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322 Chapter 5 / Integration
22. If A(x) is the area under the graph of a nonnegative contin-
uous function f over an interval [a, x], then A(x) will be a continuous function.
F O C U S O N C O N C E P TS
23. Explain how to use the formula for A(x) found in the solution to Example 2 to determine the area between the graph of y = x 2 and the interval [ 3 , 6 ]. 24. Repeat Exercise 23 for the interval [− 3 , 9 ]. 25. Let A denote the area between the graph of f(x) =
x and the interval [ 0 , 1 ], and let B denote the area between the graph of f(x) = x^2 and the interval [ 0 , 1 ]. Explain geometrically why A + B = 1.
26. Let A denote the area between the graph of f(x) = 1 /x and the interval [ 1 , 2 ], and let B denote the area be- tween the graph of f and the interval
[ 1
2 ,^1
]
. Explain geometrically why A = B.
27–28 The area A(x) under the graph of f and over the interval [a, x] is given. Find the function f and the value of a.!
27. A(x) = x^2 − 4 28. A(x) = x^2 − x 29. Writing Compare and contrast the rectangle method and the antiderivative method. 30. Writing Suppose that f is a nonnegative continuous func- tion on an interval [a, b] and that g(x) = f(x) + C, where C is a positive constant. What will be the area of the region between the graphs of f and g?
"QUICK CHECK ANSWERS 5.
1. (a)
π 2
(b) 1 +
2. 2 3. 9 4. A(x) =
x^2 2
; A
′ (x) =
2 x 2
= x = f(x) 5. ex^ + 1
5.2 THE INDEFINITE INTEGRAL
In the last section we saw how antidifferentiation could be used to find exact areas. In this section we will develop some fundamental results about antidifferentiation.
ANTIDERIVATIVES
5.2.1 definition A function F is called an antiderivative of a function f on a given open interval if F ′ (x) = f(x) for all x in the interval.
For example, the function F (x) = 1 3 x
3 is an antiderivative of f(x) = x 2 on the interval (−!, +!) because for each x in this interval
F
′ (x) =
d
dx
[
1 3 x
3
]
= x 2 = f(x)
However, F (x) = 1 3 x
(^3) is not the only antiderivative of f on this interval. If we add any
constant C to 1 3 x
3 , then the function G(x) = 1 3 x
3
- C is also an antiderivative of f on (−!, +!), since G ′ (x) =
d
dx
[
1 3 x^3 + C
]
= x 2
In general, once any single antiderivative is known, other antiderivatives can be obtained by adding constants to the known antiderivative. Thus,
1 3 x
3 , 1 3 x
3
3 − 5 , 1 3 x
3
are all antiderivatives of f(x) = x^2.
5.2 The Indefinite Integral 323
It is reasonable to ask if there are antiderivatives of a function f that cannot be obtained by adding some constant to a known antiderivative F. The answer is no —once a single antiderivative of f on an open interval is known, all other antiderivatives on that interval are obtainable by adding constants to the known antiderivative. This is so because Theorem 4.8.3 tells us that if two functions have the same derivative on an open interval, then the functions differ by a constant on the interval. The following theorem summarizes these observations.
5.2.2 theorem If F (x) is any antiderivative of f(x) on an open interval, then for any constant C the function F (x) + C is also an antiderivative on that interval. Moreover, each antiderivative of f(x) on the interval can be expressed in the form F (x) + C by choosing the constant C appropriately.
THE INDEFINITE INTEGRAL
The process of finding antiderivatives is called antidifferentiation or integration. Thus, if
E x t r a ct f r o m the m a nusc r ipt o f L eibni z
d a ted Octobe r 29 , 1675 in w hich the integ r a l sign f i r st a ppe a r ed (see y ello w
highlight).
