Lesson 26: Evaluating Definite Integrals
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Lesson 26: Evaluating Definite Integrals Lesson 26: Evaluating Definite Integrals Presentation Transcript

  • Section 5.3 Evaluating Definite Integrals V63.0121.002.2010Su, Calculus I New York University June 21, 2010Announcements Final Exam is Thursday in class
  • Announcements Sections 5.3–5.4 today Section 5.5 tomorrow Review and Movie Day Wednesday Final exam Thursday roughly half-and-half MC/FR FR is all post-midterm MC might have some pre-midterm stuff on itV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 2 / 44
  • Resurrection Policy If your final score beats your midterm score, we will add 10% to its weight, and subtract 10% from the midterm weight.Image credit: Scott Beale / Laughing SquidV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 3 / 44
  • Objectives Use the Evaluation Theorem to evaluate definite integrals. Write antiderivatives as indefinite integrals. Interpret definite integrals as “net change” of a function over an interval.V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 4 / 44
  • Outline Last time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the Integral Evaluating Definite Integrals Examples The Integral as Total Change Indefinite Integrals My first table of integrals Computing Area with integralsV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 5 / 44
  • The definite integral as a limit 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 n→∞ i=1 b−a where ∆x = , and for each i, xi = a + i∆x, and ci is a point in n [xi−1 , xi ]. 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 and is the same for any choice of ci . aV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 6 / 44
  • Notation/Terminology 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 integrationV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 7 / 44
  • Example 1 4 Estimate dx using the midpoint rule and four divisions. 0 1 + x2V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 8 / 44
  • Example 1 4 Estimate dx using the midpoint rule and four divisions. 0 1 + x2 Solution Dividing up [0, 1] into 4 pieces gives 1 2 3 4 x0 = 0, x1 = , x2 = , x3 = , x4 = 4 4 4 4 So the midpoint rule gives 1 4 4 4 4 M4 = 2 + 2 + 2 + 4 1 + (1/8) 1 + (3/8) 1 + (5/8) 1 + (7/8)2V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 8 / 44
  • Example 1 4 Estimate dx using the midpoint rule and four divisions. 0 1 + x2 Solution Dividing up [0, 1] into 4 pieces gives 1 2 3 4 x0 = 0, x1 = , x2 = , x3 = , x4 = 4 4 4 4 So the midpoint rule gives 1 4 4 4 4 M4 = 2 + 2 + 2 + 4 1 + (1/8) 1 + (3/8) 1 + (5/8) 1 + (7/8)2 1 4 4 4 4 = + + + 4 65/64 73/64 89/64 113/64V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 8 / 44
  • Example 1 4 Estimate dx using the midpoint rule and four divisions. 0 1 + x2 Solution Dividing up [0, 1] into 4 pieces gives 1 2 3 4 x0 = 0, x1 = , x2 = , x3 = , x4 = 4 4 4 4 So the midpoint rule gives 1 4 4 4 4 M4 = 2 + 2 + 2 + 4 1 + (1/8) 1 + (3/8) 1 + (5/8) 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, 465V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 8 / 44
  • Properties of the integral 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 aV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 9 / 44
  • More Properties of the Integral 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 bV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 10 / 44
  • Definite Integrals We Know So Far If the integral computes an area and we know the area, we can use that. For y instance, 1 π 1 − x 2 dx = 0 2 By brute force we computed x 1 1 1 1 x 2 dx = x 3 dx = 0 3 0 4V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 11 / 44
  • Comparison Properties of the Integral 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 aV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 12 / 44
  • Integral of a nonnegative function is nonnegative Proof. If f (x) ≥ 0 for all x in [a, b], then for any number of divisions n and choice of sample points {ci }: n n Sn = f (ci ) ∆x ≥ 0 · ∆x = 0 i=1 ≥0 i=1 Since Sn ≥ 0 for all n, the limit of {Sn } is nonnegative, too: b f (x) dx = lim Sn ≥ 0 a n→∞ ≥0V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 13 / 44
  • Comparison Properties of the Integral 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 aV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 14 / 44
