divergent/convergent integral,simple Q icant solve

[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.

 

[skeeter]

$\displaystyle u = f(x)e^{-x/2}$

$\displaystyle du = \left(-\frac{1}{2} f(x)e^{-x/2} + f'(x)e^{-x/2}\right)dx$

$\displaystyle \int_{f(3)e^{-\frac{3}{2}}}^0 du = \int_3^{\infty} \left(-\frac{1}{2} f(x)e^{-x/2} + f'(x)e^{-x/2}\right)dx$

$\displaystyle -10e^{-\frac{3}{2}} = -\frac{1}{2} \int_3^{\infty} f(x)e^{-x/2} \, dx + \int_3^{\infty} f'(x)e^{-x/2} \, dx$

$\displaystyle -10e^{-\frac{3}{2}} = 2 + \int_3^{\infty} f'(x)e^{-x/2} \, dx$

$\displaystyle -\left(2 + 10e^{-\frac{3}{2}}\right) = \int_3^{\infty} f'(x)e^{-x/2} \, dx$

 

[galactus]

Try integration by parts.

Let $\displaystyle u=f(x), \;\ dv=e^{\frac{-x}{2}}dx, \;\ du=f'(x)dx, \;\ v=-2e^{\frac{-x}{2}}$

$\displaystyle \int_{3}^{\infty}f(x)e^{\frac{-x}{2}}dx=-2e^{\frac{-x}{2}}f(x)+2\int_{3}^{\infty}e^{\frac{-x}{2}}f'(x)dx$

Making the subs you were given and we get:

$\displaystyle -4+20e^{\frac{-x}{2}}=2\int_{3}^{\infty}f'(x)e^{\frac{-x}{2}}dx$

$\displaystyle \boxed{-2+10e^{\frac{-3}{2}}=\int_{3}^{\infty}f'(x)e^{\frac{-x}{2}}dx}$

 

[skeeter]

well pardner, ... looks like at least one of us is wrong.

HellifIknow which one of us it is ... I can't find my mistake, nor can I find one in galactus' work.

 

[monokill]

skeeter is right. thanks guys very much!

 

[galactus]

We only differ by a sign. Where did I go wrong?. Something rather subtle.

 

[skeeter]

We only differ by a sign. Where did I go wrong?. Something rather subtle.

I think I found it ...

$\displaystyle -4 = \left[-2e^{-\frac{x}{2}} f(x)\right]_3^{\infty} + 2\int_3^{\infty} f'(x) e^{-\frac{x}{2}} \, dx$

$\displaystyle -4 = \left[0 - \left(-2e^{-\frac{3}{2}} f(3)\right) \right] + 2\int_3^{\infty} f'(x) e^{-\frac{x}{2}} \, dx$

 

[galactus]

There ya' go, skeet. Thanks. I knew it had to be there somewhere.