Unit D Overview

1. Unit D Overview

2. Antiderivatives and the Indefinite Integral

3. Supposed you were asked to find a function F whose derivative is f(x) = 3x². • The function F is the antiderivative of f. Antiderivatives

4. You can represent a family of antiderivatives of a function by adding a constant to a known antiderivative. • For example: • Given that f(x) = 2, find its antiderivative. • Is that the only one? • So the family of ALL antiderivatives of f(x) = 2 can be represented by • G(x) = 2x + C, where C is a constant. Family of antiderivatives

5. C is called the constant of integration • The family of functions represented by G is called the general antiderivative • G(x) = x² + C is the general solution of the differential equation C

6. A differential equation in x and y is an equation that involves x, y, and derivatives of y. For instance y’ = 3x is an example of a differential equation. A Differential Equation

7. Find the general solution of the differential equation y’ = 2 • To begin you find the function whose derivative is 2. One such function is y = 2x • Now you can use our theorem about C to conclude that the general solutions to the differential equation is • y = 2x + C Example 1 Solving a Differential Equation

8. The operation of finding all solutions of an equation is called antidifferentiation (or indefinite integration) • This is denoted with an integral sign • The term indefinite integral is a synonym for antiderivative. • Basic Integration Rule Variable of integration Integrand Constant of Integration Notation for Antiderivatives

9. Rule for powers. Normally when you take the derivative we bring the power down to the front and multiply, then subtract one from the power. For antiderivative you do the opposite of that. Add one to the power then divide by that number. Basic Integration Rules

10. Describe the antiderivative of 3x • x² + C where C is any constant Example 2

11. Sometimes it is easier to re-write the integral before using rules. • - + C Example 3 (Re-write)

12. + 2x + C Example 4 (Integrating a polynomial function)

13. You have already seen that an antiderivative can have many solutions. • Example • dx • The solution could be 2x + 3 or 2x – 5 or 2x + 167,000 • This is when we put the constant of integration to show that there is a family of functions with that derivative. General Solution

14. Find the general solution of F’(x) = , x > 0 • dx • - + C , x > 0 • And find the particular solution that satisfies the initial condition F(1) = 0. • F(1) = - + C = 0 • Therefore C must equal 1. • So the particular solution is F(x) = - + 1 Initial conditions and Particular Solutions Example 5

15. A ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 ft. • A) Find the position function giving the height s as a function of time t. • B) When does the ball hit the ground? Example 6 Vertical Motion

16. Let t = 0 represent the initial time. The two given initial conditions can be written as: • s(0) = 80 • s’(0) = 64 • Using -32 ft/sec² as acceleration due to gravity, you can write • s’’(t) = -32 • s’(t) = = -32t + • Since we know our initial velocity s’(0) = 64 we can find . • -32(0) + = 64 • Therefore = 64 A ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 ft. A) Find the position function giving the height s as a function of time t.

17. To find position we integrate velocity • We get -16t² + 64t + • We know our initial condition s(0) = 80 • So 80 = -16(0)² + 64(0) + • Therefore = 80 • So our position function is s(t) = 16t² + 64t + 80 Now we have our velocity function s’(t) = -32t + 64

18. Using the position function found in part A, you can find when the ball hits the ground by setting it equal to 0. • s(t) = 16t² + 64t + 80 = 0 • This happens at t = -1, and 5 • Because t must be positive, you can conclude the ball hits the ground 5 seconds after it was thrown. s(t) = 16t² + 64t + 80

19. Pg. 255 # 5, 12, 15, 17, 19, 25, 35 HW

20. Estimating Finite Sums- Rectangular Approximation Method

21. Figure 1: A train moves along a track at a steady rate of 75 mph from 7 AM to 9 AM. What is the total distance traveled by the train? • Can be represented by determining the area of the rectangle above. • The distance traveled would be 75 mph (2 hours ) = 150 miles • Looking at the rectangle, A = length x width = 150 miles 75 mph Velocity (mph) Distance Traveled Time 7 9

22. Solution: Graph and find the area of the shape. • Area = ½ base (height) • = ½ (11)(88) • = 484 feet A car accelerates from 0 to 88 feet in 11 seconds. If the acceleration is CONSTANT, how far will the car have traveled in that time?

23. Solution: Graph and find the area of the shape. • The shape is a trapezoid. • A = ½ h (b1 + b2 ) • h = 20 -8 = 12 • b1 = v(8) =544 • b2 = v(20) = 160 • A = ½ (12)(544 + 160) • = 4224 units v(t) = 800 – 32t. Find the distance traveled between 8 and 20 seconds.