  • The definite integral is “increasing” Proof. Let h(x) = f (x) − g (x). If f (x) ≥ g (x) for all x in [a, b], then h(x) ≥ 0 for all x in [a, b]. So by the previous property b h(x) dx ≥ 0 a This means that b b b b f (x) dx − g (x) dx = (f (x) − g (x)) dx = h(x) dx ≥ 0 a a a a So b b f (x) dx ≥ g (x) dx a aV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 15 / 44
  • Comparison Properties of the Integral 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) aV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 16 / 44
  • Bounding the integral using bounds of the function Proof. If m ≤ f (x) ≤ M on for all x in [a, b], then by the previous property b b b m dx ≤ f (x) dx ≤ M dx a a a By Property 1, the integral of a constant function is the product of the constant and the width of the interval. So: b m(b − a) ≤ f (x) dx ≤ M(b − a) aV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 17 / 44
  • Estimating an integral with inequalities Example 2 1 Estimate dx using Property 8. 1 xV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 18 / 44
  • Estimating an integral with inequalities Example 2 1 Estimate dx using Property 8. 1 x Solution Since 1 1 1 1 ≤ x ≤ 2 =⇒ ≤ ≤ 2 x 1 we have 2 1 1 · (2 − 1) ≤ dx ≤ 1 · (2 − 1) 2 1 x or 2 1 1 ≤ dx ≤ 1 2 1 xV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 18 / 44
  • Outline Last time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the Integral Evaluating Definite Integrals Examples The Integral as Total Change Indefinite Integrals My first table of integrals Computing Area with integralsV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 19 / 44
  • Socratic proof The definite integral of velocity measures displacement (net distance) The derivative of displacement is velocity So we can compute displacement with the definite integral or the antiderivative of velocity But any function can be a velocity function, so . . .V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 20 / 44
  • Theorem of the Day Theorem (The Second Fundamental Theorem of Calculus) Suppose f is integrable on [a, b] and f = F for another function F , then b f (x) dx = F (b) − F (a). aV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 21 / 44
  • Theorem of the Day Theorem (The Second Fundamental Theorem of Calculus) Suppose f is integrable on [a, b] and f = F for another function F , then b f (x) dx = F (b) − F (a). a Note In Section 5.3, this theorem is called “The Evaluation Theorem”. Nobody else in the world calls it that.V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 21 / 44
  • Proving the Second FTC b−a Divide up [a, b] into n pieces of equal width ∆x = as usual. For n each i, F is continuous on [xi−1 , xi ] and differentiable on (xi−1 , xi ). So there is a point ci in (xi−1 , xi ) with F (xi ) − F (xi−1 ) = F (ci ) = f (ci ) xi − xi−1 Or f (ci )∆x = F (xi ) − F (xi−1 )V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 22 / 44
  • Proof continued We have for each i f (ci )∆x = F (xi ) − F (xi−1 ) Form the Riemann Sum: n n Sn = f (ci )∆x = (F (xi ) − F (xi−1 )) i=1 i=1 = (F (x1 ) − F (x0 )) + (F (x2 ) − F (x1 )) + (F (x3 ) − F (x2 )) + · · · · · · + (F (xn−1 ) − F (xn−2 )) + (F (xn ) − F (xn−1 )) = F (xn ) − F (x0 ) = F (b) − F (a)V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 23 / 44
  • Proof continued We have for each i f (ci )∆x = F (xi ) − F (xi−1 ) Form the Riemann Sum: n n Sn = f (ci )∆x = (F (xi ) − F (xi−1 )) i=1 i=1 = (F (x1 ) − F (x0 )) + (F (x2 ) − F (x1 )) + (F (x3 ) − F (x2 )) + · · · · · · + (F (xn−1 ) − F (xn−2 )) + (F (xn ) − F (xn−1 )) = F (xn ) − F (x0 ) = F (b) − F (a) See if you can spot the invocation of the Mean Value Theorem!V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 23 / 44
  • Proof Completed We have shown for each n, Sn = F (b) − F (a) so in the limit b f (x) dx = lim Sn = lim (F (b) − F (a)) = F (b) − F (a) a n→∞ n→∞V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 24 / 44
  • Computing area with the Second FTC Example Find the area between y = x 3 and the x-axis, between x = 0 and x = 1.V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 25 / 44
  • Computing area with the Second FTC Example Find the area between y = x 3 and the x-axis, between x = 0 and x = 1. Solution 1 1 x4 1 A= x 3 dx = = 0 4 0 4V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 25 / 44
  • Computing area with the Second FTC Example Find the area between y = x 3 and the x-axis, between x = 0 and x = 1. Solution 1 1 x4 1 A= x 3 dx = = 0 4 0 4 Here we use the notation F (x)|b or [F (x)]b to mean F (b) − F (a). a aV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 25 / 44
