# Thread: divergent/convergent integral,simple Q icant solve

1. ## divergent/convergent integral,simple Q icant solve

Let f(x) be a continuous function defined on the interval [2, infinity) such that

determine the value of

it seems like an easy Q but I can't find the trick to it.

2. Originally Posted by monokill
Let f(x) be a continuous function defined on the interval [2, infinity) such that

determine the value of

it seems like an easy Q but I can't find the trick to it.

Here's the trick:

$\displaystyle \int f(x)e^{-x/2}~dx = -f(x)e^{-x/2}+2\int f'(x)e^{-x/2}~dx$
(Integration by parts )

3. hmm i still dont haf the right answer.

from what u suggested:

$\displaystyle \int f(x)e^{-x/2}~dx = -f(x)e^{-x/2}+2\int f'(x)e^{-x/2}~dx$ = -4

so

{ -f(x)e^(-x/2) } evaluated from 3 to infinity + 2x = -4

then i got

-infinity e^(-infinity/2) + 10e^(-3/2) = -4

but infinity x 0 = ??

4. $\displaystyle \lim_{x\to \infty} f(x)e^{-x/2}= 0$ because it says $\displaystyle \int_3^{\infty} f(x)e^{-x/2}~dx = -4$. The function can approach nowhere but 0.

So,

$\displaystyle -f(x)e^{-x/2} |^{\infty}_{3} + 2\int^{\infty}{3}f'(x)e^{-x/2}~dx = -4$

Let's call $\displaystyle \int^{\infty}{3}f'(x)e^{-x/2}~dx = t$

$\displaystyle -f(x)e^{-x/2} |^{\infty}_{3} + 2t = -4$

$\displaystyle (\lim_{x\to \infty} -f(x)e^{-x/2})-(-f(3)e^{-3/2}) + 2t = -4$

$\displaystyle f(3)e^{-3/2} + 2t = -4$...

5. That's what I did. So the answer should be

but its incorrect

6. Sorry, I mistyped the integral and went wrong..

$\displaystyle \int f(x)e^{-x/2}~dx = -2f(x)e^{-x/2}+2\int f'(x)e^{-x/2}~dx$

$\displaystyle -2f(x)e^{-x/2}+2\int f'(x)e^{-x/2}~dx~|^{\infty}_{3} = -4$

$\displaystyle t = \int^{\infty}_{3}f'(x)e^{-x/2}~dx$

$\displaystyle -2f(x)e^{-x/2}+2t~|^{\infty}_{3} = -4$

$\displaystyle \lim_{x\to \infty} f(x)e^{-x/2}= 0$

$\displaystyle (0) - (-2f(3)e^{-3/2}) + 2t = -4$

$\displaystyle 20e^{-3/2} + 2t = -4$

$\displaystyle t = -2 - 10e^{-3/2}$

I hope this one is true..

7. it is! thanks so much