How can we use the fact that $$\int_0^2 x^2 = \frac{8}{3}$$ to compute $$\int_{-2}^2 (1 - x^2)dx?$$
I do understand that there is a reflection in the $x$-axis and the original parabola is an even function. So we can double and negate the value of the integral to capture the reflection transformation.
I am unable to understand how do we use the information available to capture the positive shift in the $y$-axis.
Following up on Intelligent pauca's comment, linearity of integrals allows us to "take apart" these integrals based on the linear combination of the functions.
We have the integral: $$\int_{-2}^{2}(1-x^2)dx$$
We know that the integral $\int_0^2x^2dx = \frac{8}{3}$, Looking at the graph we can say there is symmetry from the limits of integration. We are integrating $[-2,2]$, and the area under the curve from $[0,2] = [-2,0]$. Since both sides are symmetrical, they have the same area: $$\int_{-2}^{2} x^2 dx = 2 \cdot\int_0^2x^2dx = 2 \cdot \frac{8}{3} = \frac{16}{3}$$ Now, looking at the other integral we see: $$\int_{-2}^{2}1dx = (2) - (-2) = 4$$ Putting all the information together we get: $$\int_{-2}^21dx - \int_{-2}^2x^2dx = 4 - \frac{16}{3} = \boxed{-\frac{4}{3}.}$$