## Calculus: Early Transcendentals 8th Edition

$\displaystyle\int\limits_0^4(4-t)\sqrt{t}dt=\dfrac{128}{15}$
$\displaystyle\int\limits_0^4(4-t)\sqrt{t}dt$ 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: $...=\displaystyle4\int\limits_0^4t^{1/2}-\int\limits_0^4t^{3/2}dt=(4)\dfrac{t^{1+\frac{1}{2}}}{1+\frac{1}{2}}-\dfrac{t^{1+\frac{3}{2}}}{1+\frac{3}{2}}\Big|_0^4=...$ $...=(4)\dfrac{t^{3/2}}{3/2}-\dfrac{t^{5/2}}{5/2}\Big|_0^4=\dfrac{8}{3}\sqrt{t^{3}}-\dfrac{2}{5}\sqrt{t^{5}}\Big|_0^4=...$ $...=\Big[\dfrac{8}{3}\sqrt{(4)^{3}}-\dfrac{2}{5}\sqrt{(4)^{5}}\Big]-\Big[\dfrac{8}{3}\sqrt{(0)^{3}}-\dfrac{2}{5}\sqrt{(0)^{5}}\Big]=...$ $...=\dfrac{8}{3}\sqrt{64}-\dfrac{2}{5}\sqrt{1024}=\dfrac{8}{3}(8)-\dfrac{2}{5}(32)=\dfrac{64}{3}-\dfrac{64}{5}=\dfrac{128}{15}$