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.

The lower limit of integration is \(x=0\) so, in terms of \(u,\) it becomes \(u=4-0^{2}=4\).

The upper limit of integration is \(x=2\) so the upper limit becomes \(u=4-2^{2}=0\).

Then

When using substitution to find a definite integral directly, remember to change the limits of integration.

\[ \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*} \]