  • Computing area with the Second FTC Example Find the area enclosed by the parabola y = x 2 and y = 1.V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 26 / 44
  • Computing area with the Second FTC Example Find the area enclosed by the parabola y = x 2 and y = 1. 1 −1 1V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 26 / 44
  • Computing area with the Second FTC Example Find the area enclosed by the parabola y = x 2 and y = 1. 1 −1 1 Solution 1 1 x3 1 1 4 A=2− x 2 dx = 2 − =2− − − = −1 3 −1 3 3 3V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 26 / 44
  • Computing an integral we estimated before Example 1 4 Evaluate the integral dx. 0 1 + x2V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 27 / 44
  • Example 1 4 Estimate dx using the midpoint rule and four divisions. 0 1 + x2 Solution Dividing up [0, 1] into 4 pieces gives 1 2 3 4 x0 = 0, x1 = , x2 = , x3 = , x4 = 4 4 4 4 So the midpoint rule gives 1 4 4 4 4 M4 = 2 + 2 + 2 + 4 1 + (1/8) 1 + (3/8) 1 + (5/8) 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, 465V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 28 / 44
  • Computing an integral we estimated before Example 1 4 Evaluate the integral dx. 0 1 + x2 Solution 1 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 29 / 44
  • Computing an integral we estimated before Example 1 4 Evaluate the integral dx. 0 1 + x2 Solution 1 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2 = 4 arctan(x)|1 0V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 29 / 44
  • Computing an integral we estimated before Example 1 4 Evaluate the integral dx. 0 1 + x2 Solution 1 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2 = 4 arctan(x)|1 0 = 4 (arctan 1 − arctan 0)V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 29 / 44
  • Computing an integral we estimated before Example 1 4 Evaluate the integral dx. 0 1 + x2 Solution 1 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2 = 4 arctan(x)|1 0 = 4 (arctan 1 − arctan 0) π =4 −0 4V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 29 / 44
  • Computing an integral we estimated before Example 1 4 Evaluate the integral dx. 0 1 + x2 Solution 1 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2 = 4 arctan(x)|1 0 = 4 (arctan 1 − arctan 0) π =4 −0 =π 4V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 29 / 44
  • Computing an integral we estimated before Example 2 1 Evaluate dx. 1 xV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 30 / 44
  • Estimating an integral with inequalities Example 2 1 Estimate dx using Property 8. 1 x Solution Since 1 1 1 1 ≤ x ≤ 2 =⇒ ≤ ≤ 2 x 1 we have 2 1 1 · (2 − 1) ≤ dx ≤ 1 · (2 − 1) 2 1 x or 2 1 1 ≤ dx ≤ 1 2 1 xV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 31 / 44
  • Computing an integral we estimated before Example 2 1 Evaluate dx. 1 x Solution 2 1 dx 1 xV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 32 / 44
  • Computing an integral we estimated before Example 2 1 Evaluate dx. 1 x Solution 2 1 dx = ln x|2 1 1 xV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 32 / 44
  • Computing an integral we estimated before Example 2 1 Evaluate dx. 1 x Solution 2 1 dx = ln x|2 1 1 x = ln 2 − ln 1V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 32 / 44
  • Computing an integral we estimated before Example 2 1 Evaluate dx. 1 x Solution 2 1 dx = ln x|2 1 1 x = ln 2 − ln 1 = ln 2V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 32 / 44
  • Outline Last time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the Integral Evaluating Definite Integrals Examples The Integral as Total Change Indefinite Integrals My first table of integrals Computing Area with integralsV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 33 / 44
  • The Integral as Total Change Another way to state this theorem is: b F (x) dx = F (b) − F (a), a or the integral of a derivative along an interval is the total change between the sides of that interval. This has many ramifications:V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 34 / 44
  • The Integral as Total Change Another way to state this theorem is: b F (x) dx = F (b) − F (a), a or the integral of a derivative along an interval is the total change between the sides of that interval. This has many ramifications: Theorem If v (t) represents the velocity of a particle moving rectilinearly, then t1 v (t) dt = s(t1 ) − s(t0 ). t0V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 34 / 44
