Lesson 31: Evaluating Definite Integrals

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The Second Fundamental Theorem of Calculus allows us to compute definite integrals by antidifferentiation

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Lesson 31: Evaluating Definite Integrals

  1. 1. Section 5.3 Evaluating Definite Integrals Math 1a December 10, 2007 Announcements my next office hours: Monday 1–2, Tuesday 3–4 (SC 323) MT II is graded. Come to OH to talk about it Final seview sessions: Wed 1/9 and Thu 1/10 in Hall D, Sun 1/13 in Hall C, all 7–8:30pm Final tentatively scheduled for January 17, 9:15am
  2. 2. Outline FTC2 Proof Examples Total Change Indefinite Integrals My first table of integrals Examples “Negative Area”
  3. 3. Theorem (The Second Fundamental Theorem of Calculus, Strong Form) Suppose f is integrable on [a, b] and f = F for another function f , then b f (x) dx = F (b) − F (a). a
  4. 4. Proof. We will choose Riemann sums which converge to the right-hand side. Let n be given. On the interval [xi−1 , xi ] there is a point ci such that F (xi ) − F (xi−1 ) f (ci ) = F (ci ) = xi − xi−1 Then n n F (xi ) − F (xi−1 ) Rn = f (ci )∆x = i=1 i=1 = F (xn ) − F (x0 ) = F (b) − F (a). So Rn → F (b) − F (a) as n → ∞.
  5. 5. Examples Find the following integrals: 1 1 1 2 1 x 2 dx, x 3 dx, x n dx (n = −1), dx x 0 0 0 1 π 1 π/4 e x dx, sin θ dθ, tan θ dθ 0 0 0
  6. 6. Outline FTC2 Proof Examples Total Change Indefinite Integrals My first table of integrals Examples “Negative Area”
  7. 7. 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:
  8. 8. 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 ). t0
  9. 9. 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. 0
  10. 10. 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. 0
  11. 11. Example If oil leaks from a tank at a rate of r (t) gallons per minute at time 120 t, what does r (t) dt represent? 0
  12. 12. Example If oil leaks from a tank at a rate of r (t) gallons per minute at time 120 t, what does r (t) dt represent? 0 Solution The amount of oil lost in two hours.
  13. 13. Outline FTC2 Proof Examples Total Change Indefinite Integrals My first table of integrals Examples “Negative Area”
  14. 14. 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).
  15. 15. 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 = 3 x 3 + C . 1
  16. 16. My first table of integrals [f (x) + g (x)] dx = f (x) dx + g (x) dx x n+1 x n dx = cf (x) dx = c f (x) dx + C (n = −1) n+1 1 e x dx = e x + C dx = ln x + C x ax ax dx = +C sin x dx = − cos x + C ln a csc2 x dx = − cot x + C cos x dx = sin x + C sec2 x dx = tan x + C csc x cot x dx = − csc x + C 1 √ dx = arcsin x + C sec x tan x dx = sec x + C 1 − x2 1 dx = arctan x + C 1 + x2
  17. 17. Outline FTC2 Proof Examples Total Change Indefinite Integrals My first table of integrals Examples “Negative Area”
  18. 18. 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.
  19. 19. 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 (x − 1)(x − 2) dx. Notice the integrand is positive on Consider 0 [0, 1) and (2, 3], and negative on (1, 2). If we want the area of the region, we have to do 1 2 3 (x − 1)(x − 2) dx − (x − 1)(x − 2) dx + (x − 1)(x − 2) dx A= 0 1 2 1 2 3 13 − 2 x 2 + 2x 13 − 2 x 2 + 2x 13 − 2 x 2 + 2x 3 3 3 − = 3x 3x + 3x 0 1 2 5 1 5 11 −− = + =. 6 6 6 6

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