If \(\int^3_0(px^2 + 16)dx\) = 129. Find the value of p.

Explanation

\(\int^3_0(px^2 + 16)\) = 129

\(\frac{px^2 + 1}{0 + 1} + 16x|^3_0 = 129\)

\(\frac{px^3}{3} + 16x|^3_0 = 129\)

(\(\frac{p(3)^3}{3} + 16(3)\)) - 0 = 129

9p + 48 = 129

9p = 129 - 48 

\(\frac{9p}{9} = \frac{81}{9}\) 

p = 9