

2x*

Answer 10

to what power?

Answer 11

ok sorry wio...contin ue

Answer 12

Anyway, since we want anti derivative of \[
x^{-1/2}
\]We can use:\[
\left(\frac{x^{m+1}}{m+1}\right)' = x^{m}
\]with \(m=-1/2\).

Answer 13

This gives: \[
\left(\frac{x^{1/2}}{1/2}\right)' = x^{-1/2}
\]

Answer 14

So anti derivative is \[
2\sqrt x+C
\]

Answer 15

\[\int\limits_{ }^{ } x^{-1/2}~dx=~\frac{x^{-1/2~~~+1}}{-1/2~~~~+1}=\frac{x^{1/2}}{1/2}=2x^{1/2}=2\sqrt{x}\color{red}{+C}\]

Answer 16

thank you guys very much!