Suppose f(x) is such that f''(x) is continuous on [0, π], f(π) = 1 and
{Integral with upper bound π and lower bound 0 (f(x) + f''(x)) sin(x)dx = 2.}
Find f(0).

I know that I should break the integral into two separate integrals as in:
{(Integral with upper bound π and lower bound 0) f(x) sin(x)dx + (Integral with upper bound π and lower bound 0) f''(x) sin(x)dx. Don't know what to do next.

Question

Suppose f(x) is such that f''(x) is continuous on [0, π], f(π) = 1 and 
{Integral with upper bound π and lower bound 0 (f(x) + f''(x)) sin(x)dx = 2.}
Find f(0).

I know that I should break the integral into two separate integrals as in:
{(Integral with upper bound π and lower bound 0)  f(x) sin(x)dx + (Integral with upper bound π and lower bound 0) f''(x) sin(x)dx. Don't know what to do next.

tylerd · Answer

havent done math in awhile, ill take a crack at it though

tylerd · Answer

what section of diffy q is this

dillonmerv · Answer

calc 2

zzr0ck3r · Answer

$\int uv'dx=uv-\int u'vdx$

$\int_0^\pi f(x)\sin(x)dx+\int_0^\pi f^{''}(x)\sin(x)dx=2$

On the first integral we let $v=-\cos(x)$ and $u=f(x)$

$\int_0^\pi f(x)\sin(x)dx=\ \int_0^\pi f(x)(\cos(x))'dx=\-f(x)\cos(x)|_o^\pi-\int_0^\pi -f(x)\cos(x)dx=\-1(-1)+f(0)+\int_0^\pi f^{'}(x)\cos(x)dx $

Do something similar on the second integral  (let $v=f^{'}(x)$ and $u=\sin(x)$)  and get 

$\int_0^\pi \sin(x)f^{''}(x)dx=-\int_0^\pi \cos(x)f^{'}(x)dx$


Conclude $f(0) = 1$

dillonmerv · Answer

Looks very good. Thanks!

dillonmerv · Answer

If I have any issues I'll get back to you.

zzr0ck3r · Answer

Sounds good. I might be sleeping, but someone should be around.

holsteremission · Answer

http://openstudy.com/study#/updates/5783cd72e4b000d65a2557c4