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  1. AP Calculus
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Glossary

C

Constant of Integration (+C)

Criticality: 3

An arbitrary constant added to the result of an indefinite integral. It represents the family of all possible antiderivatives of a function.

Example:

When integrating f(x)=2xf(x) = 2xf(x)=2x, the result is x2+Cx^2 + Cx2+C, where +C is the constant of integration.

L

Linear Factors

Criticality: 3

Factors of a polynomial that are of the first degree (e.g., $ax+b$). In partial fraction decomposition, the denominator is often factored into these terms.

Example:

For the denominator x2−5x+6x^2 - 5x + 6x2−5x+6, the linear factors are (x−2)(x-2)(x−2) and (x−3)(x-3)(x−3).

Long Division (for rational functions)

Criticality: 2

A polynomial division technique used when the degree of the numerator is greater than or equal to the degree of the denominator in a rational function. It simplifies the function into a polynomial plus a proper rational function.

Example:

If you need to integrate x3+1x−1\frac{x^3+1}{x-1}x−1x3+1​, you would first use long division to rewrite it as a polynomial plus a remainder term over x−1x-1x−1.

P

Partial Fraction Decomposition (PFD)

Criticality: 3

A technique used to break down a complex rational function into a sum of simpler fractions, making it easier to integrate. It's particularly useful when the denominator can be factored into linear terms.

Example:

To integrate 5x−1(x−1)(x+4)\frac{5x-1}{(x-1)(x+4)}(x−1)(x+4)5x−1​, you would use Partial Fraction Decomposition to rewrite it as Ax−1+Bx+4\frac{A}{x-1} + \frac{B}{x+4}x−1A​+x+4B​.

R

Rational Function

Criticality: 2

A function that can be expressed as the ratio of two polynomials, where the denominator is not zero. Integration of these functions often requires specific techniques like partial fraction decomposition.

Example:

The function f(x)=x2+3xx3−8f(x) = \frac{x^2 + 3x}{x^3 - 8}f(x)=x3−8x2+3x​ is a rational function because both the numerator and denominator are polynomials.

U

Undetermined Coefficients

Criticality: 3

A method used in partial fraction decomposition to find the unknown constants (A, B, C, etc.) in the numerators of the simpler fractions. This involves setting up and solving a system of equations.

Example:

After setting up 2x+1(x−1)(x+2)=Ax−1+Bx+2\frac{2x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}(x−1)(x+2)2x+1​=x−1A​+x+2B​, we use the method of undetermined coefficients by substituting specific x-values to solve for A and B.