Answer
$\displaystyle\int\limits_0^4(4-t)\sqrt{t}dt=\dfrac{128}{15}$
Work Step by Step
$\displaystyle\int\limits_0^4(4-t)\sqrt{t}dt$
Rewrite the integral like this:
$\displaystyle\int\limits_0^4(4-t)t^{1/2}dt$
Evaluate the product present inside the integral:
$\displaystyle\int\limits_0^4(4-t)t^{1/2}dt=\int\limits_0^4(4t^{1/2}-t^{3/2})dt=...$
Integrate each term separately and apply the second part of the fundamental theorem of calculus:
$...=\displaystyle4\int\limits_0^4t^{1/2}-\int\limits_0^4t^{3/2}dt=(4)\dfrac{t^{1+\frac{1}{2}}}{1+\frac{1}{2}}-\dfrac{t^{1+\frac{3}{2}}}{1+\frac{3}{2}}\Big|_0^4=...$
$...=(4)\dfrac{t^{3/2}}{3/2}-\dfrac{t^{5/2}}{5/2}\Big|_0^4=\dfrac{8}{3}\sqrt{t^{3}}-\dfrac{2}{5}\sqrt{t^{5}}\Big|_0^4=...$
$...=\Big[\dfrac{8}{3}\sqrt{(4)^{3}}-\dfrac{2}{5}\sqrt{(4)^{5}}\Big]-\Big[\dfrac{8}{3}\sqrt{(0)^{3}}-\dfrac{2}{5}\sqrt{(0)^{5}}\Big]=...$
$...=\dfrac{8}{3}\sqrt{64}-\dfrac{2}{5}\sqrt{1024}=\dfrac{8}{3}(8)-\dfrac{2}{5}(32)=\dfrac{64}{3}-\dfrac{64}{5}=\dfrac{128}{15}$
