Evaluate the given definite integral.

Question

Evaluate the given definite integral. ​  ​

Quizplus · Accepted Answer

Using the substitution $x=3\sin^2	heta$, we have $dx=6\sin	heta\cos	heta d	heta$ and the limits of integration become $	heta=0$ to $	heta=\frac{\pi}{2}$. Thus, the integral becomes 
\begin{align*}
\int_0^3x\sqrt{3-x}\,dx &= \int_0^{\frac{\pi}{2}} 3\sin^2	heta\sqrt{3-3\sin^2	heta}\cdot 6\sin	heta\cos	heta\,d	heta\
&=54\int_0^{\frac{\pi}{2}} \sin^3	heta\cos^2	heta\,d	heta\
&=54\int_0^{\frac{\pi}{2}} \sin^3	heta(1-\sin^2	heta)\,d	heta\
&=54\int_0^1 u^3(1-u^2)\frac{du}{2\sqrt{u}}\
&=\frac{27}{2}\int_0^1 u^\frac{5}{2}-u^\frac{7}{2}\,du\
&=\frac{27}{2}\left[\frac{2}{7}u^\frac{7}{2}-\frac{2}{9}u^\frac{9}{2}ight]_0^1\
&=\frac{27}{2}\left(\frac{2}{7}\cdot 1^\frac{7}{2}-\frac{2}{9}\cdot 1^\frac{9}{2}ight)\
&=\frac{27}{7}.
\end{align*}

Evaluate the Given Definite Integral. ​ ​

Evaluate the Given Definite Integral.