Challenging Integral Problem

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grapz

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Jan 13, 2007
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Suppose that f is a function such that f( 0 ) = y and f ( pi ) = 2 and that the integral from 0 --> pi of [ ( f (x) + f''(x) ) sin x dx ] exists. Find the value of the integral from 0 --> pi of [( f(x) + f''(x)) sin x dx

I multiply the integral out.

Int [ f (x) sinx + f''(x) sin x ]

Then i subtract f' (x) cos x and add f'(cos x) to obtain

Int [ f sin x - f'cosx + f'cos x + f''sinx ]
= Int [ f(x) sinx - f'(x) (cosx) ] + int [ f'(x) cos x + f''(x) sinx) ]
= int [f(x) (-cosx) ]' + int ( f'(x) (sinx) ]' - using product rule backward

But then i am stuck because i can integrate both functions, however i am not given a value for f'(x) so how do i finish this?
 
Did you try integration by parts?.

The f'(x) should cancel out and make it a moot point.

\(\displaystyle \int_{0}^{\pi}f(x)sin(x)dx+\int_{0}^{\pi}f''(x)sin(x)dx\)

Try using parts on each one. For the left one let, say, \(\displaystyle u=f(x), \;\ dv=sin(x)dx, \;\ du=f'(x)dx, \;\ v=-cos(x)\)

Can you continue?.
 
Then let \(\displaystyle u=sin(x), \;\ dv=f''(x)dx, \;\ du=cos(x)dx, \;\ v=f'(x)\)

You get \(\displaystyle -f(x)cos(x)+\int_{0}^{\pi}f'(x)cos(x)dx+f'(x)sin(x)-\int_{0}^{\pi}f'(x)cos(x)dx\)
 
Yes but now we're back to step one hehe. We dont' have an value for f ' (x), unless i'm missing something
 
galactus said:
Then let \(\displaystyle u=sin(x), \;\ dv=f''(x)dx, \;\ du=cos(x)dx, \;\ v=f'(x)\)

You get \(\displaystyle -f(x)cos(x)+\int_{0}^{\pi}f'(x)cos(x)dx+f'(x)sin(x)-\int_{0}^{\pi}f'(x)cos(x)dx\)
Click to expand...
\(\displaystyle -f(x)cos(x)+f'(x)sin(x)+\int_{0}^{\pi}f'(x)cos(x)dx-{\int_{0}^{\pi}f'(x)cos(x)dx\)
 
Use your given values and you'll see. Don't focus so much on f'(x).
 
grapz said:
Yes, but how do i deal with the f'(x) sin ( x)
Click to expand...

How much is sin(0) and sin(pi)?

Sorry pka - I did not see your response - now I cannot delete the post.
 
Ah thank you. I was to focused on the f'(x) and didn't realize what i was given
 
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