I - Antiderivatives Lesson

Antiderivatives

The process of recovering a function F(x) from one of its known values and its rate of change (derivative) f(x) is of paramount importance in applications such as predicting future behavior or output from previous behavior or input. A function F is an antiderivative of f on an interval I if F'(x) = f(x) for all x in I. A formula that produces all of the functions that could possibly have f as a derivative on an interval I is the general antiderivative of f and is denoted as F(x) + C, where C is an arbitrary constant. By assigning specific values to the constant C, a family of functions whose graphs are vertical translations of one another results.

graph of family of functions with vertical translations of one another results

View the interactive applet relating the graph of a function and its antiderivative by CLICKING HERE. Links to an external site.

Many antiderivatives are found by reversing familiar derivative formulas. Once a particular antiderivative F is known, then the general antiderivative is F plus an arbitrary constant. Consider the examples in the table below.

Function	Reversed Derivative Formula	General Antiderivative
cos x	$LaTeX: \frac{d}{dx}\left(\sin x\right)=\cos x$ $\frac{d}{dx}\left(\sin x\right)=\cos x$	$LaTeX: \sin x+C$ $\sin x+C$
sin x	$LaTeX: \frac{d}{dx}\left(-\cos x\right)=\sin x$ $\frac{d}{dx}\left(-\cos x\right)=\sin x$	$LaTeX: -\cos x+C$ $-\cos x+C$
$3x^2$	$LaTeX: \frac{d}{dx}\left(x^3\right)=3x^2$ $\frac{d}{dx}\left(x^3\right)=3x^2$	$x^3+C$
$LaTeX: \frac{1}{x^2}$ $\frac{1}{x^2}$	$LaTeX: \frac{d}{dx}\left(-\frac{1}{3}\right)=\left(\frac{1}{x^2}\right)$ $\frac{d}{dx}\left(-\frac{1}{3}\right)=\left(\frac{1}{x^2}\right)$	$LaTeX: -\frac{1}{x}+C$ $-\frac{1}{x}+C$

View the presentation below on finding antiderivatives.

Indefinite Integration and Notation

If the function f(x) is a derivative, then the set of all antiderivatives of f is the indefinite integral of f, and is denoted by equation image indicator $LaTeX: \int f\left(x\right)dx=F\left(x\right)+C$ $\int f\left(x\right)dx=F\left(x\right)+C$ . The symbol ∫ is called an integral sign; the function f is the integrand of the integral; the differential dx identifies x as the variable of integration; and C is the constant of integration. The expression equation image indicator $LaTeX: \int f\left(x\right)dx$ $\int f\left(x\right)dx$ is read as "the indefinite integral (or antiderivative) of f with respect to x is F(x) + C." The term indefinite integral is a synonym for antiderivative and represents a function or family of functions.

View the presentation below on an introduction to indefinite integrals.

Basic Integration Rules

It is often instructive to reinforce the inverse relationship between differentiation and integration more formally by substituting F'(x) for f(x) in the indefinite integration definition equation image indicator $LaTeX: \int f\left(x\right)dx=F\left(x\right)+C$ $\int f\left(x\right)dx=F\left(x\right)+C$ . The resulting equation $LaTeX: \int F'\left(x\right)dx=F\left(x\right)+C$ $\int F'\left(x\right)dx=F\left(x\right)+C$ clearly establishes integration as the inverse of differentiation. Similarly, conceiving of differentiation as the inverse of integration is symbolized by equation image indicator $LaTeX: \frac{d}{dx}\left[\int f\left(x\right)dx\right]=f\left(x\right)$ $\frac{d}{dx}\left[\int f\left(x\right)dx\right]=f\left(x\right)$ . It is then possible to obtain integration formulas/rules directly from differentiation formulas. An example of how to apply the integration rules follows.

text annotation indicator $LaTeX: \int kdx=kx+C\:\:\:\:\:\:\:\:\:\:\:\int x^ndx=\frac{x^{n+1}}{n+1}+C\left(n\ne-1\right)\\ \int\sin xdx=-\cos x + C\:\:\:\:\:\int \cos xdx=\sin x+C\\ \int \sec^2xdx=\tan x+C\:\:\:\:\:\int \csc^2 xdx=-\cot x +C\\ \int \sec x \tan xdx=\sec x+C\:\:\:\:\: \int \csc x \cot xdx=-\csc x+C$ $\int kdx=kx+C\:\:\:\:\:\:\:\:\:\:\:\int x^ndx=\frac{x^{n+1}}{n+1}+C\left(n\ne-1\right)\\ \int\sin xdx=-\cos x + C\:\:\:\:\:\int \cos xdx=\sin x+C\\ \int \sec^2xdx=\tan x+C\:\:\:\:\:\int \csc^2 xdx=-\cot x +C\\ \int \sec x \tan xdx=\sec x+C\:\:\:\:\: \int \csc x \cot xdx=-\csc x+C$

View the indefinite integral presentation below.