24. Find the area of this…

25. If the velocity varies over the timer interval [a,b], does the shaded region give the distance traveled? • Yes, but because it is an irregular region, we would need to partition it into narrow rectangles and add the areas of each rectangle. Distance Traveled

26. We use the left-handed endpoints for LRAM. • We use the right-handed endpoints for RRAM. • We use the midpoints for MRAM. Choosing values of t:

27. RAM

28. Approximate area: We could estimate the area under the curve by drawing rectangles touching at their left corners.

29. Approximate area: We could also use a Right-hand Rectangular Approximation Method (RRAM).

30. Approximate area: Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM). In this example there are four subintervals. As the number of subintervals increases, so does the accuracy.

31. A particle starts at x = 0 and moves along the x-axis with velocity v(t) = 3t2 for time t> 0. Where is the particle at t = 6? Solution: Graph the function of v and partition the time interval into subintervals of length Δt. We will define Δt to be 1, which means the base of each smaller rectangle will be 1. The height of each rectangle will be v(t), where t is represented by some point within each interval along the x-axis. Our intervals will be broken down as follows: 0 1 2 3 4 5 6 Example 1 Finding Distance Traveled when Velocity Varies

32. A particle starts at x = 0 and moves along the x-axis with velocity v(t) = 3t2 for time t> 0. Where is the particle at t = 6? LRAM Velocity Example 1 Finding Distance Traveled when Velocity Varies Time

33. A particle starts at x = 0 and moves along the x-axis with velocity v(t) = 3t2 for time t> 0. Where is the particle at t = 6? Velocity RRAM Example 1 Finding Distance Traveled when Velocity Varies Time

34. A particle starts at x = 0 and moves along the x-axis with velocity v(t) = 3t2 for time t> 0. Where is the particle at t = 6? MRAM Velocity Example 1 Finding Distance Traveled when Velocity Varies Time

35. You may see either of these written, they mean the same thing. RAM and Reimann Sum

36. Dye Concentration Data • Estimate the cardiac output of the patient whose data appears above. • Cardiac output = amount of dye/area under the curve • Assume our patient’s amount of dye = 5.6 mg Example 3

37. LRAM RRAM 2(3.8 + 8 + 6.1 + 3.6 + 2.3 + 1.45 + .91 + .57 + .36 + .23 + .14 + .09 + 0) 2(27.55) = 55.1 mg/L · sec • 2( 0 + 3.8 + 8 + 6.1 + 3.6 + 2.3 + 1.45 + .91 + .57 + .36 + .23 + .14 + .09) • 2(27.55) = 55.1 mg/L · sec Cardiac OutputTo compute MRAM, we would actually need to draw the rectangles and determine the height of each rectangle. 5.6 mg/ 55.1 mg/L · sec = .102 L/sec

38. The values of a continuous function f for selected values of x are given in the table below What is the left Reimann Sum approximation using subintervals given. Using rectangles of different width

39. Paper HW

40. Evaluate each sum • 12 + 22 + 32 + 42 + 52 =55 • [3(0) -2] + [3(1) -2] + [3(2) -2] + [3(3) -2] + [3(4) -2] = 20 • 100 (0 + 1)2 + 100 (1 + 1)2 + 100 (2 + 1)2 + 100(3 + 1)2 + 100 (4 + 1)2 =5500

41. Definite Integrals

42. A Riemann sum, Rn, for function f on the interval [a, b] is a sum of the form where the interval [a, b] is partitioned into n subintervals of widths Δxk, and the numbers {ck} are sample points, one in each subinterval. Key Concept: Riemann Sum

43. Example: Calculating Riemann Sums Upper = using right endpoints: ¼ (1/64+ 1/8 + 27/64 + 1) = 25/64 Lower = using left endpoints: ¼ ( 1/64 + 1/8 + 27/64) = 9/64

44. More Rectangles More accuracy

45. Let f be continuous on [a,b] and be partitioned into n subintervals of equal length Δx = (b – a)/n ck is some point in the kth subinterval. When you find the value of the integral, you have evaluated the integral The function is the integrand Upper limit of integration Integral sign x is the variable of integration lower limit of integration Definite Integral

46. Express each limit as a definite integral

47. If y = f (x) is nonnegative and integrable on [a, b], and if Rn is any Riemann sum for f on [a, b], then Key Concept: Area Under a Curve

48. + 3) dx 5 2 Use the graph of the integrand and areas to evaluate the integral 6 A=1/2(6)(2+5)= 21

49. Write the definite integral and determine the area under the curve over the interval [–4, 4]. Example Area Under a Curve 4 -4

50. If y = f(x) is a nonnegative and integrable over a closed [a,b], then the area under the curve y = f(x) from a to b is the intergral of f from a to b. • A = Area under a curve (as a definite integral)