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 ⋮ Calculate the definite integral of the function $4 - x^2$ from $-2$ to $2$. ‖ ‖ step 2 ⋮ The antiderivative of $4 - x^2$ is $4x - \frac{x^3}{3}$. ‖ ‖ step 3 ⋮ Evaluate the antiderivative at the upper limit of integration: $4(2) - \frac{2^3}{3} = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3}$. ‖ ‖ step 4 ⋮ Evaluate the antiderivative at the lower limit of integration: $4(-2) - \frac{(-2)^3}{3} = -8 - \frac{-8}{3} = -8 + \frac{8}{3} = -\frac{24}{3} + \frac{8}{3} = -\frac{16}{3}$. ‖ ‖ step 5 ⋮ Subtract the value of the antiderivative at the lower limit from the value at the upper limit: $\frac{16}{3} - (-\frac{16}{3}) = \frac{16}{3} + \frac{16}{3} = \frac{32}{3}$. ‖ ∻Answer∻ ⚹ $\frac{32}{3}$ or approximately $10.667$ ⚹ ∻Key Concept∻ ⚹Definite Integration⚹ ∻Explanation∻ ⚹The area under the curve $y = 4 - x^2$ from $x = -2$ to $x = 2$ is found by evaluating the definite integral of the function over the given interval.⚹◊What is the formula for calculating the area under a curve between two points using definite integrals?⍭ Generate me a similar question◊

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