1. ## Another integration problems

Ok, I solved it.
I got f(0) = 3. Is that right?

I then have another problem:

f(x) = definite integral from 0 to x : sin(t) / (t+1) dt.

1) Find where f(x) has local maximum and minimum values (I should not find the function values).

A more interesting one:
2) Show that f(x) > 0, when x > 0. (Here I should use integration by parts...)

Do anyone want to help? :-)

2. Originally Posted by expresstrain
Ok, I solved it.
I got f(0) = 3. Is that right?
i split your topic into two ones.

don't know if that is the correct answer, i have no time to do the problem.

Originally Posted by expresstrain
f(x) = definite integral from 0 to x : sin(t) / (t+1) dt.

2) Show that f(x) > 0, when x > 0. (Here I should use integration by parts...)

Do anyone want to help? :-)
Let $\displaystyle\int_0^x g(t)\sin(t)\,dt$ where $g(t)$ is class 1, positive and decreasing for $[0,\infty)$, then integrate by parts and get $\displaystyle\int_{0}^{x}{g(t)\sin (t)\,dt}=g(x)(1-\cos x)-\int_{0}^{x}{g'(t)(1-\cos t)\,dt}.$

since $g(x)>0,$ $g'(t)<0$ and $1-\cos t\ge0,$ the whole RHS is positive and we're done.

try to do part a) by yourself it's just a matter of FTC application.

3. Originally Posted by expresstrain
Ok, I solved it.
I got f(0) = 3. Is that right?
I have no idea! What was the question?

I then have another problem:

f(x) = definite integral from 0 to x : sin(t) / (t+1) dt.

1) Find where f(x) has local maximum and minimum values (I should not find the function values).
Max and min will occur where f'(x)= 0. And you find f' from the Fundamental Theorem of Calculus.

A more interesting one:
2) Show that f(x) > 0, when x > 0. (Here I should use integration by parts...)

Do anyone want to help? :-)
Jo, you should NOT use "integration by parts. You should not integrate at all. If f'(x)> 0 for x> 0, then f is increasing. and f(0)= ?

4. Originally Posted by Krizalid
since $g(x)>0,$ $g'(t)<0$ and $1-\cos t<0,$ the whole RHS is positive and we're done.
Hi, and thank you for helping me!
But shouldn't it be $1-\cos t$ is equal or greater than $0$, and not <0. ?? Then it makes sense to me...

Originally Posted by HallsofIvy
I have no idea! What was the question?
Haha, I'm sorry! :-)
The question I was referring to is:
http://www.mathhelpforum.com/math-he...em-162684.html

And I got f(0) = 3. I was just wondering if anyone else got the same... :-)

And again, thank you all for helping me!

5. Originally Posted by expresstrain
Hi, and thank you for helping me!
But shouldn't it be $1-\cos t$ is equal or greater than $0$, and not <0. ?? Then it makes sense to me...
i got a typo, of course it is as you say.