Calculate the area of the region bounded by the curve (x^2)/9+(y^2)/4=1 and the x -axis using integration.

\[\begin{aligned} & A=2 \times \frac{2}{3} \int_{0}^{3} \sqrt{9-x^{2}} d x \\ & =\frac{4}{3}\left[\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1}\left(\frac{x}{3}\right)\right]_{0}^{3} \\ & =\frac{4}{3}\left[\left(0+\frac{9}{2} \sin ^{-1} 1\right)-0\right] \\ & =3 \pi \end{aligned} \] ( $\displaystyle \frac{\mathbf{1}}{\mathbf{2}}$ for correct figure)