Math 351 Fundamental Theorem of Calculus II

Math 351 Fundamental Theorem of Calculus II write-up Fundamental Theorem of Calculus II: If f is integrable on [a, b], then F (x) = Z x f (x) dx a is continuous on [a, b]. Moreover, if f is continuous at x0 ∈ (a, b), then F is differentiable at x0 and F 0 (x0 ) = f (x0 ). Proof: We already proved on the last homework that F (x) is continuous on [a, b]. If x > x0 , we have already shown that F (x) − F (x0 ) = Z x f (t)dt. x0 We adopt the notation that if g is integrable on [a, b], then x < x0 , then Z x0 f (t)dt F (x0 ) − F (x) = Ra b g(x)dx = − Rb a g(x)dx. If x so F (x) − F (x0 ) = − (F (x0 ) − F (x)) = − Z x f (t)dt = x0 Z x f (t)dt. x0 Therefore, in both cases, F (x) − F (x0 ) −1 Z x0 f (t) dt. = x − x0 x − x0 x Also, recall that Z x x0 f (x0 )dt = (x − x0 )f (x0 ) and thus  F (x) − F (x0 ) 1  − f (x0 ) = F (x) − F (x0 ) − (x − x0 )f (x0 ) x − x0 x − x0 Z x  Z x 1 = f (t)dt − f (x0 )dt x − x 0 x0 x0  1 Z x = f (t) − f (x0 ) dt x − x 0 x0 (Notice that in the last step we applied the subtraction law for integration, which we will prove on the following sheet.) 1 Fix  > 0. Now, since f (x) is continuous at x0 , there must exist a δ > 0 so that if t ∈ [a, b] and |t − x0 | < δ, then |f (t) − f (x0 )| < 2 . In particular,   − < f (t) − f (x0 ) < 2 2 Now suppose that x ∈ (x0 , b] and |x−x0 | < δ. For all t ∈ [x0 , x], we have |t−x0 | ≤ |x−x0 | < δ, so   − < f (t) − f (x0 ) < 2 2 and thus Z x    − (x − x0 ) ≤ f (t) − f (x0 ) dt ≤ (x − x0 ). 2 2 x0 Hence Z x   f (t) − f (x0 dt ≤ (x − x0 ) < (x − x0 ) 2 x0 so  F (x) − F (x0 ) 1 Z x 1 − f (x0 ) = f (t) − f (x0 ) dt < · |(x − x0 )| = . x − x0 x − x 0 x0 x − x0 By a very similar argument, we see that if x ∈ [a, x0 ) and |x − x0 | < δ, we also have: F (x) − F (x0 ) − f (x0 ) < . x − x0 Therefore, for all  > 0 there exists δ > 0 so that if x ∈ [a, b] and 0 < |x − x0 | < δ, then F (x) − F (x0 ) − f (x0 ) < , x − x0 which implies that lim x→x0 F 0 (x0 ) = f (x0 ). F (x) − F (x0 ) = f (x0 ). Therefore, F is differentiable at x0 and x − x0 2 Addition Law for Integration: If f and g are both integrable on [a, b], then f + g is also integrable on [a, b] and Z b (f + g)(x)dx = a Z b f (x)dx + Z b a g(x)dx a Proof: We first notice that since f and g are integrable, they are both bounded, which implies that f + g is also bounded. Let P = {a = t0 < t1 < · · · < tn = b} be a partition of [a, b]. Notice that if x ∈ [tk−1 , tk ], then (f + g)(x) = f (x) + g(x) ≤ sup{f ([tk−1 , tk ])} + sup{g([tk−1 , tk ])} so sup{f ([tk−1 , tk ])} + sup{f ([tk−1 , tk ])} is an upper bound for {(f + g)([tk−1 , tk ])}, which implies that sup{(f + g)([tk−1 , tk ])} ≤ sup{f ([tk−1 , tk ])} + sup{g([tk−1 , tk ])}. Therefore, U (f + g, P ) = ≤ = n X k=1 n X k=1 n X sup{(f + g)([tk−1 , tk ])}(tk − tk−1 ) (sup{f ([tk−1 , tk ])} + sup{g([tk−1 , tk ])}) (tk − tk−1 ) sup{f ([tk−1 , tk ]) (tk − tk−1 ) + n X sup{g([tk−1 , tk ])}(tk − tk−1 ) k=1 k=1 = U (f, P ) + U (g, P ) We can similarly prove that L(f, P ) + L(g, P ) ≤ L(f + g, P ), so if P is a partition of [a, b], then L(f, P ) + L(g, P ) ≤ L(f + g, P ) ≤ U (f + g, P ) ≤ U (f, P ) + U (g, P ). Fix  > 0. Since f is integrable on [a, b], there exists a partition P1 of [a, b] so that U (f, P1 ) < L(f, P1 ) +  2 and since g is integrable on [a, b], there exists a partition P2 of [a, b] so that  U (g, P2 ) < L(g, P2 ) + . 2 Let P = P1 ∪ P + 2, since P1 ⊂ P and P2 ⊂ P , 3 L(f, P1 ) + L(g, P2 ) ≤ L(f, P ) + L(g, P ) ≤ L(f + g, P ) ≤ U (f + g, P ) ≤ U (f, P ) + U (g, P ) ≤ U (f, P1 ) + U (g, P2 ) < L(f, P1 ) +   + L(g, P2 ) + = L(f, P1 ) + L(g, P2 ) + . 2 2 Therefore, U (f )+U (g)− ≤ U (f, P )+U (g, P )− < L(f, P1 )+L(g, P2 ) ≤ L(f +g, P ) ≤ L(f +g) ≤ U (f +g) ≤ U (f + g, P ) ≤ U (f, P ) + U (g, P ) < L(f, P1 ) + L(g, P2 ) +  ≤ L(f ) + L(g) + . So, since U (f ) = L(f ) = Rb a f (x)dx and U (g) = L(g) = U (f + g, P ) − L(f + g, P ) <  and U (f + g) − Rb a g(x)dx, Z b f (x)dx + Z b ! g(x)dx < 2. (1) a a Thus, for all  > 0 there exists a partition P of [a, b] such that U (f + g, P ) − L(f + g, P ) < , so the Cauchy Criterion for integrabilty implies that f + g is integrable on [a, b]. Therefore, Rb U (f + g) = a (f + g)(x)dx, so, for all  > 0, we see from equation (1) that Z b (f + g)(x)dx − Z b f (x)dx + a a Z b ! g(x)dx < 2. a It is easily checked that this implies that Z b (f + g)(x)dx = Z b f (x)dx + g(x)dx a a a Z b which completes the proof. We have already established that if f is integrable on [a, b], then −f is integrable on [a, b] Rb Rb and a −f (x)dx = − a f (x)dx, so we immediately obtain: Subtraction Law for Integration: If f and g are both integrable on [a, b], then f − g is also integrable on [a, b] and Z b a (f − g)(x)dx = Z b a 4 f (x)dx − Z b a g(x)dx