Reproduced from C. I. Gerhardt's "Briefwechsel von G. W. Leibniz mit Mathematikern (1899)."
d
dx
[F (x)] = f(x) (1)
then integrating (or antidifferentiating ) the function f(x) produces an antiderivative of the form F (x) + C. To emphasize this process, Equation (1) is recast using integral notation , ∫
f(x) dx = F (x) + C (2)
where C is understood to represent an arbitrary constant. It is important to note that (1) and (2) are just different notations to express the same fact. For example, ∫
x 2 dx = 1 3 x
3
d
dx
[
1 3 x
3
]
= x 2
Note that if we differentiate an antiderivative of f(x), we obtain f(x) back again. Thus,
d
dx
[∫
f(x) dx
]
= f(x) (3)
The expression
f(x) dx is called an indefinite integral. The adjective “indefinite” emphasizes that the result of antidifferentiation is a “generic” function, described only up to a constant term. The “elongated s” that appears on the left side of (2) is called an integral sign ,
∗ the function f(x) is called the integrand , and the constant C is called the constant of integration. Equation (2) should be read as:
The integral of f(x) with respect to x is equal to F (x) plus a constant.
The differential symbol, dx, in the differentiation and antidifferentiation operations
d
dx
[ ] and
[ ] dx
∗ This notation was devised by Leibniz. In his early papers Leibniz used the notation “omn.” (an abbreviation for the Latin word “omnes”) to denote integration. Then on October 29, 1675 he wrote, “It will be useful to write∫ for omn., thus
∫ l for omn. l... .” Two or three weeks later he refined the notation further and wrote
∫ [ ] dx rather than
∫ alone. This notation is so useful and so powerful that its development by Leibniz must be regarded as a major milestone in the history of mathematics and science.
5.2 The Indefinite Integral 325
See Exercise 72 for a justification of For-^ Example 1^ The second integration formula in Table 5.2.1 will be easier to remember mula 14 in Table 5.2.1. if you express it in words:
To integrate a power of x ( other than − 1 ) , add 1 to the exponent and divide by the new exponent.
Here are some examples: Although Formula 2 in Table 5.2.1 is not applicable to integrating x−^1 , this func-
tion can be integrated by rewriting the integral in Formula 11 as ∫ 1 x
dx =
∫ x−^1 dx = ln |x| + C
x 2 dx =
x 3
∫ x 3 dx =
x 4
∫ 1
x^5
dx =
x − 5 dx =
x−^5 +^1
− 5 + 1
+ C = −
4 x^4
∫ √ x dx =
x
1 (^2) dx =
x
1 2 +^1 1 2
+ C =
2 3 x^
3 (^2) + C = 2 3 (
x )^3 + C r = (^12)
PROPERTIES OF THE INDEFINITE INTEGRAL
Our first properties of antiderivatives follow directly from the simple constant factor, sum, and difference rules for derivatives.
5.2.3 theorem Suppose that F (x) and G(x) are antiderivatives of f(x) and g(x), respectively, and that c is a constant. Then:
( a ) A constant factor can be moved through an integral sign; that is,
∫
cf(x) dx = cF (x) + C
( b ) An antiderivative of a sum is the sum of the antiderivatives; that is,
∫
[f(x) + g(x)] dx = F (x) + G(x) + C
( c ) An antiderivative of a difference is the difference of the antiderivatives; that is,
∫
[f(x) − g(x)] dx = F (x) − G(x) + C
proof In general, to establish the validity of an equation of the form ∫
h(x) dx = H (x) + C
one must show that (^) d
dx
[H (x)] = h(x)
We are given that F (x) and G(x) are antiderivatives of f(x) and g(x), respectively, so we know that (^) d
dx
[F (x)] = f(x) and
d
dx
[G(x)] = g(x)
326 Chapter 5 / Integration
Thus, d
dx
[cF (x)] = c
d
dx
[F (x)] = cf(x)
d
dx
[F (x) + G(x)] =
d
dx
[F (x)] +
d
dx
[G(x)] = f(x) + g(x)
d
dx
[F (x) − G(x)] =
d
dx
[F (x)] −
d
dx
[G(x)] = f(x) − g(x)
which proves the three statements of the theorem.!
