21. Antiderivatives, Areas and the FTC

b. Antiderivative Rules

In general, a derivative rule produces an antiderivative rule: \[ \text{If} \qquad \dfrac{d}{dx}[F(x)]=f(x) \qquad \text{then} \qquad F(x) \quad \text{is an antiderivative of} \quad f(x)\text{.} \]

1. Algebraic Operations

We have learned how to differentiate many kinds of functions including powers of \(x\), exponentials, logarithms, trig functions and inverse trig functions and how to combine these functions to produce other functions using the Sum, Constant Multiple, Product, Quotient, Extended Power and Chain Rules. All of these can be reversed to give antiderivative rules. These are shown in this table about algebraic operations and the ones on the next three pages about special functions. In each of the tables, \(F(x)\) is an antiderivative of \(f(x)\), \(P(x)\) is an antiderivative of \(p(x)\) and \(Q(x)\) is an antiderivative of \(q(x)\). You can check each line of the table by differentiating the quantity on the right to see you get the quantity on the left.

Antiderivative Rules for Algebraic Operations
If the Function is then the General Antiderivative is
Constant \(f(x)=0\) \(F(x)=C\)
Antiderivatives are only determined up to a constant.
Identity \(f(x)=1\) \(F(x)=x+C\)
Power \(f(x)=x^n\) \(F(x)=\dfrac{x^{n+1}}{n+1}+C\)
Sum \(f(x)=p(x)+q(x)\) \(F(x)=P(x)+Q(x)+C\)
Constant Multiple \(f(x)=c p(x)\) \(F(x)=c P(x)+C\)
Product \(f(x)=P(x) q(x)+Q(x) p(x)\) \(F(x)=P(x) Q(x)+C\)
Quotient \(f(x)=\dfrac{Q(x) p(x)- P(x) q(x)}{Q(x)^2}\) \(F(x)=\dfrac{P(x)}{Q(x)}+C\)
Extended Power \(^*\) \(f(x)=Q(x)^n q(x)\) \(F(x)=\dfrac{Q(x)^{n+1}}{n+1}+C\)
Chain \(f(x)=p(Q(x)) q(x)\) \(F(x)=P(Q(x))+C\)

\(^*\) The Extended Power Rule is the special case of the Chain Rule where \(p(u)=u^n\) and so \(P(u)=\dfrac{u^{n+1}}{n+1}\).

The Product Rule, Quotient Rule and Chain Rule, as stated here for antiderivatives, are not very useful. In a later chapter, we will see another reversal of the Chain Rule as Integration by Substitution and in Calculus 2 we will see a reversal of the Product Rule as Integration by Parts. There is no nice antiderivative rule corresponding to the Quotient Rule.

Find the general antiderivative of \(f(x)=12x^5-9x^2+4x+5\).

\(F=2x^6-3x^3+2x^2+5x+C\)

If \(f(x)=12x^5-9x^2+4x+5\), then by the Sum, Constant Multiple and Power Rules, the general antiderivative is: \[\begin{aligned} F(x)&=12\left(\dfrac{x^6}{6}\right)-9\left(\dfrac{x^3}{3}\right) +4\left(\dfrac{x^2}{2}\right)+5x+C \\ &=2x^6-3x^3+2x^2+5x+C \end{aligned}\]

We check by differentiating. If \(F=2x^6-3x^3+2x^2+5x\), then: \[ F'(x)=2\cdot6x^5-3\cdot3x^2+2\cdot2x+5=12x^5-9x^2+4x+5 \]

The Extended Power Rule is frequently used with trig functions:

Find the general antiderivative of \(g(x)=\tan^{11}x\sec^2x\).

\(G(x)=\dfrac{\tan^{12}x}{12}+C\)

Recall \(\dfrac{d}{dx}\tan x=\sec^2x\). So by the Extended Power Rule, the general antiderivative of \(g(x)=\tan^{11}x\sec^2x\) is: \[ G(x)=\dfrac{\tan^{12}x}{12}+C \]

We check by differentiating. If \(G(x)=\dfrac{\tan^{12}x}{12}\), then by the Chain Rule: \[ G'(x)=\dfrac{1}{12}12\tan^{11}x\dfrac{d}{dx}\tan x =\tan^{11}x\sec^2x \]

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