# 5.1: The Indefinite Integral

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Derivatives appear in many physical phenomena, such as the motion of objects. Recall, for example, that given the position function \(s(t)\) of an object moving along a straight line at time \(t\), you could find the velocity \(v(t)=s'(t)\) and the acceleration \(a(t)=v'(t)\) of the object at time \(t\) by taking derivatives. Suppose the situation were reversed: given the velocity function how would you find the position function, or given the acceleration function how would you find the velocity function?

In this case calculating a derivative would not help, since the reverse process is needed: instead of differentiation you need a way of performing **antidifferentiation**, i.e. you would calculate an **antiderivative**.

Differentiation is relatively straightforward. You have learned the derivatives of many classes of functions (e.g. polynomials, trigonometric functions, exponential and logarithmic functions), and with the various rules for differentiation you can calculate derivatives of complicated expressions involving those functions (e.g. sums, powers, products, quotients). Antidifferentiation, however, is a different story.

To see some of the issues involved, consider a simple function like \(f(x)=2x\). Of course you know that \(\ddx(x^2) = 2x\), so it seems that \(F(x)=x^2\) is *the* antiderivative of \(f(x)=2x\). But is it the *only* antiderivative of \(f(x)\)? No. For example, if \(F(x)=x^2+1\) then \(F'(x)=2x=f(x)\), and so \(F(x)=x^2+1\) is another antiderivative of \(f(x)=2x\). Likewise, so is \(F(x)=x^2+2\). In fact, any function of the form \(F(x)=x^2 + C\), where \(C\) is some constant, is an antiderivative of \(f(x)=2x\).

Another potential issue is that functions of the form \(F(x)=x^2 + C\) are just the most *obvious* antiderivatives of \(f(x)=2x\). Could there be some other completely different function—one that cannot be simplified into the form \(x^2 + C\)—whose derivative also turns out to be \(f(x) =2x\)? The answer, luckily, is no:

To prove this, consider the function \(H(x) = F(x) - G(x)\), defined for all \(x\) in the common domain \(I\) of \(F\) and \(G\). Since \(F'(x) = G'(x) = f(x)\), then

\[H'(x) ~=~ F'(x) ~-~ G'(x) ~=~ f(x) ~-~ f(x) ~=~ 0 \nonumber \]

for all \(x\) in \(I\), so \(H(x)\) is a constant function on \(I\), as was shown in Section 4.4 on the Mean Value Theorem. Thus, there is a constant \(C\) such that

\[H(x) ~=~ C \quad\Rightarrow\quad F(x) ~-~ G(x) ~=~ C \quad\Rightarrow\quad F(x) ~=~ G(x) ~+~ C \nonumber \]

for all \(x\) in \(I.\quad\checkmark\)

The practical consequence of the above result can be stated as follows:

So for the function \(f(x) = 2x\), since \(F(x) = x^2\) is *one* antiderivative then *all* antiderivatives of \(f(x)\) are of the form \(F(x) = x^2 + C\), where \(C\) is a generic constant. Thus, functions do not have just one antiderivative but a whole *family* of antiderivatives, all differing only by a constant. The following notation makes all this easier to express:

The large S-shaped symbol before \(f(x)\) is called an **integral sign**. Though the indefinite integral \(\int f(x)~\dx\) represents *all* antiderivatives of \(f(x)\), the integral can be thought of as a single object or function in its own right, whose derivative is \(f'(x)\):

You might be wondering what the integral sign in the indefinite integral represents, and why an infinitesimal \(\dx\) is included. It has to do with what an infinitesimal represents: an infinitesimal “piece” of a quantity. For an antiderivative \(F(x)\) of a function \(f(x)\), the infinitesimal (or differential) \(d\!F\) is given by \(d\!F = F'(x)\,\dx = f(x)\,\dx\), and so

\[F(x) ~=~ \int\,f(x)~\dx ~=~ \int\,d\!F ~. \nonumber \]

The integral sign thus acts as a summation symbol: it sums up the infinitesimal “pieces” \(d\!F\) of the function \(F(x)\) at each \(x\) so that they add up to the entire function \(F(x)\). Think of it as similar to the usual summation symbol \(\Sigma\) used for *discrete* sums; the integral sign \(\int\) takes the sum of a *continuum* of infinitesimal quantities instead.

Finding (or **evaluating**) the indefinite integral of a function is called **integrating** the function, and **integration** is antidifferentiation.

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**Solution**

Evaluate \(\displaystyle\int\,0~\dx\).

