Lecture 22: The Fundamental Theorem of Calculus, Integration by Parts, and Change of Variable Formula

18.100A: Complete Lecture Notes Lecture 22: The Fundamental Theorem of Calculus, Integration by Parts, and Change of Variable Formula Theorem 1 (Additivity) If f ∈ C([a, b]) and a < c < b, then b Z c Z Z b f. f+ f= c a a Proof : Let {(y(r), ζ(r))}r and {(z(r), η(r))}r be tagged partitions of [a, c] and [c, b] respectively such that ky(r)k → 0 and kz(r)k → 0. Define x(r) = y(r) ∪ z(r) ξ(r) = ζ(r) ∪ η(r), a sequence of tagged partitions of [a, b]. Then, kx(r)k ≤ ky(r)k + kz(r)k → 0. Thus, Z b f = lim Sf (x(r), ξ(r)) a t→∞ = lim (Sf (y(r), ζ(r)) + Sf (z(r), η(r))) r→∞ Z c Z b = f+ f. a c Theorem 2 Let f ∈ C([a, b]), and mf = inf{f (x) | x ∈ [a, b]} ∈ R Mf = sup{f (x) | x ∈ [a, b]} ∈ R. Then, Z b mf (b − a) ≤ f ≤ Mf (b − a). a Proof : Let {(x(r), ξ(r))}r be a sequence of tagged partitions with kx(r)k → 0. Then, Sf (x(r), ξ(r)) = n X f (ξk (r))(xk (r) − xk−1 (r)) ≥ mf k=1 n X (xk (r) − xk−1 (r)) = mf (b − a). k=1 1 Similarly, Sf (x(r), ξ(r)) = n X f (ξk (r))(xk (r) − xk−1 (r)) ≤ Mf k=1 n X (xk (r) − xk−1 (r)) = Mf (b − a). k=1 Therefore, for all r, b Z mf (b − a) ≤ Sf (x(r), ξ(r)) ≤ Mf (b − a) =⇒ mf (b − a) ≤ f ≤ Mf (b − a). a Theorem 3 Suppose f ∈ C([a, b]) and g ∈ C([a, b]). 1. If ∀x ∈ [a, b] f (x) ≤ g(x), then b Z b Z f≤ g. a 2. (Triangle Inequality for integrals): | Rb a Rb f| ≤ a |f |. a Proof : 1. Let {(x(r), ξ(r)}r be a sequence of tagged partitions such that kx(r)k → 0. Then, for all r ∈ N, Sf (x(r), ξ(r)) = ≤ n X j=1 n X f (ξj (r))(xj (r) − xj−1 (r)) g(ξj (r))(xj (r) − xj−1 (r)) j=1 = Sg (x(r), ξ(r)). Then, letting r → ∞, we get that Z b b Z f≤ g. a a 2. Notice that ±f (x) ≤ |f (x)| for all x, and thus Z ± Rb a f| ≤ Rb a Z f≤ a Therefore, | b b Z |f | =⇒ − a b Z f≤ a b Z f≤ a b |f |. a |f |. Remark 4. There are some conventions that are worth noting: Ra Rb 1. a f := 0. This is consistent with our definitions and theorems thus far as limb→a | a f | = 0. Rb Ra 2. a f = − b f. Fundamental Theorem of Calculus 2 Theorem 5 (Fundamental Theorem of Calculus) Suppose f ∈ C([a, b]). 