Section 4.1 Antiderivatives and Indefinite Integrals
Now we want to, in a sense, reverse the process of differentiation. In other words, given a function \(f(x)\text{,}\) we want to find another function \(g(x)\) such that \(g'(x) = f(x)\text{.}\) In this case, \(g(x)\) is called an antiderivative of \(f(x)\text{.}\) Observe that if \(g(x)\) is an antiderivative of \(f(x)\text{,}\) then so is \(g(x) + c\) for every constant \(c\text{.}\) Indeed: \(\dfrac{d}{dx}\left(g(x) + c \right) = g'(x) + 0 = g'(x) = f(x)\text{.}\)
For example, if \(f(x) = 2x\) then \(g(x) = x^2\) is an antiderivative of \(f(x)\) since \(g'(x) = 2x = f(x)\text{.}\) Notice that \(x^2 + 1\text{,}\) \(x^2 - 6\text{,}\) \(x^2 + \dfrac{2}{3}\) are also antiderivatives of \(f(x) = 2x\text{.}\) So, the most general antiderivative of \(f(x) = 2x\) is \(g(x) = x^2 + c\) where \(c\) is an arbitrary constant.
The indefinite integral (which looks like this: \(\int\)) is the notation used for antiderivatives. Let \(\) be an arbitrary constant. Then
\begin{equation*}
\int f(x) \: dx = g(x) + c
\end{equation*}
is read "the antiderivative (or integral) of \(f(x)\) equals \(g(x) + c\)" and it means \(g'(x) = f(x)\text{.}\) The \(dx\) will indicate the variable we are integrating with respect to (\(x\) in this case). The process of finding antiderivatives may be referred to as integration and \(f(x)\) is called the integrand. We can now express our above example as follows:
\begin{equation*}
\int 2x \: dx = x^2 + c.
\end{equation*}
We will make repeated use of the following rule find antiderivatives:
Fact 4.1.1.
The Power Rule For Antiderivatives: For every real number \(n \neq -1\)
\begin{equation*}
\int x^n \: dx = \dfrac{1}{n+1}x^{n+1} + c
\end{equation*}
Indeed: \(\dfrac{d}{dx}\left[\dfrac{1}{n+1}x^{n+1}\right] = \dfrac{n+1}{n+1}x^{(n+1) - 1} = x^n.\)
Example 4.1.2.
Find the following antiderivatives using Fact 4.1.1. Identify the value of \(n\) to start.
\(\displaystyle \int x \: dx \) Solution.With
\(n = 1 \)
\begin{equation*}
\displaystyle \int x \: dx = \dfrac{1}{1+1}x^{1+1} + c = \dfrac{1}{2}x^{2} + c
\end{equation*}
\(\displaystyle \int x^3 \: dx \) Solution.With
\(n = 3 \)
\begin{equation*}
\displaystyle \int x^3 \: dx = \dfrac{1}{3+1}x^{3+1} + c = \dfrac{1}{4}x^{4} + c
\end{equation*}
\(\displaystyle \int \dfrac{1}{x^5} \: dx \) Solution.With
\(n = -5 \)
\begin{align*}
\displaystyle \int \dfrac{1}{x^5} \: dx \amp = \int x^{-5} \: dx = \dfrac{1}{-5+1}x^{-5+1} + c \\
\amp = -\dfrac{1}{4}x^{-4} + c = -\dfrac{1}{4x^4} + c
\end{align*}
\(\displaystyle \int \sqrt[3]{x} \: dx \) Solution.With
\(n = \dfrac{1}{3} \)
\begin{align*}
\displaystyle \int \sqrt[3]{x} \: dx \amp = \int x^{1/3} \: dx = \dfrac{1}{1/3+1}x^{1/3+1} + c \\
\amp = \dfrac{3}{4}x^{4/3} + c = \dfrac{3}{4}\sqrt[3]{x^4} + c
\end{align*}
\(\displaystyle \int \: dx \) Solution.With
\(n = 0 \)
\begin{equation*}
\displaystyle \int \: dx = \int 1 \: dx = \int x^0 \: dx = \dfrac{1}{0+1}x^{0+1} + c = x+c
\end{equation*}
\(\displaystyle \int \dfrac{1}{x} \: dx \) Solution.Notice that
\(n= -1 \text{:}\)
\begin{equation*}
\displaystyle \int \dfrac{1}{x} \: dx = \int x^{-1} \: dx
\end{equation*}
So,
Fact 4.1.1 CANNOT be used.
Fact 4.1.3.
Properties of Antiderivatives: Suppose \(f\) and \(g\) have antiderivatives and \(k\) is a real number. Then:
Example 4.1.4.
Find the following antiderivatives using Fact 4.1.1 and Fact 4.1.3.
\(\displaystyle \int 3 \: dx \) Solution.
\begin{equation*}
\displaystyle \int 3 \: dx = 3 \int \: dx = 3x + c
\end{equation*}
\(\displaystyle \int (x^2 + x + 1) \: dx \) Solution.
\begin{align*}
\displaystyle \int (x^2 + x + 1) \: dx \amp = \int x^2 \: dx + \int x \: dx + \int 1 \: dx \\
\\
\amp = \dfrac{1}{3}x^3 + \dfrac{1}{2}x^2 + x + c
\end{align*}
\(\displaystyle \int \left( \dfrac{4\sqrt{x}}{5} - \dfrac{5}{4\sqrt{x}} \right) \: dx \) Solution.
