![]() ![]() However, the real power of the Fundamental Theorem of Calculus is that this link between areas and antiderivatives is true every single time. The fact that we get the same answer in this case might not be too surprising. Therefore, by geometry we get Area = (1/2)× b× b = b 2/2. In this case, the base and height are the same: both are equal to b. You should know how to find the area of any triangle, using Area = (1/2)× (base)× (height). Notice that the region under this particular function is simply a right triangle. By working the problem out using a different method, I hope to show you the remarkable connection between areas and antiderivatives. Now let’s try this same example (area under f( x) = x on ) but in a different way. You can verify that F( x) = x 2/2 does the trick. Then, according to the Fundamental Theorem of Calculus, we just need to find an antiderivative for f( x) = x. We will find the area under y = f( x) = x between x = 0 and x = b.įirst set up the definite integral that computes the area. The graph is a diagonal line through the origin. Let’s see how it works in one of the simplest cases, f( x) = x. The bottom boundary is the x-axis, and the top boundary is the graph of f( x) itself. The limits of integration, a and b, specify the left and right boundaries of the region. If a function f( x) is nonnegative on an interval, then the area of the region under the curve can be computed by a definite integral. One of the most common applications you’ll see on the AP Calculus exams is area under a curve. So sin x is the antiderivative of cos x.Īny time a definite integral needs to be evaluated, the Fundamental Theorem of Calculus can come to the rescue. Remember, cos x is the derivative of sin x. In the following example, we work out a definite integral using the FTC.
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