Answer
$\frac{128}{15}$
Work Step by Step
Evaluate the integral: $\int^{4}_{0}(4-t)\sqrt t dt$
Recall the 2nd part of the Fundamental Theorem of Calculus:
$∫^b_af(x)dx=F(b)−F(a)$
Find $F(t)$:
$F(t)=\int(4-t)\sqrt t dt$
$F(t)=\int(4t^{1/2} - t^{3/2}) $
$F(t)=\frac{2}{3}4t^{3/2} -\frac{2}{5}t^{5/2}$
$F(t)=\frac{8}{3}t^{3/2} -\frac{2}{5}t^{5/2}$
Now evaluate $F(b)−F(a)$:
$F(4)−F(0)$
$(\frac{8}{3}(4)^{3/2} -\frac{5}{2}(4)^{5/2}) -(\frac{8}{3}(0)^{3/2} -\frac{5}{2}(0)^{5/2})$
$(\frac{8}{3}(\sqrt 4)^{3} -\frac{2}{5}(\sqrt 4)^{5}) -0$
$\frac{8(8)}{3} -\frac{2(32)}{5}$
$\frac{64}{3} - \frac{64}{5}$
$\frac{320}{15}-\frac{192}{15} = \frac{128}{15}$
