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Home Page34.An tiderivative

34.1.

In troduction

Instead of starting with a function and asking what its derivative is, we turn things around in this section: ?Find a functionFthat has derivativex4. Since we know that the derivative of a power ofxreduces the power by one, we take as an initial guessF(x) =x5. Check:F0(x) = 5x4. Because of the factor of 5, this is not quite what we wanted, but we now know how to adjust. LetF(x) =15 x5. Check:F0(x) =x4 (yes). IfFandfare functions andF0(x) =f(x), thenFis called anantiderivativeoff.

For instance,F(x) =15

x5is an antiderivative off(x) =x4. The relationship between derivatives and antiderivatives can be represented schematically:For instance,

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Home Page34.2.Indenite in tegral

Letf(x) = 2x. The functionF(x) =x2is an antiderivative off. But so isF(x) =x2+ 1, andF(x) =x2+ 2. In fact,F(x) =x2+Cis an antiderivative offforanyconstantC. The graph ofF(x) =x2+Cis the graph ofF(x) =x2shifted vertically byCunits, so we have the following picture:

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Home PageEach function pictured is an antiderivative of 2x, that is, each function has the same derivative (= general slope function) 2x. This agrees with the observation that at anyx the tangents to the graphs are all parallel, which implies that their slopes are the same. A question remains: Are the functionsF(x) =x2+C(Cany real number)allof the possible antiderivatives off(x) = 2x? Since the graph of any antiderivative has to share with the graphs drawn above the stated property of parallel tangent lines, it is reasonable to expect that there can be no further antiderivatives. This is in fact, the case. The the main step in the verication is the following result, which says roughly that the only way a function can have a derivative that is constantly 0 is if it is a constant function (so that its graph is a horizontal line).

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Home PageFunction with zero derivative is constant.Iff0(x) = 0, then

f(x) =Cfor some constantC.The theorem says that iffhas general slope function 0 (that is, every tangent is horizontal),

thenfmust be a constant function (that is, its graph must be a horizontal line), and this seems reasonable. Here is the careful verication: Assume thatf0(x) = 0. Supposef(x) is not constant. Thenf(a)6=f(b) for somea6=b. By the mean value theorem, there existscbetweena andbsuch that f

0(c) =f(b)f(a)ba:

But the expression on the right is nonzero sincef(a)6=f(b) so we havef0(c)6= 0, in violation of our assumption thatf0(x) = 0 for allx. We conclude thatf(x) is constant, that is,f(x) =Cfor some constantC.Two antiderivatives differ by constant.IfFis any antiderivative off, then every antiderivative offis of the formF(x) +Cfor some constantC.Here is the verication: LetFbe an antiderivative offand suppose thatGis another antiderivative off. We haveF0(x) =f(x) and alsoG0(x) =f(x), so thatF0(x) =G0(x). LetH(x) =G(x)F(x). ThenH0(x) =G0(x)F0(x) = 0. By the previous theorem (with Hplaying the role off),H(x) =Cfor some constantC. ThenG(x) =F(x) +H(x) =

F(x) +C, as claimed.

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Home PageThe result just veried establishes the earlier claim that the functionsF(x) =x2+C(C

any real number) are the only antiderivatives off(x) = 2x.Indefinite integral.Iffis a function andFis any antiderivative of

f, we write Z f(x)dx=F(x) +C(C, arbitrary constant) and call it the(indenite) integraloff.For example, sincex2is an antiderivative of 2x, we have Z

2xdx=x2+C:

Saying thatCis an \arbitrary" constant, is saying that it can be any real number. So in a sense,Z 2xdx simultaneously represents x

2+ 0; x2+ 1; x2+12

; :::; x2+ any number and these are precisely all of the possible antiderivatives of 2x(according to the previous theorem). For this reason, the indenite integral offis often called the most general antiderivative off.

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Home PageThe reason for the notation

Rf(x)dxwill be given later, but for now it can be regarded as a Leibniz notation for the most general antiderivative off. The functionf(x) between the symbolsRanddxis called theintegrand. If an independent variable other thanxis used, thendxis changed accordingly. For instance, we would writeRt4dt=15 t5+C. 34.3.

