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Learning Objectives for Section 13.1 Antiderivatives and Indefinite Integrals. The student will be able to formulate problems involving antiderivatives. The student will be able to use the formulas and properties of antiderivatives and indefinite integrals.
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Learning Objectives for Section 13.1 Antiderivatives and Indefinite Integrals • The student will be able to formulate problems involving antiderivatives. • The student will be able to use the formulas and properties of antiderivatives and indefinite integrals. • The student will be able to solve applications using antiderivatives and indefinite integrals. Barnett/Ziegler/Byleen Business Calculus 11e
The Antiderivative Many operations in mathematics have inverses. For example, division is the inverse of multiplication. The inverse operation of finding a derivative, called the antiderivative, will now command our attention. A function F is an antiderivative of a function f if F ’(x) = f (x). Barnett/Ziegler/Byleen Business Calculus 11e
Examples Find a function that has a derivative of 2x. Find a function that has a derivative of x. Find a function that has a derivative of x2. Barnett/Ziegler/Byleen Business Calculus 11e
Examples Find a function that has a derivative of 2x. Answer:x2, since d/dx (x2) = 2x. Find a function that has a derivative of x. Answer:x2/2, since d/dx (x2/2) = x. Find a function that has a derivative of x2. Answer:x3/3, since d/dx (x3/3) = x2. Barnett/Ziegler/Byleen Business Calculus 11e
Examples(continued) Find a function that has a derivative of 2x. Answer: We already know that x2 is such a function. Other answers are x2 + 2 or x2 – 5. The above functions are all antiderivatives of 2x. Note that the antiderivative is not unique. Barnett/Ziegler/Byleen Business Calculus 11e
Uniqueness of Antiderivatives The following theorem says that antiderivatives are almost unique. Theorem 1: If a function has more than one antiderivative, then the antiderivatives differ by at most a constant. Barnett/Ziegler/Byleen Business Calculus 11e
Indefinite Integrals Let f (x) be a function. The family of all functions that are antiderivatives of f (x) is called the indefinite integral and has the symbol The symbol is called an integral sign, and the function f (x) is called the integrand. The symbol dx indicates that anti-differentiation is performed with respect to the variable x. By the previous theorem, if F(x) is any antiderivative of f, then The arbitrary constant C is called the constant of integration. Barnett/Ziegler/Byleen Business Calculus 11e
Example Find the indefinite integral of x2. Barnett/Ziegler/Byleen Business Calculus 11e
Example Find the indefinite integral of x2. Answer: , because Barnett/Ziegler/Byleen Business Calculus 11e
Indefinite Integral Formulas and Properties (power rule) It is important to note that property 4 states that a constant factor can be moved across an integral sign. A variable factorcannotbe moved across an integral sign. Barnett/Ziegler/Byleen Business Calculus 11e
Examples for Power Rule • 444 dx = 444x + C (power rule with n = 0) • x3dx = x4/4 + C (n = 3) • 5 x-3dx = -(5/2) x-2 + C (n = -3) • x2/3dx = (3/5) x5/3 + C (n = 2/3) • (x4 + x + x1/2 + 1 + x -1/2) dx = x5/5 + x2/2 + (2/3) x3/2 + x + 2x1/2 + C But you cannot apply the power rule for n = -1: x-1dx is not x0/0 + C (which is undefined). The integral of x-1 is the natural logarithm. Barnett/Ziegler/Byleen Business Calculus 11e
More Examples • 4 ex dx = 4 ex + C • 2 x-1dx = 2 ln |x| + C Barnett/Ziegler/Byleen Business Calculus 11e
Application In spite of the prediction of a paperless computerized office, paper and paperboard production in the United States has steadily increased. In 1990 the production was 80.3 million short tons, and since 1970 production has been growing at a rate given by f ’(x) = 0.048x + 0.95, where x is years after 1970. Find f (x), and the production levels in 1970 and 2000. Barnett/Ziegler/Byleen Business Calculus 11e
Application(continued ) We need the integral of f ’(x), or Noting that f (20) = 80.3, we calculate 80.3 = (0.024)(202) + (0.95)(20) + C 80.3 = 28.6 + C C = 51.7 f (x) = 0.024 x2 + 0.95 x + 51.7 Barnett/Ziegler/Byleen Business Calculus 11e
Application(continued ) The years 1970 and 2000 correspond to x = 0 and x = 30. f (0) = (0.024)(02) + (0.95)(0) + 51.7 = 51.7 f (30) = (0.024)(302) + (0.95)(30) + 51.7 = 101.8 The production was 51.7 short tons in 1970, and 101.8 short tons in 2000. Barnett/Ziegler/Byleen Business Calculus 11e