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.
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 = ??
$\displaystyle \lim_{x\to \infty} f(x)e^{-x/2}= 0$ 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 $\displaystyle \int^{\infty}{3}f'(x)e^{-x/2}~dx = t$
$\displaystyle -f(x)e^{-x/2} |^{\infty}_{3} + 2t = -4$
$\displaystyle (\lim_{x\to \infty} -f(x)e^{-x/2})-(-f(3)e^{-3/2}) + 2t = -4$
$\displaystyle f(3)e^{-3/2} + 2t = -4$...
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..