Question

Integration question below! Any help would be appreciated :)

Answer 1

Supposed that f(x) os such that f''(x) is continuous between [0,pi], and f(pi)=1.

Also, \[\int\limits_{0}^{\pi} f(x)+f''(x)\sin(x)dx=2\]

Find f(0).

Answer 2

what happens if you played with your integral there
write as two
and then have some fun with integration by parts
have you tried this?

Answer 3

Correction on the question by the way: it's (f(x)+(f''(x))sin(x)** @freckles

I'll try it out. :)

Answer 4

yep it should totally work now with that correction :)
i'm talking about writing as two integrals and using integration by parts

Answer 5

you want to see NO more f'' stuff
only f stuff

Answer 6

\[(I)\quad\int_0^\pi \bigl(f(x)+(f''(x)\sin(x))\bigr)dx\]
or
\[(II)\quad \int_0^\pi \bigl((f(x)+f''(x))\sin(x))\bigr)dx\]

Answer 7

II

Answer 8

he had unmatched delimiters
but i'm like 99% sure he means II

Answer 9

( f+ f'' ) * sin
right?

Answer 10

yeah

Answer 11

I also messed up the parens..

Answer 12

lol didn't count

Answer 13

I meant II, sorry haha

Answer 14

how is going salt

Answer 15

like are you able to see where the integration by parts will help?

Answer 16

still pretty salty, i think i got it now. will check up again in a few, if that's ok

Answer 17

k

Answer 18

thanks again

Answer 19

np

Answer 20

Hmm this is a cool problem :)

Answer 21

This is me being picky, but there's no reason integrating by parts to remove all instances of \(f\) (instead of \(f''\)) wouldn't work.

Answer 22

except we are given more information about f than f'' or f'
such as the f(pi)=1 thing

Answer 23

@mortonsalt you start with splitting the integral into two:
\[\int_0^\pi (f(x)+f''(x))\sin(x) dx = \int_0^\pi f(x)\sin(x)dx + \int_0^\pi f''(x)\sin(x)dx =(I) + (II)\]
And apply the IPP differently on both integrals. =>(I)= (a+integral1) and (II)=(b+integral2).
Do you see that integral1 and integral2 cancel out?

Answer 24

Seems to be a problem displaying the equation you typed out @reemii :(

Answer 25

\[\int_0^\pi (f(x)+f''(x))\sin(x) dx\\
= \int_0^\pi f(x)\sin(x)dx + \int_0^\pi f''(x)\sin(x)dx\\
=(I) + (II)\]

Answer 26

Yep, I've done that. :) I think I've got it sorted @reemii

Answer 27

good job

Answer 28

@freckles The nice thing about this problem is that the result is independent of that information in the sense that the info is used whichever way IBP is carried out (and we don't need any additional facts about \(f\) or its derivatives).

Let
\[\color{red}I+\color{blue}J=\color{red}{\int_0^\pi f(x)\sin x\,\mathrm{d}x}+\color{blue}{\int_0^\pi f''(x)\sin x\,\mathrm{d}x}\]
Integrating by parts by differentiating \(f\) in \(I\) twice gives the relation
\[\begin{align*}I&=1+f(0)+\int_0^\pi f'(x)\cos x\,\mathrm{d}x\\[1ex]
&=1+f(0)+\left(0-\int_0^\pi f''(x)\sin x\,\mathrm{d}x\right)\\[1ex]
&=1+f(0)-J
\end{align*}\]Meanwhile, differentiating \(\sin x\) in \(J\) twice gives
\[\begin{align*}
J&=0-\int_0^\pi f'(x)\cos x\,\mathrm{d}x\\[1ex]
&=-\left(-1-f(0)+\int_0^\pi f(x)\sin x\,\mathrm{d}x\right)\\[1ex]
&=1+f(0)-I
\end{align*}\]In both cases, \(I+J=2=1+f(0)\) and the result follows.

Answer 29

touche'

I think I still prefer the other way though as it only requires two rounds of IBP. :p