The statements in Theorem 5.2.3 can be summarized by the following formulas:
∫
cf(x) dx = c
f(x) dx (4)
[f(x) + g(x)] dx =
f(x) dx +
g(x) dx (5)
[f(x) − g(x)] dx =
f(x) dx −
g(x) dx (6)
However, these equations must be applied carefully to avoid errors and unnecessary com- plexities arising from the constants of integration. For example, if you use (4) to integrate 2 x by writing ∫
2 x dx = 2
x dx = 2
x^2
2
+ C
= x 2
then you will have an unnecessarily complicated form of the arbitrary constant. This kind of problem can be avoided by inserting the constant of integration in the final result rather than in intermediate calculations. Exercises 65 and 66 explore how careless application of these formulas can lead to errors.
Example 2 Evaluate
(a)
4 cos x dx (b)
(x + x 2 ) dx
Solution ( a ). Since F (x) = sin x is an antiderivative for f(x) = cos x (Table 5.2.1), we
obtain
4 cos x dx = 4
cos x dx = 4 sin x + C
(4)
Solution ( b ). From Table 5.2.1 we obtain
(x + x 2 ) dx =
x dx +
x 2 dx =
x^2
2
x^3
3
+ C
(5)
Parts (b) and (c) of Theorem 5.2.3 can be extended to more than two functions, which in combination with part (a) results in the following general formula:
∫
[c 1 f 1 (x) + c 2 f 2 (x) + · · · + cnfn(x)] dx
= c 1
f 1 (x) dx + c 2
f 2 (x) dx + · · · + cn
fn(x) dx (7)
328 Chapter 5 / Integration
Since the curve passes through ( 2 , 1 ), a specific value for C can be found by using the fact In Example 5, the requirement that the graph of f pass through the point ( 2 , 1 ) selects the single integral curve y = 13 x^3 − 53 from the family of curves y = 13 x^3 + C (Figure 5.2.2).
that y = 1 if x = 2. Substituting these values in the above equation yields
1 = 1 3 (^2
3 ) + C or C = − 5 3
so an equation of the curve is y = 1 3 x
3
(Figure 5.2.2).
x
y
− 3 3
− 4
4
(2, 1)
1 3
5 y = x 3 (^3) −
Figure 5.2.
INTEGRATION FROM THE VIEWPOINT OF DIFFERENTIAL EQUATIONS
We will now consider another way of looking at integration that will be useful in our later work. Suppose that f(x) is a known function and we are interested in finding a function F (x) such that y = F (x) satisfies the equation
dy
dx
= f(x) (8)
The solutions of this equation are the antiderivatives of f(x), and we know that these can be obtained by integrating f(x). For example, the solutions of the equation
dy
dx
= x 2 (9)
are
y =
x 2 dx =
x^3
3
+ C
Equation (8) is called a differential equation because it involves a derivative of an unknown function. Differential equations are different from the kinds of equations we have encountered so far in that the unknown is a function and not a number as in an equation such as x 2
- 5 x − 6 = 0. Sometimes we will not be interested in finding all of the solutions of (8), but rather we will want only the solution whose graph passes through a specified point (x 0 , y 0 ). For example, in Example 5 we solved (9) for the integral curve that passed through the point ( 2 , 1 ). For simplicity, it is common in the study of differential equations to denote a solution of dy/dx = f(x) as y(x) rather than F (x), as earlier. With this notation, the problem of finding a function y(x) whose derivative is f(x) and whose graph passes through the point (x 0 , y 0 ) is expressed as dy
dx
= f(x), y(x 0 ) = y 0 (10)
This is called an initial-value problem , and the requirement that y(x 0 ) = y 0 is called the initial condition for the problem.
Example 6 Solve the initial-value problem
dy
dx
= cos x, y( 0 ) = 1
Solution. The solution of the differential equation is
y =
cos x dx = sin x + C (11)
The initial condition y( 0 ) = 1 implies that y = 1 if x = 0; substituting these values in (11) yields
1 = sin( 0 ) + C or C = 1
Thus, the solution of the initial-value problem is y = sin x + 1.
