Area Between Curves Math Example 3

Follow the full solution, then compare it with the other examples linked below.

easy

Find the area between

y = 4-x^2

and the

x

-axis where the parabola is above the axis.

1
Zeros: $x = \pm 2$ . Parabola is above $x$ -axis on $[-2,2]$ .
2
$A = \int_{-2}^{2}(4-x^2)\,dx = \left[4x-\frac{x^3}{3}\right]_{-2}^{2} = (8-\frac{8}{3})-(-8+\frac{8}{3}) = \frac{32}{3}$ .

Answer

\frac{32}{3}

The parabola opens downward, is above the axis between

x = \pm 2

. Integrate the function directly over that interval.

About Area Between Curves

The area of the region enclosed between two functions $f(x)$ and $g(x)$ from $x = a$ to $x = b$ , computed as $A = \int_a^b |f(x) - g(x)|\,dx$ .

Find the area between [formula] and [formula] from [formula] to [formula].

Find the total area enclosed between [formula] and the [formula]-axis.

Find the area between [formula] and [formula].