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 ⋮ We need to find the area under the curve $y = 4 - x^2$ from $x = -2$ to $x = 2$. This requires evaluating the definite integral $\int_{-2}^{2} (4 - x^2) \, dx$. ‖ ‖ step 2 ⋮ First, we find the antiderivative of the integrand $4 - x^2$. The antiderivative is $4x - \frac{x^3}{3} + C$. ‖ ‖ step 3 ⋮ Next, we evaluate the antiderivative at the bounds $x = 2$ and $x = -2$. ‖ ‖ step 4 ⋮ Evaluating at $x = 2$: $4(2) - \frac{(2)^3}{3} = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3}$. ‖ ‖ step 5 ⋮ Evaluating at $x = -2$: $4(-2) - \frac{(-2)^3}{3} = -8 - \frac{-8}{3} = -8 + \frac{8}{3} = -\frac{24}{3} + \frac{8}{3} = -\frac{16}{3}$. ‖ ‖ step 6 ⋮ Subtract the value at $x = -2$ from the value at $x = 2$: $\frac{16}{3} - (-\frac{16}{3}) = \frac{16}{3} + \frac{16}{3} = \frac{32}{3}$. ‖ ‖ step 7 ⋮ Therefore, the area under the curve $y = 4 - x^2$ from $x = -2$ to $x = 2$ is $\frac{32}{3} \approx 10.667$. ‖ ∻Answer∻ ⚹ The area under the curve is $\frac{32}{3} \approx 10.667$ ⚹ ∻Key Concept∻ ⚹Definite Integral⚹ ∻Explanation∻ ⚹The definite integral of a function over an interval gives the area under the curve of that function within the specified interval.⚹◊What is the formula for calculating the area under a curve using definite integrals in calculus?⍭ Generate me a similar question◊

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