Question

Math

Posted 4 months ago

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

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Answer from Sia

Posted 4 months ago

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.

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