5.2 The Indefinite Integral 329
SLOPE FIELDS
If we interpret dy/dx as the slope of a tangent line, then at a point (x, y) on an integral curve of the equation dy/dx = f(x), the slope of the tangent line is f(x). What is interesting about this is that the slopes of the tangent lines to the integral curves can be obtained without actually solving the differential equation. For example, if
dy
dx
x^2 + 1
then we know without solving the equation that at the point where x = 1 the tangent line to an integral curve has slope
2; and more generally, at a point where x = a, the tangent line to an integral curve has slope
a^2 + 1. A geometric description of the integral curves of a differential equation dy/dx = f(x) can be obtained by choosing a rectangular grid of points in the xy -plane, calculating the slopes of the tangent lines to the integral curves at the gridpoints, and drawing small por- tions of the tangent lines through those points. The resulting picture, which is called a slope field or direction field for the equation, shows the “direction” of the integral curves at the gridpoints. With sufficiently many gridpoints it is often possible to visualize the integral curves themselves; for example, Figure 5.2.3 a shows a slope field for the differential equa- tion dy/dx = x^2 , and Figure 5.2.3 b shows that same field with the integral curves imposed on it—the more gridpoints that are used, the more completely the slope field reveals the shape of the integral curves. However, the amount of computation can be considerable, so computers are usually used when slope fields with many gridpoints are needed. Slope fields will be studied in more detail later in the text.
Figure 5.2.
x
y
Slope field with integral curves
− 5
− 4
− 3
− 2
− 1
1
2
3
4
5
− 5 − 4 − 3 − 2 − 1 1 2 3 4 5
x
y
Slope field for d y/ d x = x^2
− 5
− 4
− 3
− 2
− 1
1
2
3
4
5
− 5 − 4 − 3 − 2 − 1 1 2 3 4 5
( a) ( b)
"QUICK CHECK EXERCISES 5.2 ( See page 332 for answers. )
1. A function F is an antiderivative of a function f on an in- terval if for all x in the interval. 2. Write an equivalent integration formula for each given derivative formula.
(a)
d dx
[
x] =
x
(b)
d dx
[e 4 x ] = 4 e 4 x
3. Evaluate the integrals.
(a)
[x^3 + x + 5 ] dx (b)
[sec^2 x − csc x cot x] dx
4. The graph of y = x^2 + x is an integral curve for the func-
tion f(x) =. If G is a function whose graph is also an integral curve for f , and if G( 1 ) = 5, then G(x) =.
5. A slope field for the differential equation
dy
dx
2 x
x^2 − 4
has a line segment with slope through the point ( 0 , 5 ) and has a line segment with slope through the point (− 4 , 1 ).
5.2 The Indefinite Integral 331
41. Use a graphing utility to generate some representative in-
tegral curves of the function f(x) = 5 x^4 − sec^2 x over the interval (−π/ 2 , π/ 2 ).
42. Use a graphing utility to generate some representative inte-
gral curves of the function f(x) = (x − 1 )/x over the in- terval ( 0 , 5 ).
43–46 Solve the initial-value problems.!
43. (a)
dy dx
x, y( 1 ) = 2
(b)
dy dt
= sin t + 1, y
π 3
(c)
dy dx
x + 1 √ x
, y( 1 ) = 0
44. (a)
dy
dx
( 2 x)^3
, y( 1 ) = 0
(b)
dy dt
= sec 2 t − sin t, y
π 4
(c)
dy dx
= x 2
x^3 , y( 0 ) = 0
45. (a)
dy
dx
= 4 ex^ , y( 0 ) = 1 (b)
dy
dt
t
, y(− 1 ) = 5
46. (a)
dy
dt
1 − t^2
, y
(b)
dy
dx
x^2 − 1
x^2 + 1
, y( 1 ) =
π
2
47–50 A particle moves along an s-axis with position function
s = s(t) and velocity function v(t) = s′(t). Use the given in-
formation to find s(t).!
47. v(t) = 32 t; s( 0 ) = 20 48. v(t) = cos t; s( 0 ) = 2 49. v(t) = 3
t; s( 4 ) = 1 50. v(t) = 3 et^ ; s( 1 ) = 0
51. Find the general form of a function whose second derivative
is
x. [ Hint: Solve the equation f ′′ (x) =
x for f(x) by integrating both sides twice.]
52. Find a function f such that f ′′(x) = x + cos x and such
that f( 0 ) = 1 and f ′( 0 ) = 2. [ Hint: Integrate both sides of the equation twice.]
53–57 Find an equation of the curve that satisfies the given
conditions.!
