Question

For all real numbers \(x\), the function \(f(x)\) satisfies $$ 6 +f(x)=2f(-x)+3x^2\left(\int_{-1}^1 f(t)dt \right) $$ It follows that \(\int_{-1}^1 f(x) dx\) equals:

(a) 4, (b) 6, (c) 11, (d) 27/2 , (e) 23.

( i think it is (a))

Answer 1

@Zarkon

Answer 2

did you try to take the derivative of both sides

Answer 3

lets call A the integral f(x) , x = -1..1
f ' (x) = 2* f ' (-x) * -1 + 6x * A

Answer 4

I could give a hint, but I think it will make it too easy

Answer 5

( 6 + f(x) - 2f(-x) ) / (3x^2 ) = integral f(t) , t = -1..1

Answer 6

Hint: Do the opposite of what perl suggested in his first post

Answer 7

take the antiderivative?

Answer 8

integrate from -1 to 1 both sides

Answer 9

ok one sec

Answer 10

how do you integrate f(-x)

Answer 11

u-sub

Answer 12

right ok

Answer 13

another hint, treat the integral at the end as a constant

Answer 14

\[\int\limits_{-1}^{1}6 + f(x) = \int\limits_{-1}^{1}(f(-x)+3x^2\int\limits_{-1}^{1}f(x))\]

Answer 15

the last integral should be smaller

Answer 16

u=-x ... du = -1 dx

if the integral from -1 to 1 is A, we have:\[12+A=-2A+\left[ x^3 \right]_{-1}^1 A\]

Answer 17

why do you have -2A

Answer 18

be careful , you are using the old x limits (you need to change limits when applying u substitution)

Answer 19

\[6 +f(x)=2f(-x)+3x^2\left(\int_{-1}^1 f(t)dt \right)\]

ohhh .... its' not from -1 to 1 any more ... thanks

Answer 20

f(-x) is just f(x) flipped over the y-axis

so for any \(a\) where the integral exists we have

\[\int_{-a}^af(-x)dx=\int_{-a}^{a}f(x)dx\]

Answer 21

12 + A=2A +2A ??

Answer 22

yes

Answer 23

ok nvm ... A=4 ...cool

Answer 24

for fun:\[\large 6+f(x)=2f(-x)+3x^2 \int\limits_{-1}^1f(t)dt\] since x is a variable, justplug in -x to this\[\large 6+f(-x)=2f(x)+3x^2 \int\limits_{-1}^1f(t)dt\] subtract the second from the first equation: \[\large f(x)-f(-x)=2f(-x)-2f(x)\] a little more algebra \[\large f(x)=f(-x)\]

Answer 25

nice argument there

Answer 26

Thanks a lot everyone ! ... not enough medals to go around.

I took a different approach and it took forever to get started.

I simply assumed f(x) was quadratic...

Answer 27

i think this way also makes sense ... just wanted to make sure it's right

Answer 28

Allowing \[\large K = \int\limits_{-1}^1f(t)dt = \int\limits_{-1}^1f(x)dx\] since x and k are just dummy variables here and allowing f(x)=f(-x) in the original equation we have: \[\large 6+ f(x)=2f(x)+3x^2K\] solving for f(x) we have: \[\large f(x)=-3Kx^2+6\] integrate this from -1 to 1\[\large K=2 \int\limits_0^1-3Kx^2+6dx\] we get \[\large K = 2(-K+6)\] simplifies to \[\large K=4\], which is a, the right answer =P

Answer 29

I think you make too many assumptions you don't need to make @PaxPolaris =P

Answer 30

yeah ... that was what made me uncomfortable

Answer 31

Does my response make sense? I'll explain anything you want since I did skip several steps but the main stuff is there I think. It's sort of cool how you have f(x) depends on the integral and the integral of it is the integral itself you know? XD

Answer 32

yes i understand now .. how you can do this without making the assumptions i made...

Answer 33

the assumption that you made, f(x) was quadratic, is actually unjustified. f could be a non polynomial function

Answer 34

Just, can't f(x) be written as a Taylor series? .... then to satisfy the given equation all the co-efficients except for \(x^2\) and \(x^0\) will have to be 0 .

And we still end up with \(f(x)=-12x^2+6\) ... without making any assumptions

Answer 35

I suppose that would work.

Answer 36

Thanks again. you guys are the best. :)