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Barnett/Ziegler/Byleen Business Calculus 11e1 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.

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Barnett/Ziegler/Byleen Business Calculus 11e2 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).

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Barnett/Ziegler/Byleen Business Calculus 11e3 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 x 2.

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Barnett/Ziegler/Byleen Business Calculus 11e4 Examples Find a function that has a derivative of 2x. Answer: x 2, since d/dx (x 2 ) = 2x. Find a function that has a derivative of x. Answer: x 2 /2, since d/dx (x 2 /2) = x. Find a function that has a derivative of x 2. Answer: x 3 /3, since d/dx (x 3 /3) = x 2.

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Barnett/Ziegler/Byleen Business Calculus 11e5 Examples (continued) Find a function that has a derivative of 2x. Answer: We already know that x 2 is such a function. Other answers are x 2 + 2 or x 2 – 5. The above functions are all antiderivatives of 2x. Note that the antiderivative is not unique.

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Barnett/Ziegler/Byleen Business Calculus 11e6 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.

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Barnett/Ziegler/Byleen Business Calculus 11e7 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. 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

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Barnett/Ziegler/Byleen Business Calculus 11e8 Find the indefinite integral of x 2. Example

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Barnett/Ziegler/Byleen Business Calculus 11e9 Find the indefinite integral of x 2. Answer:, because Example

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Barnett/Ziegler/Byleen Business Calculus 11e10 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 factor cannot be moved across an integral sign.

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Barnett/Ziegler/Byleen Business Calculus 11e11 Examples for Power Rule 444 dx = 444x + C (power rule with n = 0) x 3 dx = x 4 /4 + C (n = 3) 5 x -3 dx = -(5/2) x -2 + C (n = -3) x 2/3 dx = (3/5) x 5/3 + C (n = 2/3) (x 4 + x + x 1/2 + 1 + x -1/2 ) dx = x 5 /5 + x 2 /2 + (2/3) x 3/2 + x + 2x 1/2 + C But you cannot apply the power rule for n = -1: x -1 dx is not x 0 /0 + C (which is undefined). The integral of x -1 is the natural logarithm.

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Barnett/Ziegler/Byleen Business Calculus 11e12 More Examples 4 e x dx = 4 e x + C 2 x -1 dx = 2 ln |x| + C

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Barnett/Ziegler/Byleen Business Calculus 11e13 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.

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Barnett/Ziegler/Byleen Business Calculus 11e14 Application (continued ) Noting that f (20) = 80.3, we calculate 80.3 = (0.024)(20 2 ) + (0.95)(20) + C 80.3 = 28.6 + C C = 51.7 f (x) = 0.024 x 2 + 0.95 x + 51.7 We need the integral of f ’(x), or

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Barnett/Ziegler/Byleen Business Calculus 11e15 Application (continued ) The years 1970 and 2000 correspond to x = 0 and x = 30. f (0) = (0.024)(0 2 ) + (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.

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