1. If F : [a, b] → R is differentiable and F 0 = f , then b Z f = F (b) − F (a). a 2. The function G(x) := Rx a f is differentiable on [a, b] and  G0 = f G(a) = 0 . Remark 6. We sometimes abbreviate the Fundamental Theorem of Calculus to FTC. Proof : 1. Let {(x(r))}r be a sequence of partitions with kxk → 0. Then, by the Mean Value Theorem, ∀r∀j, there exists a ξj (r) ∈ [xj−1 (r), xj (r)] such that F (xj (r)) − F (xj−1 (r)) = F 0 (ξj (r))(xj (r) − xj−1 (r)) = f (ξj (r))(xj (r) − xj−1 (r)). Thus, Z n(r) b f = lim a r→∞ X f (ξj (r))(xj (r) − xj−1 (r)) j=1 n(r) = lim r→∞ X F (xj (r)) − F (xj−1 (r)) j=1 = lim (F (b) − F (a)) = F (b) − F (a). r→∞ 2. Let c ∈ [a, b]. We wish to show that Rx Rc f− af = f (c). lim x→c x−c Let  > 0. Then, since f is continuous at c, ∃δ0 > 0 such that a |t − c| < δ0 =⇒ |f (t) − f (c)| < /2. Choose δ = δ0 . Suppose 0 < x − c < δ. If t ∈ [c, x], then |t − c| ≤ |x − c| < δ = δ0 . 3 Thus, 1 x−c x Z f (t) dt − f (c) = c = ≤ ≤ = Z t Z c 1 1 f (t) dt − f (c) dt x−c c x−c x Z x 1 (f (t) − f (c)) dt x−c c Z x 1 |f (t) − f (c)| dt x−c c Z x 1 /2 dt x−c c 1   · (x − c) = . x−c 2 2 A similar argument holds for 0 < c − x < δ. Thus, Rx 0 < |x − c| < δ =⇒ a Rc f− af  − f (c) ≤ < . x−c 2 Theorem 7 (Integration by Parts) Suppose f, g ∈ C([a, b]) and f 0 , g 0 ∈ C([a, b]). Then, Z b f 0 g = (f (b)g(b) − f (a)g(a)) − intba f g 0 . a Proof : We have (f g)0 = f 0 g + f g 0 . Therefore, by the Fundamental Theorem of Calculus, Z f (b)g(b) − f (a)g(a) = b f 0g + a Z b f g0 . a Remark 8. We sometimes abbreviate Integration By Parts as IBP. Lemma 9 (Riemann-Lebesgue) Suppose f ∈ C([−π, π]) and f 0 ∈ C([−π, π]) with f 2π-periodic with f (−π) = f (π). For n ∈ N ∪ {0}, let Z 1 π an = f (x) sin(nx) dx π −π Z 1 π bn = f (x) cos(nx) dx. π −π Then, lim an = lim bn = 0. n→∞ n→∞ Definition 10 (Fourier Coefficients) The an , bn defined in the above lemma are referred to as the Fourier coefficients of f . 4 Proof : Using IBP, we have Z π 1 f (x) cos(nx) dx π −π Z π 1 1 = ( sin(nx))0 f (x) dx π −π n Z 1 1 π = (f (π) sin(nπ) − f (−pi) sin(n(−π))) − sin(nx)f 0 (x) dx . n n −π |bn | = Notice that sin(nπ) = sin(n(−π)) = 0 for all n ∈ N. Hence, Z 1 π |bn | ≤ | sin(nx)||f 0 (x)| dx n −π Z 1 π 0 ≤ |f | → 0. n −π By the Squeeze Theorem, |bn | → 0. A similar arguments works for an . Theorem 11 (Change of Variables) Let ϕ : [a, b] → [c, d] be continuously differentiable with ϕ0 > 0 on [a, b], ϕ(a) = c, and ϕ(b) = d. Then, Z d Z b f (ϕ(x))ϕ0 (x) dx. f (u) du = c a Proof : Let F : [a, b] → R such that F 0 = f . Then, F (ϕ(x))0 = f (ϕ(x)). Hence, by the FTC, b Z Z 0 f (ϕ(x))ϕ (x) dx = a b F (ϕ(x))0 dx a = F (ϕ(b)) − F (ϕ(a)) = F (d) − F (c). Furthermore, by the FTC, Z d Z f (u) du = c d F (u)0 du = F (d) − F (c). c 5