The Fundamental Theorem of Calculus
                                          (A Heuristic Argument)
                  As already discussed, the basic geometrical question of integral calculus is that of
               finding the area under a curve between two endpoints. The Fundamental Theorem
               of Calculus tells us how to do this. Here’s how to think about this. We consider
               an “area function,” A(x), which measures the area under the curve y = f(t) between
               t = a and t = x:
                                y ✻
                                                  y = f(t)
                                                                ∆A(x)
                                               A(x)           ✟✟
                                                             ✟✙
                              ✛                                             ✲
                                      t = a             t = x t = x+∆x  t = b t
                                   ❄
               Thus, ∆A(x) = A(x+∆x)−A(x); moreover, we see that
                                                 ∆A(x)≈f(x)∆x.
               Therefore, we may apply the definition of the derivative of the area function:
                                     ′           ∆A(x)           f(x)∆x
                                   A(x)= lim             = lim           =f(x).
                                            ∆x→0 ∆x        ∆x→0    ∆x
               In other words, we have argued that the derivative of the area function is the
               original function! Put differently, this says that the area function is an antideriva-
               tive of the original function.
                  Next, suppose that F is any antiderivative of the function f. Therefore, we con-
               clude that
                                                d (A(x)−F(x)) = 0,
                                               dx
               which implies that F(x) − A(x) is a constant, say F(x) − A(x) = c. Since A(a) = 0
               we see that, in fact, F(a) = c. Therefore, we may conclude that
                                   Area Under The Curve = A(b) = F(b)−F(a).
               Notation. We typically use the integral sign to denote the area under the curve
               of y = f(x) between x = a and x = b (where negative areas are contributed when the
               graph goes below the x-axis):
                                                         1
                        Z b
                            f(x)dx = Area Under The Curve y = f(x) between x = a and x = b.
                          a
                     Putting all the above information together, we obtain
                                                                                                        1
                  Fundamental Theorem of Calculus. Let f be a continuous function defined
                  on the interval [a,b] and let F be any antiderivate of f. Then
                                             Z b                   b
                                                 f(x)dx = [F(x)] = F(b)−F(a).
                                                                   a
                                              a
                     Some people like to state the Fundamental Theorem of Calculus as follows:
                  Fundamental Theorem of Calculus. Let f be a continuous function defined
                  on the interval [a,b]. Then
                                                       d Z xf(t)dt = f(x).
                                                      dx a
                  Example. Let’s return to the example from the notes on l’Hˇopital’s Rule concerning
                  the computation of the area under the curve y = 4−x2 over the interval [0,2]. Since
                  4x−x3/3isanantiderivate of 4−x2, we apply the Fundamental Theorem of Calculus
                  and obtain
                                          Z 2(4−x2)dx = 4x− x32 = 8−8/3,
                                            0                           3 0
                  in agreement with our earlier computation!
                                                                                                      d 1      2      

                  Exercisesfrom(5.2): 29– 38 ,39(Wouldithelpformetoobservethat dx 2 ln(x +4) =
                   x ?), 43, 45, 46.
                  x2+4
                  Exercises from (5.3): 7– 14 , 15–20, 21–24, 25–28.
                  Exercises from (5.4): 1–14, 17–20, 25 –28, 37–42, 52 (can you prove Archimedes’
                  result in general?), 53, 57–59.
                    1Actually, continuity is not required, but we won’t try to state this theorem in its most generality.
                                                                 2