divergent/convergent integral,simple Q icant solve

**monokill** · March 2nd 2008, 12:12 PM

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.

**wingless** · March 2nd 2008, 12:23 PM

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:

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

)

**monokill** · March 2nd 2008, 01:18 PM

hmm i still dont haf the right answer.

from what u suggested:

$<br /> \int f(x)e^{-x/2}~dx = -f(x)e^{-x/2}+2\int f'(x)e^{-x/2}~dx<br />$ = -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 = ??

**wingless** · March 2nd 2008, 01:55 PM

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

So,

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

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

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

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

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

**monokill** · March 2nd 2008, 03:33 PM

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

but its incorrect

**wingless** · March 3rd 2008, 07:13 AM

Sorry, I mistyped the integral and went wrong..

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

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

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

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

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

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

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

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

I hope this one is true..

**monokill** · March 3rd 2008, 08:34 AM

it is! thanks so much

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