LECTURE 13: DEFINITE INTEGRALS AND ANTIDERIVATIVES MINGFENG ZHAO February 02, 2015 Evaluating definite integrals If f is integrable on [a, b], then b Z f (x) = lim a n→∞ n X f (x∗k )∆x, k=1 where x0 = a < x1 < · · · < xn = b is a regular partition of [a, b]. For the choice of x∗k , it can be arbitrary. In particular, n X the Riemann sum f (x∗k )∆x can be left, right or midpoint Riemann sum. k=1 Example 1. Assume that lim ∆→0 n X (3x∗k 2 + 2x∗k + 1)∆xk is the limit of a Riemann sum for a function f on [1, 3]. Then k=1 f (x) = 3x2 + 2x + 1 and lim ∆→0 n X (3x∗k 2 + 2x∗k + 1)∆xk = Z 3 (3x2 + 2 + 1) dx. 1 k=1 Z 4p Example 2. Compute 1 − (x − 3)2 dx. 3 p p Let y = 1 − (x − 3)2 , then y 2 + (x − 3)2 = 1, that is, the graph of y = 1 − (x − 3)2 is the upper semi-circle centered at (3, 0) with radius 1. So we get 4 Z p 1 − (x − 3)2 dx = 3 Z 1 π · π · 12 = . 2 2 4 Example 3. Compute (2x + 3) dx. Z 4 Approach I: The integral (2x + 3) dx is just the area of the trapezoid bounded by the x-axis and the line y = 2x + 3 2 2 from x = 2 to x = 4. The width of its base is 4 − 2 = 2, and the lengths of its two parallel sides are f (2) = 7 and f (4) = 11. So we get Z 4 (2x + 3) dx = 2 1 · 2 · [7 + 11] = 18. 2 1 2 MINGFENG ZHAO Approach II: Since f (x) := 2x + 3 is continuous on [2, 4], then f is integrable on [2, 4]. So we can use right Riemann Z 4 sum to compute (2x + 3) dx. For any n, the right Riemann sum of the regular partition of [2, 4] with n, we have 2 4−2 2 ∆x = = and n n xk = 2 + k∆x = 2 + 2k , n ∀k = 0, 1, 2, ·, n Then the right Riemann sum is: x∗k = xk for all k = 1, · · · , n and n X f (x∗k )∆x = k=1 n X f (xk )∆x k=1 = ∆x n X f (xk−1 ) k=1 = = = = = By taking n → ∞, we get Z 4 (2x + 3) dx = lim 2 n→∞ n X n 2k 2 X · +3 2· 2+ n n k=1 n 2 X 4k · 7+ n n k=1 # " n n 2 X 4 X k · 7+ · n n k=1 k=1 2 4 n(n + 1) · 7n + · n n 2 n+1 n 14 + 4 · f (x∗k )∆x = lim n→∞ k=1 n+1 14 + 4 · = 18. n Theorem 1. If f is continuous on [a, b] or bounded on [a, b] with a finite number of discontinuities, then f is integrable on [a, b]. 1, if 0 ≤ x ≤ 1, , then f is a bounded with a finite number of discontinuities. So f is Example 4. Let f (x) = −2, if 1 < x ≤ 3. integrable on [0, 3]. And we have Z 3 f (x) dx = 1 · 1 − 2 · 2 = −3. 0 Definition 1. For a < b, if f is integrable on [a, b], we define Z a Z f (x) dx = − b a b f (x) dx. LECTURE 13: DEFINITE INTEGRALS AND ANTIDERIVATIVES 3 Also we define Z a f (x) dx = 0. a Theorem 2. Suppose f (x) and g(x) are integrable, for any constants a, b, c, p, then Z b Z b Z b [f (x) + g(x)] dx = f (x) dx + g(x) dx a a b Z a Z = c· cf (x) dx f (x) dx a Z b a b Z f (x) dx p Z p a a b f (x) dx. f (x) dx + = Z b |f (x)| dx is the sum of the Theorem 3. Let f (x) be integrable on [a, b], then |f (x)| is also integrable on [a, b] and a areas of the region bounded by the graph of f and the x-axis on [a, b]. Example 5. The following picture shows the graph of a continuous function f (x) with the areas of the regions bounded by its graph and the x-axis given. Figure 1. The graph of f (x) on [a, d] Then Z b f (x) dx = 12 f (x) dx = −10 a Z c b Z c Z f (x) dx a = b Z f (x) dx + a c f (x) dx b By Theorem 2 4 MINGFENG ZHAO Z = 12 − 10 = 2 Z d f (x) dx b Z c = Z f (x) dx b |f (x)| dx −10 + 8 = −2 Z b Z b Z 12 + 10 + 8 = 30. 5 d f (x) dx c 7 Z f (x) dx = −10, then f (x) dx = 3 and 0 Z Z f (x) dx + b = By Theorem 2 c c f (x) dx − a d |f (x)| dx b = Example 6. Assume that Z |f (x)| dx + a Z c Z |f (x)| dx + = a By Theorem 2 c = d d f (x) dx + 0 7 7 Z 2f (x) dx = 2 f (x) dx 0 By Theorem 2 0 Z = 2 · (−10) = −20 Z 0 Z f (x) dx + 7 f (x) dx = 5 5 f (x) dx By Theorem 2 0 Z = 7 5 − Z f (x) dx + 0 = −3 − 10 = −13. 7 f (x) dx 0 Antiderivatives Definition 2. Let f (x) and F (x) be two functions, we say that F (x) is an antiderivative of f (x) if F 0 (x) = f (x). Example 7. d 2 (x ) = 2x =⇒ dx d (− cos(x)) = sin(x) dx =⇒ x2 is an antiderivative of 2x − cos(x) is an antiderivative of sin(x) LECTURE 13: DEFINITE INTEGRALS AND ANTIDERIVATIVES d 1 tan−1 (x) = 2 dx x +1 =⇒ 5 1 is an antiderivative of tan−1 (x) x2 + 1 Theorem 4. Any two antiderivatives of one function only differ by a constant, that is, if F (x) and G(x) are antiderivatives of f (x), then F (x) = G(x) + C for some constant C. Indefinite integrals Definition 3. Let F (x) be an antiderivative of f (x), that is, F 0 (x) = f (x). The indefinite integral, Z f (x) dx := Z F (x) + C (where C is arbitrary constant), means find the antiderivatives of f . The function f insider the symbol is called the integrand. Example 8. For Example 7, we have Z 2x dx = x2 + C sin(x) dx = − cos(x) + C 1 dx +1 = tan−1 (x) + C. Z Z x2 Theorem 5 (Linear properties). Let a be a constant, then Z Z af (x) dx = a f (x) dx. Also we have Z Z [f (x) + g(x)] dx = Z f (x) dx + g(x) dx. Some useful indefinite integrals Theorem 6. Z xp dx Z cos(ax) dx Z sin(ax) dx p+1 x + C, p+1 = ln |x|, 1 sin(ax) + C a 1 = − cos(ax) + C a = if p 6= −1 if p = −1. 6 MINGFENG ZHAO Z Z Z sec2 (ax) dx = csc2 (ax) dx Z eax dx = 1 dx − x2 Z 1 dx 2 x + a2 √ a2 = = = 1 tan(ax) + C a 1 − cot(ax) + C a 1 ax e +C a x +C sin−1 a x 1 +C tan−1 a a Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C. Canada V6T 1Z2 E-mail address: mingfeng@math.ubc.ca