Question

Find the area between $f(x)=8-2x^2$ and $x$-axis over the interval [0,2].

Answer 1

$\frac{32}{3}$

Answer 2

To find the area between the function $f(x)=8-2x^2$ and the x-axis over the interval $[0,2]$, we need to evaluate the definite integral of $f(x)$ over this interval.

First, determine the sign of $f(x)$ in the interval $[0,2]$. The function is a downward-opening parabola. To find its x-intercepts, set $f(x)=0$: $8 - 2x^2 = 0$ $2x^2 = 8$ $x^2 = 4$ $x = \pm 2$

The x-intercepts are at $x=-2$ and $x=2$. Since $f(x)$ is a downward-opening parabola and its roots are $-2$ and $2$, $f(x) \ge 0$ for $x \in [-2, 2]$. The given interval is $[0,2]$, which is entirely within $[-2,2]$. Therefore, $f(x) \ge 0$ for all $x \in [0,2]$.

Since $f(x)$ is non-negative on the interval $[0,2]$, the area $A$ is given directly by the definite integral: $A = \int_0^2 f(x) dx = \int_0^2 (8 - 2x^2) dx$

Now, perform the integration: $A = \left[8x - \frac{2x^{2+1}}{2+1}\right]_0^2$ $A = \left[8x - \frac{2x^3}{3}\right]_0^2$

Next, evaluate the definite integral using the limits of integration: $A = \left(8(2) - \frac{2(2)^3}{3}\right) - \left(8(0) - \frac{2(0)^3}{3}\right)$ $A = \left(16 - \frac{2 \times 8}{3}\right) - (0 - 0)$ $A = \left(16 - \frac{16}{3}\right) - 0$ $A = \frac{16 \times 3}{3} - \frac{16}{3}$ $A = \frac{48}{3} - \frac{16}{3}$ $A = \frac{48 - 16}{3}$ $A = \frac{32}{3}$

The area between $f(x)=8-2x^2$ and the x-axis over the interval $[0,2]$ is $\frac{32}{3}$ square units.