Sketch and clearly indicate the area enclosed by the graph y = -2x² + 2 the lines x = -1 and x = 1. Also clearly show the representative strip that will be used to calculate the area enclosed.

The equation $y = -2x^2 + 2$ represents a parabola opening downwards.

• The y-intercept is found by setting $x=0$: $y = -2(0)^2 + 2 = 2$. So, the point is $(0, 2)$. This is also the vertex.
• The x-intercepts are found by setting $y=0$: $0 = -2x^2 + 2 \implies 2x^2 = 2 \implies x^2 = 1 \implies x = \pm 1$. So, the points are $(-1, 0)$ and $(1, 0)$.

Draw a coordinate plane. Plot the vertex at $(0, 2)$ and the x-intercepts at $(-1, 0)$ and $(1, 0)$. Draw a smooth parabola opening downwards through these points. The lines $x = -1$ and $x = 1$ are vertical lines that pass through the x-intercepts. The area enclosed is the region under the parabola and above the x-axis, between $x=-1$ and $x=1$.
To show a representative strip, draw a thin vertical rectangle within this enclosed area, from the x-axis up to the curve $y = -2x^2 + 2$. Label its width as $dx$ and its height as $y$ or $f(x)$.

4.3.2 Determine, by using integration, the magnitude of the area indicated in QUESTION 4.3.1.

The area $A$ under the curve $y = -2x^2 + 2$ from $x = -1$ to $x = 1$ is given by the integral:

$$A = \int_{-1}^{1} (-2x^2 + 2)\,dx$$

$$A = \left( -\frac{2(1)^3}{3} + 2(1) \right) - \left( -\frac{2(-1)^3}{3} + 2(-1) \right)$$ $$A = \left( -\frac{2}{3} + 2 \right) - \left( -\frac{2(-1)}{3} - 2 \right)$$ $$A = \left( -\frac{2}{3} + \frac{6}{3} \right) - \left( \frac{2}{3} - \frac{6}{3} \right)$$ $$A = \left( \frac{4}{3} \right) - \left( -\frac{4}{3} \right)$$ $$A = \frac{4}{3} + \frac{4}{3}$$ $$A = \frac{8}{3}$$

The magnitude of the area is $\frac{8}{3} \text{ square units}$.