53. At each point (x, y) on the curve the slope is 2x + 1; the
curve passes through the point (− 3 , 0 ).
54. At each point (x, y) on the curve the slope is (x + 1 )^2 ; the
curve passes through the point (− 2 , 8 ).
55. At each point (x, y) on the curve the slope is − sin x; the
curve passes through the point ( 0 , 2 ).
56. At each point (x, y) on the curve the slope equals the square
of the distance between the point and the y-axis; the point (− 1 , 2 ) is on the curve.
57. At each point (x, y) on the curve, y satisfies the condition
d^2 y/dx^2 = 6 x; the line y = 5 − 3 x is tangent to the curve at the point where x = 1.
C (^) 58. In each part, use a CAS to solve the initial-value problem.
(a)
dy
dx
= x^2 cos 3x, y(π/ 2 ) = − 1
(b)
dy
dx
x^3
( 4 + x^2 )^3 /^2
, y( 0 ) = − 2
59. (a) Use a graphing utility to generate a slope field for the dif- ferential equation dy/dx = x in the region − 5 ≤ x ≤ 5 and − 5 ≤ y ≤ 5. (b) Graph some representative integral curves of the func- tion f(x) = x. (c) Find an equation for the integral curve that passes through the point ( 2 , 1 ). 60. (a) Use a graphing utility to generate a slope field for the differential equation dy/dx = e x (^) / 2 in the region − 1 ≤ x ≤ 4 and − 1 ≤ y ≤ 4. (b) Graph some representative integral curves of the func- tion f(x) = ex^ /2. (c) Find an equation for the integral curve that passes through the point ( 0 , 1 ).
61–64 The given slope field figure corresponds to one of the differential equations below. Identify the differential equation that matches the figure, and sketch solution curves through the highlighted points.
(a)
dy
dx
= 2 (b)
dy
dx
= −x
(c)
dy dx
= x 2 − 4 (d)
dy dx
= e x/ 3 !
x
y
− 6 − 4 − 2 2 4 6
− 6
− 4
− 2
2
4
6
x
y
− 6 − 4 − 2 2 4 6
− 6
− 4
− 2
2
4
6
x
y
− 6 − 4 − 2 2 4 6
− 6
− 4
− 2
2
4
6
x
y
− 6 − 4 − 2 2 4 6
− 6
− 4
− 2
2
4
6
F O C U S O N C O N C E P TS
65. Critique the following “proof” that an arbitrary constant must be zero:
C =
0 dx =
0 · 0 dx = 0
0 dx = 0
332 Chapter 5 / Integration
66. Critique the following “proof” that an arbitrary constant must be zero:
0 =
x dx
x dx
(x − x) dx =
0 dx = C
67. (a) Show that
F (x) = tan − 1 x and G(x) = − tan − 1 ( 1 /x)
differ by a constant on the interval ( 0 , +!) by show- ing that they are antiderivatives of the same func- tion. (b) Find the constant C such that F (x) − G(x) = C by evaluating the functions F (x) and G(x) at a partic- ular value of x. (c) Check your answer to part (b) by using trigonomet- ric identities.
68. Let F and G be the functions defined by
F (x) =
x^2 + 3 x x
and G(x) =
x + 3 , x > 0 x, x < 0 (a) Show that F and G have the same derivative. (b) Show that G(x) &= F (x) + C for any constant C. (c) Do parts (a) and (b) contradict Theorem 5.2.2? Ex- plain.
69–70 Use a trigonometric identity to evaluate the integral.!
tan^2 x dx 70.
cot^2 x dx
71. Use the identities cos 2θ = 1 − 2 sin^2 θ = 2 cos^2 θ − 1 to
help evaluate the integrals
(a)
sin 2 (x/ 2 ) dx (b)
cos 2 (x/ 2 ) dx
72. Recall that
d
dx
[sec − 1 x] =
|x|
x^2 − 1
Use this to verify Formula 14 in Table 5.2.1.
73. The speed of sound in air at 0◦^ C (or 273 K on the Kelvin scale) is 1087 ft/s, but the speed v increases as the temper- ature T rises. Experimentation has shown that the rate of change of v with respect to T is
dv
dT
T
− 1 / 2
where v is in feet per second and T is in kelvins (K). Find a formula that expresses v as a function of T.
