Definite integral finding unknown

Question

Given ∫1^2 f(u) du =-5, ∫1^2 h(u) du =4, ∫2^5 f(u) du =8.

Find the value of p if ∫1^5 [f(u)-3pu] du=39

I just know few on how to solve this 39 = ∫1^5 f(u) du - ∫1^5 3pu du 39 = ∫1^2 f(u) du + ∫2^5 f(u) du - ∫1^5 pu du 39 = -5+8 -3∫1^5 pu du 36 = -3∫1^5 pu du

What to do next?

Answer 1

Assuming I've understood your question:

Since $\int_1^5 f(u) du=-5+8=3$, $3p=\frac{3-39}{\int_1^5 u du}=-\frac{36}{12}=-3$, i.e. $p=-1$.