Question

If \(f\) is a function such that \(f'(x) = -f(x)\), then \( ∫x f(x)~dx\) =
a) x f(x) + f(x) + C
b) -xf(x) - f(x) + C
c) (x^2 /2) f(x) + C
d) -(x^2 /2) f(x) + C
e) - (x^2 /2) f(x) - (x^3 /6) f(x) + C

symbolic notations makes my brain not think straight...
it's probably really easy, but I just don't see it >,<
help? :3

Answer 1

Integration by parts.

\(\int xf(x)\;dx = \int f(x)\;d\left(½x^2\right) = ½x^{2}f(x) - \int ½x^{2}f'(x)\;dx\)

Does that get us anywhere?

Answer 2

so
u = f(x) v = (1/2)x^2
du = -f'(x) dv = x dx
f(x) (1/2)(x^2) + (1/2) ∫ x^2 f'(x) dx
(x^2 /2)f(x) + (1/2) ∫x^2 f'(x) dx
it gets us closer to Options C, D, or E ...
so then should I integrate by parts again? :/ @tkhunny

Answer 3

or would it keep on going forever... ? .-. whut

Answer 4

im getting B

Answer 5

mmm could you explain how please? :3

Answer 6

yeah

Answer 7

\[f^{'}(x)=-f(x)\]
\[\int\limits_{}^{}f^{'}(x)=-\int\limits_{}^{}f(x) \rightarrow f(x)=-\int\limits_{}^{}f(x)\]
can we write it?

Answer 8

oh... that works... but how do I use that in my question?

Answer 9

You could solve the differential equation
\[f'(x)+f(x)=0\]
for \(f(x)\) and then find \(\int xf(x)~dx\)

Answer 10

f(x) = f'(x) ? is that what you mean by "solve" ?
so ∫x(f'(x)) dx then by parts? :3

Answer 11

now
\[\int\limits_{}^{}xf(x)dx=\int\limits_{}^{}-xf^{'}(x)dx\]
\[=-[x \int\limits_{}^{}f^{'}(x)-\int\limits_{}^{}f(x)]=-[xf(x)+f(x)+c]\]

Answer 12

My suggestion is probably beyond the scope of the question. Just another way to solve this. Is this for Calc II or some equivalent?

Answer 13

yes, Calc 2 / APCalcBC

Answer 14

do u got my solution @jigglypuff314

Answer 15

@sidsiddhartha, how is \(\int f(x)~dx=f(x)\)?

Answer 16

Wait nevermind, I see what you did

Answer 17

it is given
f'(x)=-f(x)
i have integrated both sides

Answer 18

Yeah I noticed just now, thanks

Answer 19

after that just used by parts then just substituted
\[-\int\limits_{}^{}f(x)=f(x)\]

Answer 20

Anyway @jigglypuff314 my suggestion involves the differential equation
\[f'(x)+f(x)=0\]
which gives the characterstic equation
\[r+1=0~~\Rightarrow~~r=-1\]
So the general solution is
\[f(x)=Ce^{-x}\]
And you can see that
\[\begin{cases}f'(x)=-Ce^{-x}\\
f(x)=Ce^{-x}\end{cases}~~\Rightarrow~~f'(x)=Ce^{-x}=-Ce^{-x}=-f(x)\]

Answer 21

Then if you integrated,
\[C\int xe^{-x}~dx=-Cxe^{-x}+\int e^{-x}~dx=-Cxe^{-x}-Ce^{-x}+C^*=-xf(x)-f(x)+C^*\]

Answer 22

.-. um yeah I try sids' way O-O thanks @SithsAndGiggles xD

thanks @sidsiddhartha I think I understand now, thank you all! <3