74. Suppose that a uniform metal rod 50 cm long is insulated laterally, and the temperatures at the exposed ends are main- tained at 25◦^ C and 85◦^ C, respectively. Assume that an x- axis is chosen as in the accompanying figure and that the temperature T (x) satisfies the equation
d 2 T dx^2
Find T (x) for 0 ≤ x ≤ 50.
x
0
25 °C
50
85 °C
Figure Ex-
75. Writing What is an initial-value problem? Describe the sequence of steps for solving an initial-value problem. 76. Writing What is a slope field? How are slope fields and integral curves related?
"QUICK CHECK ANSWERS 5.
1. F ′(x) = f(x) 2. (a)
x
dx =
x + C (b)
4 e 4 x dx = e 4 x
3. (a) 1 4 x
2 x
(^2) + 5 x + C (b) tan x + csc x + C 4. 2 x + 1; x (^2) + x + 3 5. 0; − 2 3
5.3 INTEGRATION BY SUBSTITUTION
In this section we will study a technique, called substitution , that can often be used to transform complicated integration problems into simpler ones.
u -SUBSTITUTION The method of substitution can be motivated by examining the chain rule from the viewpoint of antidifferentiation. For this purpose, suppose that F is an antiderivative of f and that g is a differentiable function. The chain rule implies that the derivative of F (g(x)) can be expressed as (^) d
dx
[F (g(x))] = F ′ (g(x))g ′ (x)
334 Chapter 5 / Integration
Guidelines for u-Substitution
Step 1. Look for some composition f(g(x)) within the integrand for which the substi- tution u = g(x), du = g ′ (x) dx
produces an integral that is expressed entirely in terms of u and its differential du. This may or may not be possible.
Step 2. If you are successful in Step 1, then try to evaluate the resulting integral in terms of u. Again, this may or may not be possible.
Step 3. If you are successful in Step 2, then replace u by g(x) to express your final answer in terms of x.
EASY TO RECOGNIZE SUBSTITUTIONS
The easiest substitutions occur when the integrand is the derivative of a known function, except for a constant added to or subtracted from the independent variable.
Example 2 ∫
sin(x + 9 ) dx =
sin u du = − cos u + C = − cos(x + 9 ) + C
u = x + 9 du = 1 · dx = dx ∫ (x − 8 )^23 dx =
u^23 du =
u 24
+ C =
(x − 8 ) 24
+ C
u = x − 8 du = 1 · dx = dx
Another easy u-substitution occurs when the integrand is the derivative of a known function, except for a constant that multiplies or divides the independent variable. The following example illustrates two ways to evaluate such integrals.
Example 3 Evaluate
cos 5x dx.
Solution.
cos 5x dx =
(cos u) ·
du =
cos u du =
sin u + C =
sin 5x + C
u = 5 x du = 5 dx or dx = 15 du
Alternative Solution. There is a variation of the preceding method that some people
prefer. The substitution u = 5 x requires du = 5 dx. If there were a factor of 5 in the inte- grand, then we could group the 5 and dx together to form the du required by the substitution. Since there is no factor of 5, we will insert one and compensate by putting a factor of 1 5 in front of the integral. The computations are as follows: ∫
cos 5x dx =
cos 5x · 5 dx =
cos u du =
sin u + C =
sin 5x + C
u = 5 x du = 5 dx
5.3 Integration by Substitution 335
More generally, if the integrand is a composition of the form f(ax + b), where f(x) is
an easy to integrate function, then the substitution u = ax + b, du = a dx will work.
Example 4 ∫ dx ( 1 3 x^ −^8
3 du
u^5
u − 5 du = −
u − 4
x − 8
+ C
u = 13 x − 8 du = 13 dx or dx = 3 du
Example 5 Evaluate
dx
1 + 3 x^2
Solution. Substituting
u =
3 x, du =
3 dx
yields ∫ dx
1 + 3 x^2
du
1 + u^2
tan−^1 u + C =
tan−^1 (
3 x) + C
With the help of Theorem 5.2.3, a complicated integral can sometimes be computed by
expressing it as a sum of simpler integrals.
