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Chapter 5: Antidifferentiation

A. Antiderivatives

  • The antiderivative or indefinite integral of a function f(x) is a function F(x) whose derivative is f(x).

  • Since the derivative of a constant equals zero, the antiderivative of f(x) is not unique; that is, if F(x) is an integral of f(x), then so is F(x) + C, where C is any constant.

  • The arbitrary constant C is called the constant of integration.

  • The indefinite integral of f(x) is written as ∫ f(x) dx; thus

  • The function f(x) is called the integrand.

  • The Mean Value Theorem can be used to show that, if two functions have the same derivative on an interval, then they differ at most by a constant; that is, if

    then

    F(x) − G(x) = C(C is a constant)

B. Basic Formulas

15. 16.

17.

C. Integration by Partial Fractions

  • The method of partial fractions makes it possible to express a rational function f(x)/g(x) as a sum of simpler fractions.

  • Here f(x) and g (x) are real polynomials in x and it is assumed that f(x)/g(x) is a proper fraction; that is, that f(x) is of lower degree than g(x).

  • If not, we divide f(x) by g(x) to express the given rational function as the sum of a polynomial and a proper rational function. Thus,

where the fraction on the right is proper

  • Theoretically, every real polynomial can be expressed as a product of (powers of) real linear factors and (powers of) real quadratic factors.

  • In the following, the capital letters denote constants to be determined.

  • We consider only nonrepeating linear factors. For each distinct linear factor (xa) of g(x) we set up one partial fraction of the type A/xa.

  • The techniques for determining the unknown constants are illustrated in the following examples.

Example:

Find ∫ x cos x dx.

SOLUTION:

We let u = x and dv = cos x dx.

Then du = dx and v = sin x. Thus, the Parts Formula yields

x cos x dx = x sin x − ∫ sin x dx = x sin x + cos x + C

The Tic-Tac-Toe Method

This method of integrating is extremely useful when repeated integration by parts is necessary. To integrate ∫ u(x)v(x) dx, we construct a table as follows:

  • Here the column at the left contains the successive derivatives of u(x).

  • The column at the right contains the successive antiderivatives of v(x) (always with C = 0); that is, v(x) is the antiderivative of v(x), v2(x) is the antiderivative of v1(x), and so on.

  • The diagonal arrows join the pairs of factors whose products form the successive terms of the desired integral; above each arrow is the sign of that term. By the tic-tac-toe method,

u(x)v(x)dx = u(x)v1(x) − u′(x)v2(x) + u″(x)v3(x) − u″′(x)v4(x) + . . . .

Example:

To integrate ∫ x4 cos x dx by the tic-tac-toe method, we let u(x) = x4 and v(x) = cos x, and get the following table:

E. Applications of Antiderivatives: Differential Equations

The following examples show how we use given conditions to determine constants of integration.

Example :

Find a curve whose slope at each point (x,y) equals the reciprocal of the x-value if the curve contains the point (e,−3).

SOLUTION:

M

Chapter 5: Antidifferentiation

A. Antiderivatives

  • The antiderivative or indefinite integral of a function f(x) is a function F(x) whose derivative is f(x).

  • Since the derivative of a constant equals zero, the antiderivative of f(x) is not unique; that is, if F(x) is an integral of f(x), then so is F(x) + C, where C is any constant.

  • The arbitrary constant C is called the constant of integration.

  • The indefinite integral of f(x) is written as ∫ f(x) dx; thus

  • The function f(x) is called the integrand.

  • The Mean Value Theorem can be used to show that, if two functions have the same derivative on an interval, then they differ at most by a constant; that is, if

    then

    F(x) − G(x) = C(C is a constant)

B. Basic Formulas

15. 16.

17.

C. Integration by Partial Fractions

  • The method of partial fractions makes it possible to express a rational function f(x)/g(x) as a sum of simpler fractions.

  • Here f(x) and g (x) are real polynomials in x and it is assumed that f(x)/g(x) is a proper fraction; that is, that f(x) is of lower degree than g(x).

  • If not, we divide f(x) by g(x) to express the given rational function as the sum of a polynomial and a proper rational function. Thus,

where the fraction on the right is proper

  • Theoretically, every real polynomial can be expressed as a product of (powers of) real linear factors and (powers of) real quadratic factors.

  • In the following, the capital letters denote constants to be determined.

  • We consider only nonrepeating linear factors. For each distinct linear factor (xa) of g(x) we set up one partial fraction of the type A/xa.

  • The techniques for determining the unknown constants are illustrated in the following examples.

Example:

Find ∫ x cos x dx.

SOLUTION:

We let u = x and dv = cos x dx.

Then du = dx and v = sin x. Thus, the Parts Formula yields

x cos x dx = x sin x − ∫ sin x dx = x sin x + cos x + C

The Tic-Tac-Toe Method

This method of integrating is extremely useful when repeated integration by parts is necessary. To integrate ∫ u(x)v(x) dx, we construct a table as follows:

  • Here the column at the left contains the successive derivatives of u(x).

  • The column at the right contains the successive antiderivatives of v(x) (always with C = 0); that is, v(x) is the antiderivative of v(x), v2(x) is the antiderivative of v1(x), and so on.

  • The diagonal arrows join the pairs of factors whose products form the successive terms of the desired integral; above each arrow is the sign of that term. By the tic-tac-toe method,

u(x)v(x)dx = u(x)v1(x) − u′(x)v2(x) + u″(x)v3(x) − u″′(x)v4(x) + . . . .

Example:

To integrate ∫ x4 cos x dx by the tic-tac-toe method, we let u(x) = x4 and v(x) = cos x, and get the following table:

E. Applications of Antiderivatives: Differential Equations

The following examples show how we use given conditions to determine constants of integration.

Example :

Find a curve whose slope at each point (x,y) equals the reciprocal of the x-value if the curve contains the point (e,−3).

SOLUTION:

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