\begin{align*}
\displaystyle \int \left( \dfrac{4\sqrt{x}}{5} - \dfrac{5}{4\sqrt{x}} \right) \: dx \amp = \int \left( \dfrac{4}{5}x^{1/2} - \dfrac{5}{4}x^{-1/2} \right) \: dx \\
\\
\amp = \dfrac{4}{5}\int x^{1/2} \: dx - \dfrac{5}{4} \int x^{-1/2} \: dx \\
\\
\amp = \dfrac{4}{5} \cdot \dfrac{2}{3} x^{3/2} - \dfrac{5}{4} \cdot 2 x^{1/2} + c \\
\\
\amp = \dfrac{8}{15}\sqrt{x^3} - \dfrac{5}{2}\sqrt{x} + c
\end{align*}
\(\displaystyle \int \dfrac{x^2 - 4x}{2x} \: dx \) Solution.
\begin{align*}
\displaystyle \int \dfrac{x^2 - 4x}{2x} \: dx \amp = \int \left( \dfrac{x^2}{2x} - \dfrac{4x}{2x} \right) \: dx = \int \left( \dfrac{1}{2}x - 2 \right) \: dx \\
\\
\amp = \dfrac{1}{2} \int x \: dx - 2 \int \: dx = \dfrac{1}{2} \cdot \dfrac{1}{2}x^2 - 2x + c \\
\\
\amp = \dfrac{1}{4}x^2 - 2x + c
\end{align*}
\(\displaystyle \int 2x(x+1)^2 \: dx \) Solution.
\begin{align*}
\displaystyle \int 2x(x+1)^2 \: dx \amp = \int 2x(x^2 + 2x + 1) \: dx = \int \left( 2x^3 + 4x^2 + 2x \right) \: dx \\
\\
\amp = 2\int x^3 \: dx + 4 \int x^2 \: dx + 2 \int x \: dx \\
\\
\amp = 2 \cdot \dfrac{1}{4}x^4 + 4 \cdot \dfrac{1}{3}x^3 + 2 \cdot \dfrac{1}{2}x^2 + c \\
\\
\amp = \dfrac{1}{2}x^4 + \dfrac{4}{3}x^3 + x^2 + c
\end{align*}
WARNING: In general,
\(\displaystyle \displaystyle \int \dfrac{f(x)}{g(x)} \: dx \neq \dfrac{\int f(x) \: dx}{\int g(x) \: dx } \)
\(\displaystyle \displaystyle \int f(x) \cdot g(x) \: dx \neq \left( \int f(x) \: dx \right) \left( \int g(x) \: dx \right) \)
In cases where the integrand is the quotient or product of functions, you will often need to do some algebraic simplification before you find the antiderivative.
Example 4.1.5.
C. Farbman Furniture (CFF) is a small furniture outlet that designs, manufactures, and sells home furnishings. The marginal cost function for their retro seven drawer dresser is is given by
\begin{equation*}
MC(x) = 92.47
\end{equation*}
dollars per dresser, where \(x\) is the number of dressers manufactured and sold. Determine Farbman's cost function given that the cost to manufacture 8 retro seven drawer dressers is $1,337.28. Find Farbman's fixed costs. Solution.We are given
So, up to an arbitrary constant
\(c\) (which we will have to determine),
\begin{equation*}
C(x) = \int MC(x) \: dx = \int 92.47 \: dx = 92.47x + c
\end{equation*}
To find
\(c\text{,}\) we have
\begin{align*}
C(8) \amp = 92.47(8) + c \\
\amp = 739.76 + c
\end{align*}
Using the given info (see second bullet point above),
\begin{equation*}
739.76 + c = 1,337.28
\end{equation*}
so that
\(c = 597.52\) and we see that Farbman's cost function is
\begin{equation*}
C(x) = 92.47x + 597.52
\end{equation*}
from which we conclude Farbman's fixed costs are $597.52.
Example 4.1.6.
Sanalac is a progressive company with a small but prestigious group of clients. The annual marginal profit from the sale of their rest stop supplies is given by
\begin{equation*}
MP(t) = -.75t^2 +12t + 55
\end{equation*}
hundred dollars per year, where \(t = 0\) corresponds to the year 2001. Find Sanalac's profit function given that their profit in 2011 was $78,300. Use your answer to determine Sanalac's revenue in 2001 if their costs for that year were $15,000. Solution.We are given
So, up to an arbitrary constant
\(c\) (which we will have to determine),
\begin{equation*}
P(t) = \int MP(t) \: dt = \int (-.75t^2 +12t + 55) \: dt = -.25t^3 + 6t^2 +55t + c
\end{equation*}
To find
\(c\text{,}\) we have
\begin{align*}
P(10) \amp = -.25(10^3) + 6(10^2) + 55(10) + c \\
\amp = -250 + 600 + 550 + c \\
\amp= 900 + c
\end{align*}
Using the given info (see second bullet point above),
\begin{equation*}
900 + c = 783
\end{equation*}
so that
\(c = -117\) and we see that Sanalac's annual profit function in hundreds of dollars is
\begin{equation*}
P(t) = -.25t^3 + 6t^2 + 55t - 117.
\end{equation*}
Notice that in 2001 Sanalac experienced a loss of $11,700 (
\(P(0) = -117\)). Let
\(C(t)\) and
\(R(t)\) represent Sanalac's annual cost and revenue functions respectively. We now want to find
\(R(0)\) given that
\(C(0) = 150\text{.}\) We solve
\begin{align*}
P(0) \amp = R(0) - C(0) \\
-117 \amp = R(0) - 150 \\
R(0) \amp = 150 - 117\\
R(0) \amp = 33.
\end{align*}
Sanalac's revenue in 2001 was $3,300.
