Find $\int_{0}^{2}x\sqrt{4-x^{2}}\,dx$.

Solution

Method 1: Use the related indefinite integral and then apply the Fundamental Theorem of Calculus. The related indefinite integral $\int {x\sqrt{4-x^{2}}\,dx}$ can be found using the substitution $u = 4 - x^{2}$. Then $du\,=-2x\,dx$, so $x\,dx=-\dfrac{du}{2}.$ $$\begin{eqnarray*} \int {x\sqrt{4-x^{2}}\,dx}&=&\int {\sqrt{u}\left( {-}{\dfrac{{du}}{{2}}}\right) }={-}{\dfrac{{1}}{{2}}}\int u^{1/2}du=-\dfrac{1}{2}\cdot {{\dfrac{{u^{3/2}}}{{{ \dfrac{{3}}{{2}}}}}}}+C \\ &=&-\dfrac{1}{3}{{(4-x^{2})^{3/2}}}+C \end{eqnarray*}$$

Then by the Fundamental Theorem of Calculus, $$\int_{0}^{2} x\sqrt{4-x^{2}}\,dx = -\dfrac{1}{3} \Big[(4-x^{2})^{3/2}\Big]_{0}^{2} = -\dfrac{1}{3}\big[0-4^{3/2}\big] =\dfrac{8}{3}$$

Method 2: Find the definite integral directly by making a substitution in the integrand and changing the limits of integration. We let $u=4-x^{2}$; then $du=-2x\,dx.$ Now use the function $u=4-x^{2}$ to change the limits of integration.

Then

$$\begin{eqnarray*} {\int_{0}^{2}{x\sqrt{4-x^{2}}\,dx}} && \underset{\underset{{\underset{{\color{#0066A7}{\hbox{\(x dx=-\tfrac{1}{2}du\)}}}}{{\color{#0066A7}{\hbox{\(u=4-x^{2} \)}}}}}}{\color{#0066A7}{\uparrow }}}{=} \int_{4}^{0}\sqrt{u}\left( -\dfrac{du}{2}\right) =-\dfrac{1}{2}\int_{4}^{0} \sqrt{u} du=-\dfrac{1}{2}\cdot \left[ \dfrac{u^{3/2}}{\dfrac{3}{2}}\right] _{4}^{0} \\ {\;\;} &&= -{\dfrac{1}{3}}(0-8)={\dfrac{{8}}{{3}}} \end{eqnarray*}$$