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Section 4.8 Antiderivatives

Definition 4.8.1. Antiderivative.

A function \(F\) is an antiderivative of \(f\) on an interval \(I\) if \(F'(x)=f(x)\) for all \(x\) in \(I\text{.}\)

Example 4.8.2.

Find antiderivatives for:

  1. \(\displaystyle f(x)=3x^2\)

  2. \(\displaystyle g(x)=\cos x\)

  3. \(h(x)= \frac{1}{x}\) (contrast the natural domain and the domain \((0,\infty)\))

Answer.
  1. \(\displaystyle x^3\)

  2. \(\displaystyle \sin x\)

  3. \(\displaystyle \ln |x|\)

Example 4.8.4.

Find the general antiderivative of \(f(x)=xe^{x^2}\text{.}\)

Answer.

\(\frac12 e^{x^2} + C\)

Example 4.8.5.

Find an antiderivative \(F\) of \(f(x)=\sin(4x)\) that satisfies \(F(0)=-1\text{.}\)

Answer.

\(F(x) = -\frac14 \cos (4x) -\frac34\)

Subsection 4.8.1 Indefinite integrals

Definition 4.8.6. Indefinite integral.

The collection of all antiderivatives of a function \(f(x)\) is called the indefinite integral of \(f\) with respect to \(x\) and is denoted

\begin{equation*} \int f(x)\, dx \end{equation*}

The function \(f(x)\) is called the integrand and the differential \(dx\) indicates that \(x\) is the variable of integration.

Example 4.8.8.

  1. Evaluate \(\int x^3+2x+3\, dx\)

  2. Evaluate \(\int t^{-2/3} - \frac{5}{t}\, dt\)

  3. Evaluate \(\int \frac{3}{4x^2+1}\, dx\)

Answer.
  1. \(\displaystyle \frac14 x^4 + x^2 + 3x + C\)

  2. \(\displaystyle 3t^{1/3} - 5\ln |t| + C\)

  3. \(\displaystyle \frac32\tan^{-1} (2x) + C\)

Subsection 4.8.2 Initial value problems and differential equations

Definition 4.8.9. Differential equation.

A differential equation, loosely, is an equation involving the derivative of an unknown function.

Example 4.8.10.

\(\displaystyle\frac{dy}{dx} = x^2\) is a differential equation. The antiderivative \(y=\frac{1}{3}x^3+C\) is the general solution.

Adding a specific value, for example, \(y(0)=5\text{,}\) makes the problem into an initial value problem, for which \(y=\frac{1}{3}x^3+5\) is the particular solution.

Example 4.8.11.

A rocket lifts off with constant acceleration of \(20 \text{m}/\text{sec}^2\text{.}\) How fast will it be going 1 minute later?

Example 4.8.12.

Illinois requires motorcycles to be able to stop from \(44 \text{feet}/\text{sec}\) to 0 in 45 feet. What constant deceleration is required to achieve that stopping distance?

Solution.

\(\frac{d^2s}{dt^2} = -k\) for an unknown \(k\)

Integrating: \(\frac{ds}{dt} = -kt+C\text{;}\) since \(v(0)=44\text{,}\) we have

\begin{equation*} \frac{ds}{dt} = -kt+44 \end{equation*}

Integrating: \(s = -\frac{1}{2}kt^2 + 44t + C\text{;}\) since \(s(0)=0\text{,}\) we have

\begin{equation*} s = -\frac{1}{2}kt^2 + 44t \end{equation*}

The time to stop is the solution of \(-kt+44=0\text{,}\) so \(t_{\text{stop}}=\frac{44}{k}\text{.}\)

So, \(45 = -\frac{1}{2}k\left(\frac{44}{k}\right)^2 + 44 \left(\frac{44}{k}\right)\) yields \(k=\frac{90}{44^2}\text{.}\)