How to find the equation f(x), if given f(1)= 2, f(3)= 4, f(9) = 1, integral of f(x)dx from 1 to 3 = 4...

How would you find the equation of f(x) (or what might you just call this), if given f(1)= 2, f(3)= 4, f(9) = 1, integral of f(x)dx from 1 to 3 = 4, integral of f(x)dx from 1 to 9 = -2.

The question asks,

(a) the integral of f(x)dx from 9 to 3 = ?
(b) the integral of xf(x^2)dx from 1 to 3 =?
(c) the integral of xf ' (x)dx =?
(d) d/dx*[integral of sf(s^2)ds from 1 to x] =?

Im totally lost and i don't even know where to begin. this looks like calc 1 and I'm in calc 2 now and i just forgot what to do with this.

Thanks!

(a) ∫(x = 1 to 9) f(x) dx = ∫(x = 1 to 3) f(x) dx + ∫(x = 3 to 9) f(x) dx
==> -2 = 4 + ∫(x = 3 to 9) f(x) dx
==> ∫(x = 3 to 9) f(x) dx = -6
==> ∫(x = 9 to 3) f(x) dx = 6.

(b) Let u = x^2, du = 2x dx.
Bounds: x = 1 ==> u = 1 and x = 3 ==> u = 9.

So, ∫(x = 1 to 3) x f(x^2) dx
= ∫(u = 1 to 9) f(u) * du/2
= (1/2) ∫(u = 1 to 9) f(u) du
= (1/2) * (-2); the dummy variable is irrelevant
= -1.

(c) Use integration by parts.
Let u = x, dv = f '(x) dx
du = dx, v = f(x).

So, ∫ x f '(x) dx = x f(x) - ∫ f(x) dx
[If you have bounds, then you can take it from here.]

(d) By the Fundamental Theorem of Calculus, this equals x f(x^2).

I hope this helps!