In tegralrules

Any derivative rule gives rise to an integral rule (and conversely). For example, ddx [sinx] = cosx)Z cosxdx= sinx+C ddx [tanx] = sec2x)Z sec

2xdx= tanx+C

ddx [ex] =ex)Z e xdx=ex+C ddx [xn] =nxn1)Z nx n1dx=xn+C The last integral rule is not very convenient; we would prefer to have a rule for the integral of simplyxn. Such a rule follows:Power rule for integrals. Z x ndx=xn+1n+ 1+C(n6=1):

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Home PageIn words, the integral of a power ofxequalsxto the one higher power over that higher power, plus an arbitrary constant. For example, Z x

4dx=x55

+C; in agreement with what we found earlier using trial and error. The power rule is veried (as is any integral rule) by checking the validity of the corresponding derivative rule: ddx  xn+1n+ 1 =ddx h

1n+1xn+1i

=xn; as desired. The power rule excludes the casen=1 (as it must since this would produce a zero in the denominator). The omitted case is neatly handled by an earlier derivative rule:Integral of 1/x. Z x 1dx= lnjxj+CThis formula holds since ddx [lnjxj] =1x =x1.Constant multiple rule.For any constantc, Z cf(x)dx=cZ f(x)dx

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Home PageSum/difference rule.

Z (f(x)g(x))dx=Z f(x)dxZ g(x)dx34.3.1 ExampleFindZ5x3+ 7x2
dx. SolutionUsing the sum/dierence rule, the constant multiple rule, and the power rule, we getZ5x3+ 7x2
dx=Z

5x3dx+Z

7x2dx = 5 Z x

3dx+ 7Z

x 2dx = 5 x44 +C1 + 7x33 +C2 ; where we have used subscripts on the arbitrary constants since one cannot assume that the constants are equal. After distributing and collecting terms, we get

Z5x3+ 7x2
dx= 5x44

 + 7x33  + (5C1+ 7C2) =5x44 +7x33 +C; whereC= 5C1+ 7C2. AsC1andC2range through all real numbers,Cranges through all real numbers as well. Therefore, we can forget aboutC1andC2and just regardC as an arbitrary constant. In short, one can apply the sum rule, the dierence rule, and the constant multiple rule, ignoring any arbitrary constants, provided a single arbitrary constant is appended at the end.

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Home PageNow that we have seen in detail how the rules work, we can suppress steps.

34.3.2 ExampleFindZ



7 + 4x35x

2+6x + 23px 2 dx. SolutionWe do some rewriting in order to use the power rule: Z (7+4x35x 2+6x + 23px 2 dx = Z 

7x0+ 4x35x2+ 6x1+ 2x2=3

dx = 7 x11  + 4x44  5x11 + 6lnjxj+ 2x5=35=3 +C = 7x+x4+5x + 6lnjxj+63px 55
+C:Verifying a stated integral formula is dierent from nding an integral. To verify an integral formula, it is only necessary to verify the corresponding derivative formula:

34.3.3 ExampleVerify that

Zp1x2dx=12

 xp1x2+ sin1x +C:

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Home PageSolutionWe verify the corresponding derivative formula: ddx h 12  xp1x2+ sin1x
 = 12 

1
p1x2+x
12

1x2
1=2(2x) +1p1x2 = 12  p1x2x2p1x2+1p1x2 = 12 

22x2p1x2

= p1x2:34.3.4 ExampleFindZ 

4ex+ sinx8x

+31 +x2
dx. SolutionWe use the sum/dierence rule and the constant multiple rule, and then the fact pointed out above that every derivative rule gives rise to a corresponding integral rule (rewriting the second term in order to use the rule): Z (4ex+ sinx8x +31 +x2
dx = 4 Z e xdxZ (sinx)dx8Z1x dx+ 3Z11 +x2dx = 4excosx8lnjxj+ 3tan1x+C: (See 25
for a list of deriv ativerules.)