*Solution:* Since the derivative of any constant function is 0, then \(\int\,0~\dx = C\), where \(C\) is a generic constant.

Note: From now on \(C\) will simply be assumed to represent a generic constant, without having to explicitly say so every time.

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**Solution**

Evaluate \(\displaystyle\int\,1~\dx\).

*Solution:* Since the derivative of \(F(x) = x\) is \(F'(x) = 1\), then \(\int\,1~\dx = x + C\).

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**Solution**

Evaluate \(\displaystyle\int\,x~\dx\).

*Solution:* Since the derivative of \(F(x) = \frac{x^2}{2}\) is \(F'(x) = x\), then \(\int\,x~\dx = \frac{x^2}{2} + C\).

Since \(\ddx\,\left(\frac{x^{n+1}}{n+1}\right) = x^n\) for any number \(n \ne -1\), and \(\ddx\,(\ln\,\abs{x}) = \frac{1}{x} = x^{-1}\), then any power of \(x\) can be integrated:

The following rules for indefinite integrals are immediate consequences of the rules for derivatives:

The above rules are easily proved. For example, the first rule is a simple consequence of the Constant Multiple Rule for derivatives: if \(F(x) = \int\,f(x)~\dx\), then

\[\ddx(k\,F(x)) ~=~ k\,\ddx(F(x)) ~=~ k\,f(x) \quad\Rightarrow\quad \int\,k\;f(x)~\dx ~=~ k\,F(x) ~=~ k\,\int\,f(x)~\dx ~.\quad\checkmark \nonumber \]

The other rules are proved similarly and are left as exercises. Repeated use of the above rules along with the Power Formula shows that any polynomial can be integrated term by term—in fact any finite sum of functions can be integrated in that manner:

[antideriv4] Evaluate \(\displaystyle\int\,(x^7 - 3x^4)~\dx\).

*Solution:* Integrate term by term, pulling constant multiple outside the integral:

\[\int\,(x^7 - 3x^4)~\dx ~=~ \int\,x^7~\dx ~-~ 3\int\,x^4~\dx ~=~ \frac{x^8}{8} ~-~ \frac{3x^5}{5} ~+~ C \nonumber \]

[antideriv5] Evaluate \(\displaystyle\int\,\sqrt{x}~\dx\).

*Solution:* Use the Power Formula:

\[\int\,\sqrt{x}~\dx ~=~ \int\,x^{1/2}~\dx ~=~ \frac{x^{3/2}}{3/2} ~+~ C ~=~ \frac{2x^{3/2}}{3} ~+~ C \nonumber \]

[antideriv6] Evaluate \(\displaystyle\int\,\left(\dfrac{1}{x^2} + \dfrac{1}{x}\right)~\dx\).

*Solution:* Use the Power Formula and integrate term by term:

\[\int\,\left(\frac{1}{x^2} + \frac{1}{x}\right)~\dx ~=~ \int\,\left(x^{-2} + \frac{1}{x}\right)~\dx ~=~ \frac{x^{-1}}{-1} ~+~ \ln\,\abs{x} ~+~ C ~=~ -\frac{1}{x} ~+~ \ln\,\abs{x} ~+~ C \nonumber \]

The following indefinite integrals are just re-statements of the corresponding derivative formulas for the six basic trigonometric functions:

Since \(\ddx(e^x) = e^x\), then:

[antideriv7] Evaluate \(\displaystyle\int\,(3\sin\,x ~+~ 4\cos\,x ~-~ 5e^x)~\dx\).

*Solution:* Integrate term by term:

\[\begin{aligned} \int\,(3\sin\,x ~+~ 4\cos\,x ~-~ 5e^x)~\dx ~&=~ 3\int\,\sin\,x~\dx ~+~ 4\int\,\cos\,x~\dx ~-~ 5\int\,e^x~\dx\

\[10pt] &=~ -3\cos\,x ~+~ 4\sin\,x ~-~ 5e^x ~+~ C\end{aligned} \nonumber \]

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**Solution**

Recall from Section 1.1 the example of an object dropped from a height of 100 ft. Show that the height \(s(t)\) of the object \(t\) seconds after being dropped is \(s(t) = -16t^2 + 100\), measured in feet.

