Question

Use integration to find the area of the region enclosed by the curve $y = -x^2$ and the straight lines $x = -3$, $x = 2$ and $y = 0$. Sketch a rough figure to illustrate the bounded region.

Hint:

When the curve lies below the $x$-axis, use the absolute value of the function to compute the positive area between the curve and the axis.

Answer 1

The area enclosed by the curve $y = -x^2$ and the $x$-axis ($y = 0$) from $x = -3$ to $x = 2$ is determined. As the curve $y = -x^2$ is below the $x$-axis in this interval, the area is calculated as: \[ \text{Area} = \int_{-3}^{2} |y|\, dx = \int_{-3}^{2} |-x^2|\, dx = \int_{-3}^{2} x^2\, dx \] The evaluation of the integral proceeds as follows: \[ \int_{-3}^{2} x^2\, dx = \left[\frac{x^3}{3}\right]_{-3}^{2} = \left(\frac{8}{3} - \left(\frac{-27}{3}\right)\right) = \frac{8 + 27}{3} = \frac{35}{3} \] % Final Answer Final Answer: The calculated area is $\boxed{\frac{35}{3}}$ square units. % Rough Sketch \begin{center} \includegraphics[width=0.6\textwidth]{ig4.png} \end{center}