Answer
See the proof below.
Work Step by Step
Step 1. Using the figure shown, we have the function $y=h-(\frac{4h}{b^2})x^2$ and the interval $[-\frac{b}{2},\frac{b}{2}]$,
Step 2. For the area under the function, we have $A=\int_{-\frac{b}{2}}^{\frac{b}{2}}(h-(\frac{4h}{b^2})x^2)dx=(hx-(\frac{4h}{3b^2})x^3)|_{-\frac{b}{2}}^{ \frac{b}{2} }=(h( \frac{b}{2})-(\frac{4h}{3b^2})( \frac{b}{2})^3)-(h(- \frac{b}{2})-(\frac{4h}{3b^2})(- \frac{b}{2})^3)= \frac{bh}{2}- \frac{bh}{6}+ \frac{bh}{2}-\frac{bh}{6}= \frac{2}{3}bh$
Step 3. As $b$ is the length of the base and $h$ is the height of the parabolic arch, we have proved the Archimedes’ area formula.
