# Thread: find integral of this

1. ## find integral of this

integral( 1/sqrt(x),x, 0,1) = 2sqrt(x)

2. 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.

3. ## how

how to?

4. Originally Posted by szpengchao
how to?
Power Rule:

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

Does that help?

5. ## ..

i mean from definition.... is it the only way to find integral?

6. 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

7. ## 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)

8. 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.