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

We are given the curve $y = -x^2$ and we are to find the area enclosed between $x = -3$ and $x = 2$, bounded below by the $x$-axis ($y = 0$). Since the curve $y = -x^2$ lies below the $x$-axis between $x = -3$ and $x = 2$, the area is given by: \[ \text{Area} = \int_{-3}^{2} |y|\, dx = \int_{-3}^{2} |-x^2|\, dx = \int_{-3}^{2} x^2\, dx \] Evaluating the integral: \[ \int_{-3}^{2} x^2\, dx = \left[\frac{x^3}{3}\right]_{-3}^{2} = \left(\frac{8}{3} - \left(\frac{-27}{3}\right)\right) = \frac{8 + 27}{3} = \frac{35}{3} \] % Final Answer Final Answer: The required area is $\boxed{\frac{35}{3}}$ square units. % Rough Sketch \begin{center} \includegraphics[width=0.6\textwidth]{ig4.png} \end{center}