  • The Integral as Total Change Another way to state this theorem is: b F (x) dx = F (b) − F (a), a or the integral of a derivative along an interval is the total change between the sides of that interval. This has many ramifications: Theorem If MC (x) represents the marginal cost of making x units of a product, then x C (x) = C (0) + MC (q) dq. 0V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 34 / 44
  • The Integral as Total Change Another way to state this theorem is: b F (x) dx = F (b) − F (a), a or the integral of a derivative along an interval is the total change between the sides of that interval. This has many ramifications: Theorem If ρ(x) represents the density of a thin rod at a distance of x from its end, then the mass of the rod up to x is x m(x) = ρ(s) ds. 0V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 34 / 44
  • Outline Last time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the Integral Evaluating Definite Integrals Examples The Integral as Total Change Indefinite Integrals My first table of integrals Computing Area with integralsV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 35 / 44
  • A new notation for antiderivatives To emphasize the relationship between antidifferentiation and integration, we use the indefinite integral notation f (x) dx for any function whose derivative is f (x).V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 36 / 44
  • A new notation for antiderivatives To emphasize the relationship between antidifferentiation and integration, we use the indefinite integral notation f (x) dx for any function whose derivative is f (x). Thus x 2 dx = 1 x 3 + C . 3V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 36 / 44
  • My first table of integrals [f (x) + g (x)] dx = f (x) dx + g (x) dx x n+1 x n dx = + C (n = −1) cf (x) dx = c f (x) dx n+1 1 e x dx = e x + C dx = ln |x| + C x ax sin x dx = − cos x + C ax dx = +C ln a cos x dx = sin x + C csc2 x dx = − cot x + C sec2 x dx = tan x + C csc x cot x dx = − csc x + C 1 sec x tan x dx = sec x + C √ dx = arcsin x + C 1 − x2 1 dx = arctan x + C 1 + x2V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 37 / 44
  • Outline Last time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the Integral Evaluating Definite Integrals Examples The Integral as Total Change Indefinite Integrals My first table of integrals Computing Area with integralsV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 38 / 44
  • Example Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and the vertical lines x = 0 and x = 3.V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 39 / 44
  • Example Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and the vertical lines x = 0 and x = 3. Solution 3 Consider (x − 1)(x − 2) dx. 0V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 39 / 44
  • Graph y x 1 2 3V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 40 / 44
  • Example Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and the vertical lines x = 0 and x = 3. Solution 3 Consider (x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1) 0 and (2, 3], and negative on (1, 2).V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 41 / 44
  • Example Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and the vertical lines x = 0 and x = 3. Solution 3 Consider (x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1) 0 and (2, 3], and negative on (1, 2). If we want the area of the region, we have to do 1 2 3 A= (x − 1)(x − 2) dx − (x − 1)(x − 2) dx + (x − 1)(x − 2) dx 0 1 2 1 3 1 2 3 = 3x − 2 x 2 + 2x 3 0 − 1 3 3x − 3 x 2 + 2x 2 1 + 1 3 3x − 2 x 2 + 2x 3 2 5 1 5 11 = − − + = . 6 6 6 6V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 41 / 44
  • Interpretation of “negative area” in motion There is an analog in rectlinear motion: t1 v (t) dt is net distance traveled. t0 t1 |v (t)| dt is total distance traveled. t0V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 42 / 44
  • What about the constant? It seems we forgot about the +C when we say for instance 1 1 x4 1 1 x 3 dx = = −0= 0 4 0 4 4 But notice 1 x4 1 1 1 +C = +C − (0 + C ) = +C −C = 4 0 4 4 4 no matter what C is. So in antidifferentiation for definite integrals, the constant is immaterial.V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 43 / 44
  • Summary The second Fundamental Theorem of Calculus: b f (x) dx = F (b) − F (a) a where F = f . Definite integrals represent net change of a function over an interval. We write antiderivatives as indefinite integrals f (x) dxV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 44 / 44