*Solution:* When the object is dropped at time \(t=0\) the only force acting on it is gravity, causing the object to accelerate downward at the known constant rate of 32 ft/s^{2}. The object’s acceleration \(a(t)\) at time \(t\) is thus \(a(t) = -32\). If \(v(t)\) is the object’s velocity at time \(t\), then \(v'(t) = a(t)\), which means that

\[v(t) ~=~ \int a(t)~\dt ~=~ \int -32~\dt ~=~ -32t ~+~ C \nonumber \]

for some constant \(C\). The constant \(C\) here is *not* generic—it has a specific

value determined by the *initial condition* on the velocity: the object was at rest at time \(t=0\). That is, \(v(0) = 0\), which means

\[0 ~=~ v(0) ~=~ -32(0) ~+~ C ~=~ C \quad\Rightarrow\quad v(t) ~=~ -32t \nonumber \]

for all \(t \ge 0\). Likewise, since \(s'(t) = v(t)\) then

\[s(t) ~=~ \int v(t)~\dt ~=~ \int -32t~\dt ~=~ -16t^2 ~+~ C \nonumber \]

for some constant \(C\), determined by the initial condition that the object was 100 ft above the ground at time \(t=0\). That is, \(s(0) = 100\), which means

\[100 ~=~ s(0) ~=~ -16(0)^2 ~+~ C ~=~ C \quad\Rightarrow\quad s(t) ~=~ -16t^2 ~+~ 100 \nonumber \]

for all \(t \ge 0\).

The formula for \(s(t)\) in Example

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**Solution**

can be generalized as follows: denote the object’s initial position at time \(t=0\) by \(s_0\), let \(v_0\) be the object’s initial velocity (positive if thrown upward, negative if thrown downward), and let \(g\) represent the (positive) constant acceleration due to gravity. By Newton’s First Law of motion the only acceleration imparted to the object *after* throwing it is due to gravity:

\[a(t) ~=~ -g \quad\Rightarrow\quad v(t) ~=~ \int a(t)~\dt ~=~ \int -g~\dt ~=~ -gt ~+~ C \nonumber \]

for some constant \(C\): \(v_0 = v(0) = -g(0) + C = C\). Thus, \(v(t) = -gt + v_0\) for all \(t \ge 0\), and so

\[s(t) ~=~ \int v(t)~\dt ~=~ \int \left(-gt ~+~ v_0\right)~\dt ~=~ -\tfrac{1}{2}gt^2 ~+~ v_0t ~+~ C \nonumber \]

for some constant \(C\): \(s_0 = s(0) = -\tfrac{1}{2}g(0)^2 + v_0(0) + C = C\). To summarize:

Note that the units are not specified—they just need to be consistent. In metric units, \(g = 9.8\) m/s^{2}, while \(g = 32\) ft/s^{2} in English units.

Thinking of an indefinite integral as the sum of all the infinitesimal “pieces” of a function—for the purpose of retrieving that function—provides a handy way of integrating a differential equation to obtain the solution. The key idea is to transform the differential equation into an *equation of differentials*, which has the effect of treating functions as variables. Some examples will illustrate the technique.

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**Solution**

For any constant \(k\), show that every solution of the differential equation \(\dydt = ky\) is of the form \(y = Ae^{kt}\) for some constant \(A\). You can assume that \(y(t) > 0\) for all \(t\).

*Solution:* Put the \(y\) terms on the left and the \(t\) terms on the right, i.e. separate the variables:

\[\frac{\dy}{y} ~=~ k\,\dt \nonumber \]

Now integrate both sides (notice how the *function* \(y\) is treated as a *variable*):

\[\begin{aligned} \int\,\frac{\dy}{y} ~&=~ \int k\,\dt\

\[6pt] \ln\,y + C_1 ~&=~ kt + C_2 \quad\text{($C_1$ and $C_2$ are constants)}\\ \ln\,y ~&=~ kt + C \quad\text{(combine $C_1$ and $C_2$ into the constant $C$)}\\ y ~&=~ e^{kt+C} ~=~ e^{kt} \cdot e^C ~=~ A e^{kt}\end{aligned} \nonumber \]

where \(A = e^C\) is a constant. Note that this is the formula for radioactive decay from Section 2.3.

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**Solution**

Recall from Section 3.6 the equation of differentials

\[\dfrac{\dP}{P} ~+~ \dfrac{\dV}{V} ~=~ \dfrac{\dT}{T} \nonumber \]

relating the pressure \(P\), volume \(V\) and temperature \(T\) of an ideal gas. Integrate that equation to obtain the original ideal gas law \(PV = RT\), where \(R\) is a constant. .

