Consider the region enclosed by the parabola f(x)=18x-3x^2 and the x-axis. Find the area of this region via integral calculus and verify that Archimedes' are formula yields the same result.

Archimedes' formula: Area=2/3(width)(height)
*Width is the distance between the roots and the height is the vertical distance from the x=axis to the vertex.

Question

Consider the region enclosed by the parabola f(x)=18x-3x^2 and the x-axis. Find the area of this region via integral calculus and verify that Archimedes' are formula yields the same result.

Archimedes' formula: Area=2/3(width)(height) 
*Width is the distance between the roots and the height is the vertical distance from the x=axis to the vertex.

rogue · Answer

0 = 18x - 3x^2
0 = 3x(6 - x)
x = 0, x = 6
Those are your limits of integration.

rogue · Answer

$$A = \int\limits_{0}^{6} f(x)dx = \int\limits_{0}^{6} \left[ 18x - 3x^2 \right] dx = \left[ 9x^2 - x^3 \right]_{0}^{6} = 108$$

rogue · Answer

$$A = \frac {2}{3} wh$$Your parabola's max is at x = 3, where y = 27, so the height is 27. The width is just 6.$$A = \frac {2}{3} * 6 * 27 = 108$$