Example 6 ∫ (^ 1
x
dx =
dx
x
sec^2 πx dx
= ln |x| +
sec 2 πx dx
= ln |x| +
π
sec 2 u du u = πx du = π dx or dx = 1 π
du
= ln |x| +
π
tan u + C = ln |x| +
π
tan πx + C
The next four examples illustrate a substitution u = g(x) where g(x) is a nonlinear
function.
Example 7 Evaluate
sin^2 x cos x dx.
Solution. If we let u = sin x, then
du
dx
= cos x, so du = cos x dx
Thus, ∫
sin 2 x cos x dx =
u 2 du =
u 3
+ C =
sin 3 x
3
+ C
5.3 Integration by Substitution 337
Example 12 Evaluate
cos 3 x dx.
Solution. The only compositions in the integrand that suggest themselves are
cos 3 x = (cos x) 3 and cos 2 x = (cos x) 2
However, neither the substitution u = cos x nor the substitution u = cos^2 x work (verify).
In this case, an appropriate substitution is not suggested by the composition contained in
the integrand. On the other hand, note from Equation (2) that the derivative g′(x) appears
as a factor in the integrand. This suggests that we write
∫ cos^3 x dx =
cos^2 x cos x dx
and solve the equation du = cos x dx for u = sin x. Since sin 2 x + cos 2 x = 1, we then
have ∫ cos 3 x dx =
cos 2 x cos x dx =
( 1 − sin 2 x) cos x dx =
( 1 − u 2 ) du
= u −
u 3
sin 3 x + C
Example 13 Evaluate
dx
a^2 + x^2
dx, where a &= 0 is a constant.
Solution. Some simple algebra and an appropriate u-substitution will allow us to use
Formula 12 in Table 5.2.1. ∫ dx
a^2 + x^2
a(dx/a)
a^2 ( 1 + (x/a)^2 )
a
dx/a
1 + (x/a)^2
u = x/a du = dx/a
a
du
1 + u^2
a
tan − 1 u + C =
a
tan − 1 x a
+ C
The method of Example 13 leads to the following generalizations of Formulas 12, 13,
and 14 in Table 5.2.1 for a > 0:
∫ du
a^2 + u^2
a
tan − 1 u a
+ C (5)
du √ a^2 − u^2
= sin − 1 u a
+ C (6)
du
u
u^2 − a^2
a
sec − 1
u
a
∣ + C (7)
Example 14 Evaluate
dx √ 2 − x^2
Solution. Applying (6) with u = x and a =
2 yields ∫ dx √ 2 − x^2
= sin − 1 x √ 2
+ C
338 Chapter 5 / Integration
INTEGRATION USING COMPUTER ALGEBRA SYSTEMS
The advent of computer algebra systems has made it possible to evaluate many kinds of integrals that would be laborious to evaluate by hand. For example, a handheld calculator
T E C H N O LO GY M A ST E RY
If you have a CAS, use it to calculate the integrals in the examples in this sec- tion. If your CAS produces an answer that is different from the one in the text, then confirm algebraically that the two answers agree. Also, explore the effect of using the CAS to simplify the expres- sions it produces for the integrals.
evaluated the integral ∫ 5 x^2
( 1 + x)^1 / 3 dx^ =^
3 (x + 1 )^2 / 3 ( 5 x^2 − 6 x + 9 )
8
+ C
in about a second. The computer algebra system Mathematica , running on a personal computer, required even less time to evaluate this same integral. However, just as one would not want to rely on a calculator to compute 2 + 2, so one would not want to use a CAS to integrate a simple function such as f(x) = x^2. Thus, even if you have a CAS, you will want to develop a reasonable level of competence in evaluating basic integrals. Moreover, the mathematical techniques that we will introduce for evaluating basic integrals are precisely the techniques that computer algebra systems use to evaluate more complicated integrals.
"QUICK CHECK EXERCISES 5.3 ( See page 340 for answers. )
1. Indicate the u-substitution.
(a)
3 x 2 ( 1 + x 3 ) 25 dx =
u 25 du if u =
and du =.
(b)
2 x sin x^2 dx =
sin u du if u = and
du =.
