1. ## log and integrals

Hello

In this problem, need to find the area below the function:
1.
$f(x) = \frac{2-2x}{1+2x}$ in the interval [0,1].

And above the x axis.

I am told there is some logarithms involved here.

After sitting all night....another problem:

2.
Prove that the function has an absolute minimum.

$f(x) = x^2+(\ln{x})^2$

Thanks.

2. ## Re: log and integrals

Hey ryu1.

The area will involve the integral so you need to be able to solve the integral.

Hint: Try substituting in (2 - 1 + 1) and getting your integral in terms -1 + a/(1+2x) dx and consider the derivative of log(1+2x).

3. ## Re: log and integrals

Originally Posted by ryu1
Hello

2.
Prove that the function has an absolute minimum.

$f(x) = x^2+(\ln{x})^2$

Thanks.
As for this problem, can I just say that lnx is continuous for all X>0
and
$\lim_ {x \to 0^+}{x^2+(lnx)^2} = +\infty$

and also
$\lim_ {x \to +\infty}{x^2+(lnx)^2} = +\infty$

therefore an absolute minimum is guaranteed ? Thanks.

4. ## Re: log and integrals

Originally Posted by chiro
Hey ryu1.

The area will involve the integral so you need to be able to solve the integral.

Hint: Try substituting in (2 - 1 + 1) and getting your integral in terms -1 + a/(1+2x) dx and consider the derivative of log(1+2x).

Got it!

Thanks, problem solved