We have previously studied derivatives. Given a function relationship between two variables \(x \overset{f}{\mapsto} Q\), the derivative \(f'\) is the function relating \(x\) to the rate of change \(\frac{dQ}{dx}\). Because \(f'\) is itself a function, we can compute the derivative of \(f'\) to get \(f''\). This process can repeat indefinitely.

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Example6.6.1

Consider \(f(x) = x^4+2x^2-3x\). There is a sequence of functions corresponding to the derivatives: \begin{align*}
f(x) &= x^4+2x^2-3x, \\
f'(x) &= 4x^3+4x-3, \\
f''(x) &= 12x^2+4, \\
f^{(3)}(x) &= 24x, \\
f^{(4)}(x) &= 24, \\
f^{(5)}(x) &= 0, \\
f^{(6)}(x) &= 0.
\end{align*} This pattern continues with \(f^{(n)}(x) = 0\) for \(n=5,6,7,\ldots\).

As this example illustrated, given a function we can find its derivative. That is, differentiation is an operation that takes a function as input and produces a new function as its output: \begin{equation*} f(x) \overset{\frac{d}{dx}}{\mapsto} f'(x). \end{equation*}

One of the major themes of mathematics is the idea of inverse operations. Is there an inverse operation to differentiation? Yes. It is called *antidifferentiation*. However, differentiation is not one-to-one. Because the derivative of any constant is zero, we will never know exactly which constant an original function had when we look at its derivative. Infinitely many different functions have the same derivative. We call all such functions *antiderivatives*.

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Definition6.6.2Antiderivatives

Given a function \(f(x)\), we say that \(F(x)\) is an *antiderivative* of \(f(x)\) if \(f(x)\) is the derivative of \(F(x)\). That is, \(F'(x)=f(x)\).

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Example6.6.3

Compare \begin{equation*}\frac{d}{dx}[x^2+3x-1] = 2x+3\end{equation*} with \begin{equation*}\frac{d}{dx}[x^2+3x+4] = 2x+3.\end{equation*} Both functions have the same derivative. So \(x^2+3x-1\) and \(x^2+3x+4\) are both antiderivatives of \(2x+3\). In fact, \(f(x)=x^2+3x+C\) will be an antiderivative for any other constant \(C\).

The next theorem tells us that two different antiderivatives can only be different by a constant value on any intervals on which the function is defined.

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Theorem6.6.4

Given a function \(f(x)\) that is defined on any interval \(I\). Suppose that \(F(x)\) and \(G(x)\) are both antiderivatives of \(f(x)\) so that for all \(x \in I\) we have \begin{equation*}F'(x)=G'(x)=f(x).\end{equation*} Then there is a constant \(C\) so that for all \(x \in I\), \(G(x) = F(x)+C\).

##### Proof

Define \(h(x)=G(x)-F(x)\). Because \(F\) and \(G\) are antiderivatives, we know \(F'(x)=G'(x)=f(x)\) so that \(h'(x)=0\) for all values \(x \in I\). As a consequence of the Mean Value Theorem, Theorem 6.3.1 guarantees that \(h(x)\) is constant, say \(h(x)=C\), for \(x \in I\). Then \(G(x) = F(x)+C\) on \(I\).

Knowing just one antiderivative allows us to determine all possible antiderivatives by adding some constant. Suppose \(F(x)\) is an antiderivative of \(f(x)\). Then any other antiderivative must be \(F(x)+C\) for some constant \(C\). If we leave the constant as an unspecified parameter, we call this the *general antiderivative*. Graphically, different antiderivatives correspond to a vertical translation of the graph. That is, all antiderivatives have the same graph shifted up or down relative to one another.

The Fundamental Theorem of Calculus will show that every accumulation function is an antiderivative of the rate function in the integral. Consequently, there is a close connection between antiderivatives and integrals. This motivates the standard notation for finding antiderivatives.

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Definition6.6.5Indefinite Integrals

Given a function \(f(x)\), the *indefinite integral* of \(f(x)\) with respect to \(x\), written \(\displaystyle \int f(x) \, dx\), is the general antiderivative of \(f(x)\). That is, if \(F(x)\) is any antiderivative such that \(F'(x)=f(x)\), then \begin{equation*} \int f(x)\, dx = F(x) + C.\end{equation*}

For the most part, finding antiderivatives corresponds to recognizing how a function might have been computed as a derivative.

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Example6.6.6

Find \(\displaystyle \int 4x^3 - 2e^{2x} \, dx\).

SolutionWe are looking for a function \(F(x)\) for which \(F'(x)=4x^3 - 2e^{2x}\). From experience computing derivatives, we know \begin{gather*}
\frac{d}{dx}[x^4] = 4x^3, \\
\frac{d}{dx}[e^{2x}] = 2e^{2x}.
\end{gather*} We expect \(F(x) = x^4-e^{2x}\). We verify by differentiation: \begin{equation*} F'(x) = \frac{d}{dx}[x^4-e^{2x}] = 4x^3 - 2e^{2x}.\end{equation*} This verifies that \(F(x)\) is an antiderivative of \(4x^3-2e^{2x}\). The general antiderivative is written \begin{equation*} \int 4x^3-2e^{2x}\,dx = x^4 - e^{2x} + C.\end{equation*}