If \(\int^3_0(px^2 + 16)dx\) = 129. Find the value of p.
Answer Details
The value of p can be found by solving the definite integral for a given value of p and equating it to the given value of 129.
The definite integral of the function px^2 + 16 between the limits 0 and 3 can be calculated using antiderivatives. The antiderivative of px^2 is (1/3)px^3 and the antiderivative of 16 is 16x. So,
∫^3_0(px^2 + 16)dx = (1/3)p * 3^3 + 16 * 3 - (1/3)p * 0^3 - 16 * 0 = (1/3)p * 27 + 48
Now, equating this expression to 129 and solving for p, we get:
(1/3)p * 27 + 48 = 129
(1/3)p * 27 = 81
p * 27 = 243
p = 243/27
p = 9
So, the value of p is 9.