CLICK HERE to view the indefinite integral widget which provides self-check opportunities with a variety of self-entered problems Links to an external site.

Differential Equation

An equation containing one or more derivatives of a function is a differential equation. Like algebraic equations that describe relations among varying quantities, differential equations describe relations among changing quantities and the rates at which they change. Solving an equation such as equation image indicator $LaTeX: \frac{dy}{dx}=f\left(x\right)$ $\frac{dy}{dx}=f\left(x\right)$ produces functions $LaTeX: y=F\left(x\right)$ $y=F\left(x\right)$ that make the equation true. When solving a differential equation of the form $LaTeX: \frac{dy}{dx}=f\left(x\right)$ $\frac{dy}{dx}=f\left(x\right)$ , it is convenient to write it using its equivalent differential form dy = f(x)dx. The general antiderivative y = F(x) + C of the function f(x) is called the general solution of the differential equation equation image indicator $LaTeX: \frac{dy}{dx}=f\left(x\right)$ $\frac{dy}{dx}=f\left(x\right)$ . Explore the interactive applet Links to an external site.that illustrates the graphical solutions to a differential equation by CLICKING HERE. Links to an external site.

Initial-Value Problems

In many applications of integration, you need to find a particular solution of a differential equation that satisfies a condition of the form y(x₀) = y₀ . If you know the value of y = F(x) for one value of x, namely x₀, then it is possible to find a particular solution of an initial-value problem that satisfies an initial condition. For example, given the differential equation equation image indicator $LaTeX: \frac{dy}{dx}=4x^3-1$ $\frac{dy}{dx}=4x^3-1$ that passes through the point (1, -4), it is possible to find the equation of the particular curve satisfying these conditions.

equation image indicator $LaTeX: F\left(x\right)=x^4-x+C$ $F\left(x\right)=x^4-x+C$ (General solution)

F(1) = -4 (Initial condition)

equation image indicator $LaTeX: F\left(1\right)=x^4-x+C$ $F\left(1\right)=x^4-x+C$

equation image indicator $LaTeX: -4=\left(1\right)^4-1+C$ $-4=\left(1\right)^4-1+C$

equation image indicator LaTeX: -4=C $-4=C$

equation image indicator $LaTeX: F\left(x\right)=x^4-x-4$ $F\left(x\right)=x^4-x-4$ (Particular solution)

Rectilinear Motion

Antidifferentiation is particularly useful in analyzing rectilinear motion, the motion of an object moving in a straight line. Recall that if an object has position s = f(t), then the velocity function is v(t) = s'(t). This means that the position function is an antiderivative of the velocity function. Likewise, the acceleration function is a(t) = v'(t), and thus the velocity function is an antiderivative of the acceleration. If the acceleration and the initial values s(0) and v(0) are known, then the position function can be found by antidifferentiating twice.

Consider a particle that moves in a straight line and has acceleration given by a(t) = t - 2 when s(0) = 1 and v(0)= 3. Find the position function s(t).

Since v'(t) = a(t) = t - 2, antidifferentiation gives v(t) = t²/2 - 2t + C.

v(0)= 0²/2 - 2(0) + C and since v(0) = 3, C = 3.

The velocity of the particle is v(t) = t²/2 - 2t + 3.

Since s'(t) = v(t) = t²/2 - 2t + 3, s is the antiderivative of v.

s(t) = t³/6 - t² + 3t + D

s(0) = 0³/6 - 0² + 3(0) + D and since s(0) = 1, D = 1.

The position function is s(t) = t³/6 - t² + 3t + 1.

Only by solving differential equations given initial conditions can many real-life phenomena be usefully described.

Antiderivatives Practice

Antiderivatives: Even More Problems!

Complete problems from your textbook and/or online resources as needed to ensure your complete understanding of antiderivatives and indefinite integration.

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