Diese Präsentation wurde erfolgreich gemeldet.

Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
×

1 von 13 Anzeige

# Lesson 25: Evaluating Definite Integrals (Section 021 handout)

A remarkable theorem about definite integrals is that they can be calculated with antiderivatives.

A remarkable theorem about definite integrals is that they can be calculated with antiderivatives.

Anzeige
Anzeige

## Weitere Verwandte Inhalte

Anzeige

Anzeige

### Lesson 25: Evaluating Definite Integrals (Section 021 handout)

1. 1. V63.0121.021, Calculus I Section 5.3 : Evaluating Deﬁnite Integrals December 7, 2010 Section 5.3 Notes Evaluating Deﬁnite Integrals V63.0121.021, Calculus I New York University December 7, 2010 Announcements Today: Section 5.3 Thursday: Section 5.4 ”Thursday,” December 14: Section 5.5 ”Monday,” December 15: (WWH 109, 12:30–1:45pm) Review and Movie Day! Monday, December 20, 12:00–1:50pm: Final Exam Announcements Notes Today: Section 5.3 Thursday: Section 5.4 ”Thursday,” December 14: Section 5.5 ”Monday,” December 15: (WWH 109, 12:30–1:45pm) Review and Movie Day! Monday, December 20, 12:00–1:50pm: Final Exam V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 2 / 41 Objectives Notes 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. V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 3 / 41 1
2. 2. V63.0121.021, Calculus I Section 5.3 : Evaluating Deﬁnite Integrals December 7, 2010 Outline Notes 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 integrals V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 4 / 41 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 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 . a V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 5 / 41 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 6 / 41 2
3. 3. V63.0121.021, Calculus I Section 5.3 : Evaluating Deﬁnite Integrals December 7, 2010 Example 1 4 Notes 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 = + + + 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 7 / 41 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 8 / 41 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 Theorem c b c 5. f (x) dx = f (x) dx + f (x) dx for all a, b, and c. a a b V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 9 / 41 3
4. 4. V63.0121.021, Calculus I Section 5.3 : Evaluating Deﬁnite Integrals December 7, 2010 Illustrating Property 5 Notes Theorem c b c 5. f (x) dx = f (x) dx + f (x) dx for all a, b, and c. a a b y b c f (x) dx f (x) dx a b a b c x V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 10 / 41 Illustrating Property 5 Notes Theorem c b c 5. f (x) dx = f (x) dx + f (x) dx for all a, b, and c. a a b y c c f (x) dx f (x) dx = a b b − f (x) dx c a c x b V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 10 / 41 Deﬁnite Integrals We Know So Far Notes 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 4 By brute force we computed x 1 1 1 1 x 2 dx = x 3 dx = 0 3 0 4 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 11 / 41 4
5. 5. V63.0121.021, Calculus I Section 5.3 : Evaluating Deﬁnite Integrals December 7, 2010 Comparison Properties of the Integral Notes Theorem Let f and g be integrable functions on [a, b]. b 6. If f (x) ≥ 0 for all x in [a, b], then 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 12 / 41 Estimating an integral with inequalities Notes 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 x V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 13 / 41 Outline Notes 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 integrals V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 14 / 41 5
6. 6. V63.0121.021, Calculus I Section 5.3 : Evaluating Deﬁnite Integrals December 7, 2010 Socratic proof Notes 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 . . . V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 15 / 41 Theorem of the Day Notes 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 16 / 41 Proving the Second FTC Notes b−a Divide up [a, b] into n pieces of equal width ∆x = as usual. n For 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 ) V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 17 / 41 6
7. 7. V63.0121.021, Calculus I Section 5.3 : Evaluating Deﬁnite Integrals December 7, 2010 Proof continued Notes 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) V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 18 / 41 Proof Completed Notes We have shown for each n, Sn = F (b) − F (a) Which does not depend on n. So in the limit b f (x) dx = lim Sn = lim (F (b) − F (a)) = F (b) − F (a) a n→∞ n→∞ V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 19 / 41 Computing area with the Second FTC Notes 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 a V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 20 / 41 7
8. 8. V63.0121.021, Calculus I Section 5.3 : Evaluating Deﬁnite Integrals December 7, 2010 Computing area with the Second FTC Notes Example Find the area enclosed by the parabola y = x 2 and the line y = 1. 1 −1 1 Solution 1 1 x3 1 1 4 A=2− x 2 dx = 2 − =2− − − = −1 3 −1 3 3 3 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 21 / 41 Computing an integral we estimated before Notes Example 1 4 Evaluate the integral dx. 0 1 + x2 Solution V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 22 / 41 Computing an integral we estimated before Notes Example 2 1 Evaluate dx. 1 x Solution V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 25 / 41 8
9. 9. V63.0121.021, Calculus I Section 5.3 : Evaluating Deﬁnite Integrals December 7, 2010 Outline Notes 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 integrals V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 28 / 41 The Integral as Total Change Notes 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 ). t0 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 29 / 41 The Integral as Total Change Notes 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. 0 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 29 / 41 9
10. 10. V63.0121.021, Calculus I Section 5.3 : Evaluating Deﬁnite Integrals December 7, 2010 The Integral as Total Change Notes 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. 0 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 29 / 41 Outline Notes 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 integrals V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 30 / 41 A new notation for antiderivatives Notes 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 . 3 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 31 / 41 10
11. 11. V63.0121.021, Calculus I Section 5.3 : Evaluating Deﬁnite Integrals December 7, 2010 My ﬁrst table of integrals Notes [f (x) + g (x)] dx = f (x) dx + g (x) dx n x n+1 x dx = + C (n = −1) cf (x) dx = c f (x) dx n+1 x x 1 e dx = e + C dx = ln |x| + C x x ax sin x dx = − cos x + C a 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 + x2 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 32 / 41 Outline Notes 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 integrals V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 33 / 41 Computing Area with integrals Notes Example Find the area of the region bounded by the lines x = 1, x = 4, the x-axis, and the curve y = e x . Solution The answer is 4 e x dx = e x |4 = e 4 − e. 1 1 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 34 / 41 11
12. 12. V63.0121.021, Calculus I Section 5.3 : Evaluating Deﬁnite Integrals December 7, 2010 Computing Area with integrals Notes Example Find the area of the region bounded by the curve y = arcsin x, the x-axis, and the line x = 1. Solution V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 35 / 41 Example Notes 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 36 / 41 Graph Notes V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 37 / 41 12
13. 13. V63.0121.021, Calculus I Section 5.3 : Evaluating Deﬁnite Integrals December 7, 2010 Interpretation of “negative area” in motion Notes There is an analog in rectlinear motion: t1 v (t) dt is net distance traveled. t0 t1 |v (t)| dt is total distance traveled. t0 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 39 / 41 What about the constant? Notes 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. V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 40 / 41 Summary Notes 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) dx V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Deﬁnite Integrals December 7, 2010 41 / 41 13