Mastering Integration and Antiderivatives

1. 6 • Antiderivatives and the Rules of Integration • Integration by Substitution • Area and the Definite Integral • The Fundamental Theorem of Calculus • Evaluating Definite Integrals • Area Between Two Curves • Applications of the Definite Integral to Business and Economics Integration

2. 6.1 Antiderivatives and the Rules of Integration

3. Antiderivatives • Recall the Maglev problem discussed in chapter 2. • The question asked then was: • If we know the position of the maglev at any timet, can we find its velocity at that time? • The position was described by f(t), and the velocity by f′(t). • Now, in Chapters 6 and 7 we will consider precisely the opposite problem: • If we know the velocity of the maglev at any timet, can we find its position at that time? • That is, knowing its velocity functionf′(t), can we find its position functionf(t)?

4. Antiderivatives • To solve this kind of problems, we need the concept of the antiderivative of a function. • A function F is an antiderivative of f on an interval I if F′(t)=f(t) for all of t in I.

5. Examples • Let • Show that F is an antiderivative of Solution • Differentiating the function F, we obtain and the desired result follows. Example 1, page 398

6. Examples • Let F(x) = x, G(x) = x + 2, H(x) = x + C, where C is a constant. • Show that F, G, and H are all antiderivatives of the function f defined by f(x) = 1. Solution • Since we see that F, G, and H are indeed antiderivatives of f. Example 2, page 398

7. Theorem 1 • Let G be an antiderivative of a function f. • Then, every antiderivative F of f must be of the form F(x) = G(x) + C where C is a constant.

8. Example • Prove that the function G(x) = x2 is an antiderivative of the function f(x) = 2x. • Write a general expression for the antiderivatives of f. Solution • Since G′(x) = 2x = f(x), we have shown that G(x) = x2 is an antiderivative of f(x) = 2x. • By Theorem 1, every antiderivative of the function f(x) = 2x has the form F(x) = x2+ C, where C is a constant. Example 3, page 399

9. The Indefinite Integral • The process of finding all the antiderivatives of a function is called antidifferentiation or integration. • We use the symbol ∫, called an integral sign, to indicate that the operation of integration is to be performed on some function f. • Thus, where C and K are arbitrary constants.

10. Basic Integration Rules • Rule 1: The Indefinite Integral of a Constant

11. Example • Find each of the following indefinite integrals: a. b. Solution • Each of the integrals had the formf(x) = k, where k is a constant. • Applying Rule 1 in each case yields: a. b. Example 4, page 400

12. Basic Integration Rules • From the rule of differentiation, we obtain the following rule of integration: • Rule 2: The Power Rule

13. Examples • Find the indefinite integral: Example 5, page 401

14. Examples • Find the indefinite integral: Example 5, page 401

15. Examples • Find the indefinite integral: Example 5, page 401

16. Basic Integration Rules • Rule 3: The Indefinite Integral of a Constant Multiple of a Function where c is a constant.

17. Examples • Find the indefinite integral: Example 6, page 402

18. Examples • Find the indefinite integral: Example 6, page 402

19. Basic Integration Rules • Rule 4: The Sum Rule

20. Examples • Find the indefinite integral: Example 7, page 402

21. Basic Integration Rules • Rule 5: The Indefinite Integral of the Exponential Function

22. Examples • Find the indefinite integral: Example 8, page 403

23. Basic Integration Rules • Rule 6:The Indefinite Integral of the Functionf(x) = x–1

24. Examples • Find the indefinite integral: Example 9, page 403

25. Differential Equations • Given the derivative of a function, f′, can we find the functionf? • Consider the function f′(x)= 2x – 1 from which we want to find f(x). • We can find f by integrating the equation: where C is an arbitrary constant. • Thus, infinitely many functions have the derivative f′, each differing from the other by a constant.

26. Differential Equations • Equationf′(x)= 2x – 1 is called a differential equation. • In general, a differential equation involves the derivative of an unknown function. • A solution of a differential equation is any function that satisfies the differential equation. • For the case of f′(x)= 2x – 1, we find that f(x)= x2 – x + Cgives allthe solutions of the differential equation, and it is therefore called the general solution of the differential equation.

