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Analysis Homework #6 Solutions 1. Suppose ∫ 1 f is non-negative and integrable on [0, 1] with f (x) = 0 for all x ∈ Q. Show that 0 f (x) dx = 0. • Given a partition P = {x0 , x1 , . . . , xn } of [0, 1], one clearly has inf [xk ,xk+1 ] f (x) = 0 for each 0 ≤ k ≤ n − 1 because f is non-negative and every subinterval contains a rational. Thus, − S (f, P ) = n−1 ∑ k=0 inf [xk ,xk+1 ] f (x) · (xk+1 − xk ) = 0 and we may take the supremum of both sides to conclude that ∫ 1 f (x) dx = sup S − (f, P ) = 0. 0 2. Suppose f is increasing on [0, 1]. Show that f is integrable on [0, 1]. Hint: let P be the partition of [0, 1] into n subintervals of the same length and show that S + (f, P ) − S − (f, P ) = f (1) − f (0) . n • Let P = {x0 , x1 , . . . , xn } be as in the hint. Since f is increasing, we then have inf [xk ,xk+1 ] f (x) = f (xk ) for each 0 ≤ k ≤ n − 1 and this implies that − S (f, P ) = n−1 ∑ k=0 inf [xk ,xk+1 ] f (x) · (xk+1 − xk ) ) 1( f (x0 ) + f (x1 ) + . . . + f (xn−1 ) . n Using the exact same reasoning to compute the upper Darboux sum, we now get = + S (f, P ) = n−1 ∑ sup f (x) · (xk+1 − xk ) k=0 [xk ,xk+1 ] ) 1( = f (x1 ) + f (x2 ) + . . . + f (xn ) , n so we may combine the last two equations to arrive at f (1) − f (0) f (xn ) − f (x0 ) = . n n Letting n be suﬃciently large, we can certainly make this expression as small as we wish, hence f is integrable by Theorem 4.9 in your notes. S + (f, P ) − S − (f, P ) = 3. Show that there exists a unique function f , defined for all x ∈ R, such that f ′ (x) = • Since 1 1+x4 1 , 1 + x4 f (0) = 0. is continuous, the fundamental theorem of calculus ensures that ∫ f (x) = 0 x dt 1 + t4 has the desired properties. If g(x) also does, then g ′ (x) = 1 = f ′ (x) 1 + x4 =⇒ g(x) = f (x) + C and the fact that g(0) = 0 = f (0) implies that C = 0, hence g(x) = f (x). 4. Suppose f is continuous on [a, b]. Show that there exists some c ∈ (a, b) such that ∫ b f (t) dt = (b − a) · f (c). a Hint: apply the mean value theorem to the function F (x) = ∫x a f (t) dt. • According to the mean value theorem, there exists some c ∈ (a, b) such that F (b) − F (a) = F ′ (c). b−a Moreover, F ′ (x) = f (x) by the fundamental theorem of calculus, while ∫ F (a) = ∫ a f (t) dt = 0, F (b) = a b f (t) dt. a Once we now combine all these facts, we may conclude that ∫ b ′ F (b) − F (a) = (b − a) · F (c) =⇒ f (t) dt = (b − a) · f (c). a