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