27. Differential Equations f(x)= x2 – x+ 3 y • Different values of C yield different functionsf(x). • But all these functions have the same slope for any given value of x. • For example, for any value ofC, we always find thatf ′(1) =1. f(x)= x2 – x+ 2 5 4 3 2 1 –1 f(x)= x2 – x+ 1 f(x)= x2 – x+ 0 f ′(1) =1 f ′(1) =1 f(x)= x2 – x– 1 f ′(1) =1 f ′(1) =1 x –1 1 2 3 f ′(1) =1

28. Differential Equations • It is possible to obtain a particular solution by specifying the value the function must assume for a given value of x. • For example, suppose we know the function f must pass through the point(1, 2), which means f(1) = 2. • Using this condition on the general solution we can find the value of C: f(1) = 12 – 1 + C = 2 C = 2 • Thus, the particular solution is f(x) = x2 – x + 2

29. Differential Equations y • Here is the graph of the particular solution of f when C = 2. • Note that this graph does go through the point (1, 2). f(x)= x2 – x+ 2 5 4 3 2 1 –1 (1, 2) x –1 1 2 3

30. Initial Value Problems • The problem we just discussed is of a type called initial value problem. • In this type of problem we are required to find a function satisfying • A differential equation. • One or more initial conditions.

31. Example • Find the function f if it is known that Solution • Integrating the function f′, we find Example 10, page 404

32. Example • Find the function f if it is known that Solution • Using the conditionf(1) = 9, we have • Therefore, the required functionf is Example 10, page 404

33. Applied Example: Velocity of Maglev • In a test run of a maglev, data obtained from reading its speedometer indicate that the velocity of the maglev at time t can be described by the velocity function • Find the position function of the maglev. • Assume that initially the maglev is located at the origin of a coordinate line. Applied Example 11, page 404

34. Applied Example: Velocity of Maglev Solution • Let s(t) denote the position of the maglev at any given time t (0 t 30). Then, s′(t) = v(t). • So, we have the initial value problem • Integrating the function s′, we find Applied Example 11, page 404

35. Applied Example: Velocity of Maglev Solution • Using the conditions(0) = 0, we have • Therefore, the required functions is Applied Example 11, page 404

36. 6.2 Integration by Substitution

37. Integration by Substitution • The method of substitution is related to the chain rule for differentiating functions. • It is a powerful tool for integrating a large class of functions.

38. How the Method of Substitution Works • Consider the indefinite integral • One way to solve this integral is to expand the expression and integrate the resulting integrand term by term. • An alternative approachsimplifies the integral by making a change of variable. • Write u = 2x + 4 with differential du = 2dx

39. How the Method of Substitution Works • Substituteu = 2x + 4anddu = 2dxin the original expression: • Now it’s easy to integrate: • Replacingu by u = 2x + 4, we obtain:

40. How the Method of Substitution Works • We can verify the result by finding its derivative: • The derivative is indeed the original integrand expression.

41. The Method of Integration by Substitution Step 1 Let u = g(x), where g(x) is part of the integrand, usually the “inside function” of the composite function f(g(x)). Step 2 Find du =g′(x)dx. Step 3 Use the substitutionu = g(x) and du =g′(x)dx to convert the entireintegral into one involving onlyu. Step 4 Evaluate the resulting integrand. Step 5Replaceu by g(x) to obtain the final solution as a function of x.

42. Examples • Find Solution Step 1 The integrand involves the composite function with “inside function” So, we choose Example 1, page 413

43. Examples • Find Solution Step 2 Find du/dx and solve for du: Example 1, page 413

44. Examples • Find Solution Step 3Substituteu = x2 + 3 and du =2xdx, to obtain an integral involving onlyu: Example 1, page 413

45. Examples • Find Solution Step 4Evaluate the integral: Step 5 Replaceu by x2 + 3 to find the solution: Example 1, page 413

46. Examples • Find Solution • Let u = –3x, so that du = –3dx, or dx = –⅓du. • Substitute to express the integrand in terms of u: • Evaluate the integral: • Replaceu by –3x to find the solution: Example 4, page 414

47. Examples • Find Solution • Let u = 3x2+ 1, so that du = 6xdx, or xdx = du. • Substitute to express the integrand in terms of u: • Evaluate the integral: • Replaceu by 3x2+ 1 to find the solution: Example 5, page 415

48. Examples • Find Solution • Let u = ln x, so that du = 1/xdx, or dx/x = du. • Substitute to express the integrand in terms of u: • Evaluate the integral: • Replaceu by ln x to find the solution: Example 6, page 415

49. 6.3 Area and the Definite Integral

50. The Area Under the Graph of a Function • Let f be a nonnegative continuous function on [a, b]. Then, the area of the region under the graph of f is where x1, x2, … , xn are arbitrary points in the n subintervals of [a, b] of equal width Dx = (b – a)/n.