Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 02/Solution 1

Intersection of the curve $y=8-2x^{2}$ with the $x-$axis.

First, we identify the points where the curve intersects the $x-$ axis by solving $8-2x^{2}=0,$ which yields $x=\pm 2$. Thus, the region of interest lies above the x-axis on the interval $[-2,2]$. This is represented as the shaded region in the attached graph.

To determine the enclosed area, we set up a definite integral to compute the area under the curve from $x=2$ to $x=2$:

$$\int _{-2}^{2}(8-2x^{2})dx=\left[8x-{\frac {2x^{3}}{3}}\right]_{-2}^{2}={\frac {64}{3}}.$$