Answer
$\frac{13}{4}$
Work Step by Step
Step 1. Graph the function for the given interval of $-2\leq x\leq 3$ and identify a zero at $x=2$ which separates the positive and negative regions.
Step 2. The total area is the sum of the absolute value of the two separate areas $A_1$ and $A_2$.
Step 3. Use symmetry and evaluate
$A_1=2\int_{0}^2(1-\frac{x^2}{4})dx=2(x-\frac{x^3}{12})|_0^2=2(2-\frac{2^3}{12})=\frac{8}{3}$
Step 4. Evaluate
$A_2=\int_{2}^3(1-\frac{x^2}{4})dx=(x-\frac{x^3}{12})|_2^3=(3-\frac{3^3}{12})-(2-\frac{2^3}{12})=-\frac{7}{12}$
Step 5. Thus, we have the area as $A=|A_1|+|A_2|=\frac{13}{4}$
