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Lesson 23: The Definite Integral (handout)

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Lesson 23: The Definite Integral (handout)

  1. 1. V63.0121.006/016, Calculus I Section 5.2 : The Definite Integral April 15, 2010 Notes Section 5.2 The Definite Integral V63.0121.006/016, Calculus I New York University April 15, 2010 Announcements Announcements Notes V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 2 / 21 Outline Notes Recall The definite integral as a limit Estimating the Definite Integral Properties of the integral Comparison Properties of the Integral V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 3 / 21 1
  2. 2. V63.0121.006/016, Calculus I Section 5.2 : The Definite Integral April 15, 2010 Cavalieri’s method in general Notes Let f be a positive function defined on the interval [a, b]. We want to find the area between x = a, x = b, y = 0, and y = f (x). For each positive integer n, divide up the interval into n pieces. Then b−a ∆x = . For each i between 1 and n, let xi be the ith step between a n and b. So x0 = a b−a x1 = x0 + ∆x = a + n b−a x2 = x1 + ∆x = a + 2 · ... n b−a xi = a + i · ... n b−a xn = a + n · =b x n x0 x1 . . . xi . . .xn−1xn V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 4 / 21 Forming Riemann sums Notes We have many choices of representative points to approximate the area in each subinterval. left endpoints. . . right endpoints. . . midpoints. . . random points. . . n n n xi−1 + xi Ln = f (xi−1 )∆xRn = f (xi )∆xMn = f ∆x 2 i=1 i=1 i=1 x In general, choose ci to be a point in the ith interval [xi−1 , xi ]. Form the Riemann sum n Sn = f (c1 )∆x + f (c2 )∆x + · · · + f (cn )∆x = f (ci )∆x i=1 V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 5 / 21 Theorem of the (previous) Day Notes Theorem If f is a continuous function on [a, b] or has finitely many jump S3 = 8.0805 14 7.53027 13 7.88342 12 7.67683 11 7.42482 10 8.05293 4 =7.69078 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 9 8 7 6 5 7.4937 7.45966 7.51125 7.50056 7.50064 7.50656 7.49278 7.56894 7.52928 7.56194 7.48744 7.57709 7.57236 7.44841 7.54318 7.71298 7.5657 7.60515 7.42561 7.5189 7.19319 discontinuities, then n lim Sn = lim f (ci )∆x n→∞ n→∞ i=1 exists and is the same value no matter what choice of ci we made. x V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 6 / 21 2
  3. 3. V63.0121.006/016, Calculus I Section 5.2 : The Definite Integral April 15, 2010 Outline Notes Recall The definite integral as a limit Estimating the Definite Integral Properties of the integral Comparison Properties of the Integral V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 7 / 21 The definite integral as a limit Notes Definition If f is a function defined on [a, b], the definite integral of f from a to b is the number b n f (x) dx = lim f (ci ) ∆x a ∆x→0 i=1 V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 8 / 21 Notation/Terminology Notes b f (x) dx a — integral sign (swoopy S) f (x) — integrand a and b — limits of integration (a is the lower limit and b the upper limit) dx — ??? (a parenthesis? an infinitesimal? a variable?) The process of computing an integral is called integration or quadrature V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 9 / 21 3
  4. 4. V63.0121.006/016, Calculus I Section 5.2 : The Definite Integral April 15, 2010 The limit can be simplified Notes Theorem If f is continuous on [a, b] or if f has only finitely many jump discontinuities, then f is integrable on [a, b]; that is, the definite integral b f (x) dx exists. a Theorem If f is integrable on [a, b] then b n f (x) dx = lim f (xi )∆x, a n→∞ i=1 where b−a ∆x = and xi = a + i ∆x n V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 10 / 21 Outline Notes Recall The definite integral as a limit Estimating the Definite Integral Properties of the integral Comparison Properties of the Integral V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 11 / 21 Estimating the Definite Integral Notes Given a partition of [a, b] into n pieces, let xi be the midpoint of [xi−1 , xi ]. ¯ Define n Mn = f (¯i ) ∆x. x i=1 V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 12 / 21 4
  5. 5. V63.0121.006/016, Calculus I Section 5.2 : The Definite Integral April 15, 2010 Example Notes 1 4 Estimate dx using the midpoint rule and four divisions. 0 1 + x2 Solution 1 1 3 The partition is 0 < < < < 1, so the estimate is 4 2 4 1 4 4 4 4 M4 = + + + 4 1 + (1/8)2 1 + (3/8)2 1 + (5/8)2 1 + (7/8)2 1 4 4 4 4 = + + + 4 65/64 73/64 89/64 113/64 150, 166, 784 = ≈ 3.1468 47, 720, 465 V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 13 / 21 Outline Notes Recall The definite integral as a limit Estimating the Definite Integral Properties of the integral Comparison Properties of the Integral V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 14 / 21 Properties of the integral Notes Theorem (Additive Properties of the Integral) Let f and g be integrable functions on [a, b] and c a constant. Then b 1. c dx = c(b − a) a b b b 2. [f (x) + g (x)] dx = f (x) dx + g (x) dx. a a a b b 3. cf (x) dx = c f (x) dx. a a b b b 4. [f (x) − g (x)] dx = f (x) dx − g (x) dx. a a a V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 15 / 21 5
  6. 6. V63.0121.006/016, Calculus I Section 5.2 : The Definite Integral April 15, 2010 More Properties of the Integral Notes Conventions: a b f (x) dx = − f (x) dx b a a f (x) dx = 0 a This allows us to have c b c 5. f (x) dx = f (x) dx + f (x) dx for all a, b, and c. a a b V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 16 / 21 Example Notes Suppose f and g are functions with 4 f (x) dx = 4 0 5 f (x) dx = 7 0 5 g (x) dx = 3. 0 Find 5 (a) [2f (x) − g (x)] dx 0 5 (b) f (x) dx. 4 V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 17 / 21 Solution Notes We have (a) 5 5 5 [2f (x) − g (x)] dx = 2 f (x) dx − g (x) dx 0 0 0 = 2 · 7 − 3 = 11 (b) 5 5 4 f (x) dx = f (x) dx − f (x) dx 4 0 0 =7−4=3 V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 18 / 21 6
  7. 7. V63.0121.006/016, Calculus I Section 5.2 : The Definite Integral April 15, 2010 Outline Notes Recall The definite integral as a limit Estimating the Definite Integral Properties of the integral Comparison Properties of the Integral V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 19 / 21 Comparison Properties of the Integral Notes Theorem Let f and g be integrable functions on [a, b]. 6. If f (x) ≥ 0 for all x in [a, b], then b f (x) dx ≥ 0 a 7. If f (x) ≥ g (x) for all x in [a, b], then b b f (x) dx ≥ g (x) dx a a 8. If m ≤ f (x) ≤ M for all x in [a, b], then b m(b − a) ≤ f (x) dx ≤ M(b − a) a V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 20 / 21 Notes Example 2 1 Estimate dx using the comparison properties. 1 x Solution Since 1 1 ≤x ≤ 2 1 for all x in [1, 2], we have 2 1 1 ·1≤ dx ≤ 1 · 1 2 1 x V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 21 / 21 7

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