## Calculus 8th Edition

$\displaystyle \frac{256}{15}≈17.066667$
$\displaystyle \int_1^4\sqrt{t}(1+t)dt = \int_1^4t^{1/2}(1+t)dt=\int_1^4t^{1/2}+t^{3/2}dt$ Use the reverse power rule (add one and divide by it) to evaluate the integral: $\displaystyle (\frac{t^{3/2}}{3/2}+\frac{t^{5/2}}{5/2})|_1^4$ $\displaystyle (\frac{2t^{3/2}}{3}+\frac{2t^{2/5}}{5})|_1^4$ $\displaystyle [\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}]-[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}]$ $\displaystyle [\frac{2(2)^{3}}{3}+\frac{2(2)^{5}}{5}]-[\frac{2}{3}+\frac{2}{5}]$ $\displaystyle [\frac{2(8)}{3}+\frac{2(32)}{5}]-\frac{2}{3}-\frac{2}{5}$ $\displaystyle \frac{16}{3}+\frac{64}{5}-\frac{2}{3}-\frac{2}{5}$ $\displaystyle \frac{14}{3}+\frac{62}{5}$ $\displaystyle \frac{14\cdot5}{15}+\frac{62\cdot3}{15}=\frac{70+186}{15}$ $\displaystyle \frac{256}{15}≈17.066667$