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

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Answer from Sia

Posted 5 months ago

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 this antiderivative at the bounds $x = -2$ and $x = 2$

step 4

Evaluating at $x = 2$: $4(2) - \frac{(2)^3}{3} = 8 - \frac{8}{3} = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3}$

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 over that interval.

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