Week of May 6th Math Problem: Using Geometry and Precalculus to Determine the Area Under a Curve

One of the best tips to scoring well on the AP Calculus AB or BC exams is to make sure you know your antiderivatives like the back of your hand. We are trained to easily recognize the antiderivatives of linear, quadratic, trigonometric, and other functions, and are taught a series of techniques to evaluate definite integrals. However, what happens when you see an integral you do not know how to evaluate. Ironically, one tip to solving this calculus problem would be to remember your geometry and precalculus!

How would you calculate area underneath the curve for the following integral?

ʃ √(1 - x^2) dx (evaluated from x = 0 to x = 1)

You take a quick glance and realize this integral is not quickly (if at all) solved by u substitution nor integration by parts, and is not in a form which invokes the use of an inverse trigonometric function as an antiderivative. What do we do here?

Recalling our studies from precalculus, we may remember the equation for a circle of radius 1, centered at (0, 0) is:

x^2 + y^2 = 1

Note that the expression inside the integral above (first bold expression) is what one would get if we solved the second bold expression for y.

y^2 = 1 - x^2

y = ±√(1 - x^2)

Evaluated from x = 0 to x = 1, we notice the integral above is simply the area of the unit circle in the first quadrant. Therefore, since we recall from geometry the area of a circle is equal to πr^2, this means the area of the unit circle is simply equal to π. One-fourth of this area, as the integral above asks for, would be π/4

Therefore, this calculus problem was solved using geometry and precalculus. The more tools in your repertoire, the better off you will be at problem solving!

Go to Virtual Tutor Center (http://www.virtualtutorcenter.com) for all of your tutoring needs!