f ″(x) = x^−2, x > 0, f(1) = 0, f(5) = 0

Question

f ″(x) = x⁻²,    x > 0,    f(1) = 0,    f(5) = 0

find f

Answer 1

f'(x) = ∫f"(x)dx = -1/x + C

f(x) = ∫f'(x)dx = -lnx + Cx + D

Since f(1) = 0, -ln1 + C + D = 0. So, C + D = 0

Since f(5) = 0, -ln5 + 5C + D = 0.

We have the system of equations:

C + D = 0

5C + D = ln5

Subtracting the equations, we get -4C = -ln5. So, C = (1/4)ln5.

Since C = (1/4)ln5 and C + D = 0, we have D = -C = -(1/4)ln5

Therefore, f(x) = -lnx + (1/4)xln5 - (1/4)ln5