∫03(px2+16)\int\limits_{0}^{3}(px^2 + 16)0∫3(px2+16) = 129
px2+10+1+16x∣03=129\frac{px^2 + 1}{0 + 1} + 16x\Big|_{0}^{3} = 1290+1px2+1+16x03=129
px33+16x∣03=129\frac{px^3}{3} + 16x\Big|_{0}^{3} = 1293px3+16x03=129
(p(3)33+16(3)\frac{p(3)^3}{3} + 16(3)3p(3)3+16(3)) - 0 = 129
9p + 48 = 129
9p = 129 - 48
9p9=819\frac{9p}{9} = \frac{81}{9}99p=981
p = 9