*Solution:* Integrating both sides of the equation yields

\[\begin{aligned} \int\,\dfrac{\dP}{P} ~+~ \int\,\dfrac{\dV}{V} ~&=~ \int\,\dfrac{\dT}{T}\

\[6pt] \ln\,P ~+~ \ln\,V ~&=~ \ln\,T ~+~ C \quad\text{($C$ is a constant)}\\ \ln\,(PV) ~&=~ \ln\,T ~+~ C\\ PV ~&=~ e^{\ln\,T + C} ~=~ e^{\ln\,T} \cdot e^{C} ~=~ T\,e^C ~=~ RT\end{aligned} \nonumber \]

where \(R = e^C\) is a constant.

The integration formulas in this section depended on already knowing the derivatives of certain functions and then “working backward” from their derivatives to obtain the original functions. Without that prior knowledge you would be reduced to guessing, or perhaps recognizing a pattern from some derivative you have encountered. A number of integration techniques will be presented shortly, but there are many indefinite integrals for which no simple closed form exists (e.g. \(\int e^{x^2}\,\dx\) and \(\int \sin(x^2)\,\dx\)).

[sec5dot1]

For Exercises 1-15, evaluate the given indefinite integral.

3

\(\displaystyle\int\,\left(x^2 ~+~ 5x ~-~ 3\right)~\dx\)

\(\displaystyle\int\,3 \cos\,x~\dx\)

\(\displaystyle\int\,4 e^x~\dx\)

3

\(\displaystyle\int\,\left(x^5 ~-~ 8x^4 ~-~ 3x^3 ~+~ 1\right)~\dx\vphantom{\dfrac{3e^x}{5}}\)

\(\displaystyle\int\,5 \sin\,x~\dx\vphantom{\dfrac{3e^x}{5}}\)

\(\displaystyle\int\,\dfrac{3e^x}{5}~\dx\)

3

\(\displaystyle\int\,\dfrac{6}{x}~\dx\)

\(\displaystyle\int\,\dfrac{4}{3x}~\dx\)

\(\displaystyle\int\,\left(-2 \sqrt{x}\,\right)~\dx\vphantom{\dfrac{6}{x}}\)

3

\(\displaystyle\int\,\dfrac{1}{3 \sqrt{x}}~\dx\)

\(\displaystyle\int\,\left(x ~+~ x^{4/3}\right)~\dx\)

\(\displaystyle\int\,\dfrac{1}{3 \sqrt[3]{x}}~\dx\)

3

\(\displaystyle\int\,3\sec\,x\;\tan\,x~\dx\)

\(\displaystyle\int\,5 \sec^2x~\dx\)

\(\displaystyle\int\,7\csc^2x~\dx\)

Prove the sum and difference rules for indefinite integrals: \(\int (f(x) \pm g(x))\,\dx \;=\; \int f(x)\,\dx \;\pm\; \int g(x)\,\dx\)

Integrate both sides of the equation

\[\frac{\dP}{P} ~+~ \frac{d\!M}{M} ~=~ \frac{\dT}{2T} \nonumber \]

to obtain the ideal gas continuity relation: \(\dfrac{PM}{\sqrt{T}} =\) constant.

[exer:projmax0] Use the free fall motion equation for position to show that the maximum height reached by an object launched straight up from the ground with an initial velocity \(v_0\) is \(\frac{v_0^2}{2g}\).

- The function \(f\) is assumed to be differentiable at \(x\), in this case. If not then the points where \(f\) is not differentiable can be excluded without affecting the integral.↩
- For a proof and fuller discussion of all this, see Ch.1-2 in Knopp, M.I.,
*Theory of Area*, Chicago: Markham Publishing Co., 1969. The book attempts to define precisely what an “area” actually means, including that of a rectangle (showing agreement with the intuitive notion of width times height).↩ - The theorem can be proved for the weaker condition that \(f\) is merely continuous on \(\ival{a}{b}\). See p.173-175 in Parzynski, W.R. and P.W. Zipse,
*Introduction to Mathematical Analysis*, New York: McGraw-Hill, Inc., 1982.↩ - Created by the physicist P.A.M. Dirac (1902-1984), who won a Nobel Prize in physics in 1933. The function is neither real-valued nor continuous at \(x=0\). The “graph” in Figure [fig:dirac] is perhaps misleading, as \(\infty\) is not an actual point on the \(y\)-axis. One interpretation is that \(\delta\) is an abstraction of an instantaneous pulse or burst of
*something*, preceded and followed by*nothing*. To learn more about this fascinating and useful function, see §15 in Dirac, P.A.M.,*The Principles of Quantum Mechanics*, 4th ed., Oxford, UK: Oxford University Press, 1958.↩ - See pp.140-141 in Buck, R.C.,
*Advanced Calculus*, 2nd ed., New York: McGraw-Hill Book Co., 1965.↩