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Home Page34.4.Initial v alueproblem

Iffis an unknown function and we know only that it has derivativef0(x) = 2x, then we cannot hope to determinefprecisely since there are innitely many functions with derivative 2x(namely,x2+Cfor any constantC). However, if we also know the value of ffor some particular input, for instancef(2) = 5, then we can determinefprecisely:

34.4.1 ExampleFind the functionfgiven thatf0(x) = 2xandf(2) = 5.

SolutionThe equationf0(x) = 2xsays that the desired functionfis an antiderivative of

2x, so

f(x) =Z

2xdx= 2x22

 +C=x2+C; that is,f(x) =x2+C. Using the conditionf(2) = 5, we get 5 = " givenf(2) =" evaluate2 2+C: The preceding equations reveal that 5 = 4 +C, so thatC= 1. Therefore,f(x) =x2+ 1. Knowledge of the general slope functionf0(x) = 2xof the desired functionfallowed us to determine thatf(x) =x2+C. At that point we could tell that the graph offhad the shape of the parabolay=x2, but there was an unknown vertical shift ofCunits that kept us from completely determiningf. The second condition,f(2) = 5, said that the graph of fhad to go through the point (2;5), and from this we determined that the shift amount had to be 1:

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Home PageThe problem of nding a function given its derivative and its value for some particular input is called aninitial value problem. Many modeling problems in the sciences and engineering are initial value problems.

34.4.2 ExampleA car traveling northeast on I-85 (assumed to be straight) has ve-

locity at timethr given byv(t) = 20t+ 55 mph. Given that the car is at Lagrange after one hour, nd where the car began its trip. SolutionLetf(t) be the car's position, relative to Lagrange, at timet. Since velocity is the rate at which position changes, we havev(t) =f0(t), which says thatfis an antiderivative ofv, so f(t) =Z v(t)dt=Z (20t+ 55)dt= 20t22  + 55t+C= 10t2+ 55t+C;

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Home Pagethat is,f(t) = 10t2+ 55t+C. Since the car is at Lagrange after one hour, its position at timet= 1 is 0, that is,f(1) = 0. Therefore, 0 = " givenf(1) =" evaluate10(1)

2+ 55(1) +C:

The preceding equations reveal that 0 = 65 +C, so thatC=65. Therefore,f(t) =

10t2+55t65. The car began its trip at timet= 0, so its initial position wasf(0) =65.

In other words, the car was 65 miles southwest of Lagrange (and therefore around Tuskegee). The informationv(t) = 20t+55 tells us the car's speedometer reading at any timet. From this alone, we could not have hoped to determine the car's initial position (one can imagine cars at various places along I-85 always having identical speedometer readings). It was the additional information that the car was at Lagrange after one hour that allowed for the determination of its initial position (and, in fact, its position at any time). In terms of the graph, the speedometer readings allowed us to determine that the position function wasf(t) = 10t2+ 55t+C, so we could tell that the graph had the shape of the upturning parabolay= 10t2+ 55tbut with an unknown vertical shift ofCunits. The additional information told us that the graph had to go through the point (1;0) and this allowed for the determination of the shift amount.

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Home Page34{Exercises

34
{1 (a) V erifythat F(x) = sinx2is an antiderivative off(x) = 2xcosx2. (b) Find

R2xcosx2dx.

Hint:After doing part (a), part (b) should be easy if you understand what is meant by the notation. See denition of indenite integral in 34.2
34
{2V erifythe in tegralform ula Z cos

3xdx= sinx13

sin3x+C: Hint:See Example34.3.3 and the paragr aphpreceding it for what is required. 34
{3Find

Zx7+ 3x54x2+x8
dx:

34
{4Find Z  4 +7x 13 px

5+ 3px

 dx.

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Home Page34{5Find

Z 3x2+x4 +xpx x 2 dx.

Hint:First, rewrite the integrand.

34
{6Find Z  2 x+ cscxcotx1p1x2 dx:

Hint:See Example34.3.4

34
{7Find the function fgiven thatf0(x) = sinxandf(0) = 3. 34
{8A ball dropp edfrom the top of a tall building has v elocityat time tseconds given by v(t) =10tm/s. Find the height of the building given that the ball strikes the ground after 6 seconds.