Determine the area under the curve $y = 4 - x^2$ from $x = -2$ to $x = 2$.

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the area under the curve $ y = 4 - x^2 $ from $ x = -2 $ to $ x = 2 $, we need to evaluate the definite integral $ \int_{-2}^{2} (4 - x^2) \, dx $. ‖ ‖ step 2 ⋮ The antiderivative of $ 4 - x^2 $ is $ 4x - \frac{x^3}{3} $. ‖ ‖ step 3 ⋮ We evaluate the antiderivative at the upper and lower limits of the integral and subtract: $ \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2} $. ‖ ‖ step 4 ⋮ Plugging in the upper limit $ x = 2 $, we get $ 4(2) - \frac{(2)^3}{3} = 8 - \frac{8}{3} $. ‖ ‖ step 5 ⋮ Plugging in the lower limit $ x = -2 $, we get $ 4(-2) - \frac{(-2)^3}{3} = -8 + \frac{8}{3} $. ‖ ‖ step 6 ⋮ Subtracting the value at the lower limit from the value at the upper limit gives us $ \left( 8 - \frac{8}{3} \right) - \left( -8 + \frac{8}{3} \right) $. ‖ ‖ step 7 ⋮ Simplifying the expression, we get $ 8 - \frac{8}{3} + 8 - \frac{8}{3} = 16 - \frac{16}{3} = \frac{48}{3} - \frac{16}{3} = \frac{32}{3} $. ‖ ∻Answer∻ ⚹ $ \frac{32}{3} $ or approximately 10.667 ⚹ ∻Key Concept∻ ⚹Definite Integral as Area Under the Curve⚹ ∻Explanation∻ ⚹The definite integral of a function between two limits gives the area under the curve of the function between those two points, provided the function is continuous and non-negative on that interval.⚹◊What is the formula to find the area under a curve defined by $y = f(x⍭ Generate me a similar question◊

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