1. Definite Integral. Equivalency.

My question is about a). I see that:

$\int^e_1\frac{\sqrt(ln(x))}{x}\;dx=\int^1_0\sqrt(u )\;du$

those are equivalent because I graphed both of them. In the screenshots the area is listed in the lower left hand corner. Algebraically I am having a tougher time figuring out how those are equal. What keeps getting to me is how was it determined that the interval changes from 1 to e (to) 0 to 1.

Thanks in advance for any responses...

2. Re: Definite Integral. Equivalency.

Originally Posted by sepoto

My question is about a). I see that:

$\int^e_1\frac{\sqrt(ln(x))}{x}\;dx=\int^1_0\sqrt(u )\;du$

those are equivalent because I graphed both of them. In the screenshots the area is listed in the lower left hand corner. Algebraically I am having a tougher time figuring out how those are equal. What keeps getting to me is how was it determined that the interval changes from 1 to e (to) 0 to 1.

Thanks in advance for any responses...
$\int_1^e\dfrac{\sqrt{\ln(x)}}{x}dx$

$u=\ln(x)\;\;du=\dfrac{dx}{x}$ Do you understand this?

now find new integration limits given the substitution

$\ln(1)=0\;\;\ln(e)=1$

so the new integral becomes

$\int_0^1\sqrt{u}\;du$

3. Re: Definite Integral. Equivalency.

So it appears then that the new integral holds the corresponding u values of x. I think I can follow that. Thank you for posting.