Integrals?

Question

x
∫f(t)dt = xe^2x + ∫ (e^-t) f(t)dt
0
for all x, find an explicit formula for f(x)

I'm pretty confused on the layout of the function.
So its adding xe^2x to the integral? and if so how would someone integrate something like this?

(By the way, the second ∫ has the same limits: from 0 to x)

Answer 1

differentiate both sides...

f(x) = e^(2x) + 2xe^(2x) + e^(-x)f(x)
[1 - e^(-x)] f(x) = e^(2x) + 2xe^(2x)

f(x) = [e^(2x) + 2xe^(2x)] / [1 - e^(-x)]


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Answer 2

I think you intended to put the limits 0 to x on the integral on the right also. The fundamental theorem of calculus says that if you differentiate both sides of the equation, you get

f(x) = (1 + 2x)exp(2x) + exp(-x)f(x),

from which you can easily solve for f(x) explicitly. To check, you can just plug it into the equation and verify that it works.

Answer 3

we can write the equation as

x
∫f(t) (1 -e^-t) dt = xe^(2x)
0

Now differentiate wrt x to get

f(x) (1 -e^-x) + C= d / dx (xe^(2x) )

Now take the deriv of the RHS then solve for f(x)

Answer 4

Use Liebnitz's rule to differentiate both sides of the equation:

d/dx (integral[0->x] f(t)dt) = f(x)-f(0) so

f(x) = e^(2x) +2xe^(2x)+e^(-x)f(x)

f(x)*(1 -e^(-x)) = (1+2x)e^(2x)

f(x) = (1+2x)e^(2x)/(1-e^(-x))