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- 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. Outline FTC2 Proof Examples Total Change Indeﬁnite Integrals My ﬁrst table of integrals Examples “Negative Area”
- 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. 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. 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. Outline FTC2 Proof Examples Total Change Indeﬁnite Integrals My ﬁrst table of integrals Examples “Negative Area”
- 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. 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. 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. 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. 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. 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. Outline FTC2 Proof Examples Total Change Indeﬁnite Integrals My ﬁrst table of integrals Examples “Negative Area”
- 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. 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. 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. Outline FTC2 Proof Examples Total Change Indeﬁnite Integrals My ﬁrst table of integrals Examples “Negative Area”
- 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. 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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