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 $x=2$:

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