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Antiderivatives and the Rules of Integration Integration by Substitution

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. 6.1.

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Antiderivatives and the Rules of Integration Integration by Substitution

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  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.

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