Answer:
a) lnf(x)dx =[ xlnf(x)- x] + C
b) 5(ln5pq-1)
Explanation:
Given f(x) = pxq,
Y=ln(pxq)
(a). f(y) = ∫lnf(x)dx over range x
Using integration by parts
∫udv = uv - ∫vdu
lnf(x) = u, dv =1, -----> v = x
∫udv = lnf(x). x - x∫(1/pxq)pq.dx
∫udv = xlnf(x) - ∫dx
∫lnf(x)dx =[ xlnf(x)- x] + C
(b) ∫lnf(x)dx = [xln(pxq) - x], at range x= 0, and x= 5
= 5ln5pq - 5
=5(ln5pq-1)