Question

If f is a continuous function such that `int_0^xf(t)dt=xe^(2x)`+ `int_0^xe^-tf(t)dt` for all x, find an explicit formula for f(x) ?

I'm not sure how to use the fundamental theorem of calculus to solve for this problem.

I don't think I wrote the f(x)=? in the correct formal form.

I do know that you first take the derivative of the function and add and plug in the values from the bounds of the function?

Also I am unsure of what to do with the "`+e^-x`" in the first part?

Thanks so much for your help in advance

Answer 1

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

Explanation:

We have:

`int_0^x \ f(t) \ dt=xe^(2x) + int_0^x \ e^(-t)f(t) \ dt`

If we differentiate wrt `x`, we get:

`d/dx \ int_0^x \ f(t) \ dt = d/dx \ xe^(2x) + d/dx \ int_0^x \ e^(-t)f(t) \ dt`

The Fundamental Theorem of Calculus tell us that:

`d/dx \ int_0^x \ f(t) \ dt = f(x)`

Thus if we apply this theorem, we have:

`f(x) = d/dx \ xe^(2x) + e^(-x)f(x)`

And now we can apply the product rule:

`f(x) = (x) (d/dx e^(2x)) + (d/dx x)(e^(2x)) + e^(-x)f(x)`

`\ \ \ \ \ \ \ = (x) (2e^(2x)) + (1)(e^(2x)) + e^(-x)f(x)`

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

`:. f(x) - e^(-x)f(x) = 2xe^(2x) + e^(2x)`

`:. (1 - e^(-x))f(x) = (2x + 1)e^(2x)`

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