# find integral of this

• May 28th 2008, 09:52 AM
szpengchao
find integral of this
integral( 1/sqrt(x),x, 0,1) = 2sqrt(x)
• May 28th 2008, 09:54 AM
ThePerfectHacker
Quote:

Originally Posted by szpengchao
integral( 1/sqrt(x),x, 0,1) = 2sqrt(x)

$\int_0^1 \frac{dx}{\sqrt{x}} = \int_0^1 x^{-1/2} dx$.
Now use the exponent rule.
• May 28th 2008, 11:22 AM
szpengchao
how
how to?
• May 28th 2008, 11:24 AM
Chris L T521
Quote:

Originally Posted by szpengchao
how to?

Power Rule:

$\int u^n \,du = \frac{u^{n+1}}{n+1}$

Does that help?
• May 28th 2008, 11:25 AM
szpengchao
..
i mean from definition.... is it the only way to find integral?
• May 28th 2008, 11:32 AM
Moo
Quote:

Originally Posted by szpengchao
i mean from definition.... is it the only way to find integral?

Mmmm... remember the derivative of $\sqrt{t}$ ?

It's $\frac{1}{2 \sqrt{t}}$, recognize it in the integral (Wink)
• May 28th 2008, 11:34 AM
szpengchao
well
i do it by riemann integrable.

define LRS and URS, i suceed to show LRS=URS, but cant show that LRS=2sqrt(a)

m...so we can say, becoz derivative of sqrt(x) is that, so the definite integral =2sqrt(x)
• May 28th 2008, 01:30 PM
ThePerfectHacker
You cannot use the Riemann integral over here. The function $\frac{1}{\sqrt{x}}$ is even defined on $[0,1]$. If you had $[1,2]$ then it is possible to do it by definition, but not the way you posted the problem.