(c)
18 x
1 + 9 x^2
dx =
u
du if u = and
du =.
(d)
1 + 9 x^2
dx =
1 + u^2
du if u = and
du =.
2. Supply the missing integrand corresponding to the indicated u-substitution.
(a)
5 ( 5 x − 3 ) − 1 / 3 dx =
du; u = 5 x − 3
(b)
( 3 − tan x) sec 2 x dx =
du;
u = 3 − tan x
(c)
x √ x
dx =
du; u = 8 +
x
(d)
e 3 x dx =
du; u = 3 x
EXERCISE SET 5.3 Graphing Utility C^ CAS
1–12 Evaluate the integrals using the indicated substitutions.
!
1. (a)
2 x(x^2 + 1 )^23 dx; u = x^2 + 1
(b)
cos 3 x sin x dx; u = cos x
2. (a)
x
sin
x dx; u =
x
(b)
3 x dx √ 4 x^2 + 5
; u = 4 x^2 + 5
3. (a)
sec 2 ( 4 x + 1 ) dx; u = 4 x + 1
(b)
y
1 + 2 y^2 dy; u = 1 + 2 y^2
4. (a)
sin πθ cos πθ dθ; u = sin πθ
(b)
( 2 x + 7 )(x 2
- 7 x + 3 ) 4 / 5 dx; u = x 2
- 7 x + 3
5. (a)
cot x csc 2 x dx; u = cot x
(b)
( 1 + sin t) 9 cos t dt; u = 1 + sin t
6. (a)
cos 2x dx; u = 2 x (b)
x sec^2 x^2 dx; u = x^2
7. (a)
x 2
1 + x dx; u = 1 + x
(b)
[csc(sin x)] 2 cos x dx; u = sin x
8. (a)
sin(x − π) dx; u = x − π
(b)
5 x^4
(x^5 + 1 )^2
dx; u = x 5
9. (a)
dx
x ln x
; u = ln x
(b)
e−^5 x^ dx; u = − 5 x
340 Chapter 5 / Integration
69–72 Solve the initial-value problems.!
dy
dx
5 x + 1, y( 3 ) = − 2
dy dx
= 2 + sin 3x, y(π/ 3 ) = 0
dy
dt
= −e 2 t , y( 0 ) = 6
dy dt
25 + 9 t^2
, y
π 30
73. (a) Evaluate
[x/
x^2 + 1 ] dx. (b) Use a graphing utility to generate some typical integral curves of f(x) = x/
x^2 + 1 over the interval (− 5 , 5 ).
74. (a) Evaluate
[x/(x^2 + 1 )] dx. (b) Use a graphing utility to generate some typical integral curves of f(x) = x/(x^2 + 1 ) over the interval (− 5 , 5 ).
75. Find a function f such that the slope of the tangent line at
a point (x, y) on the curve y = f(x) is
3 x + 1 and the curve passes through the point ( 0 , 1 ).
76. A population of minnows in a lake is estimated to be 100,
at the beginning of the year 2010. Suppose that t years after the beginning of 2010 the rate of growth of the population
p(t) (in thousands) is given by p′(t) = ( 3 + 0. 12 t)^3 / 2
. Es- timate the projected population at the beginning of the year
77. Let y(t) denote the number of E. coli cells in a container of nutrient solution t minutes after the start of an experiment. Assume that y(t) is modeled by the initial-value problem dy
dt
= (ln 2) 2 t/ 20 , y( 0 ) = 20
Use this model to estimate the number of E. coli cells in the container 2 hours after the start of the experiment.
78. Derive integration Formula (6). 79. Derive integration Formula (7). 80. Writing If you want to evaluate an integral by u-substitution, how do you decide what part of the inte- grand to choose for u? 81. Writing The evaluation of an integral can sometimes re- sult in apparently different answers (Exercises 67 and 68). Explain why this occurs and give an example. How might you show that two apparently different answers are actually equivalent?
"QUICK CHECK ANSWERS 5.
1. (a) 1 + x^3 ; 3x^2 dx (b) x^2 ; 2x dx (c) 1 + 9 x^2 ; 18x dx (d) 3x; 3 dx 2. (a) u−^1 / 3 (b) −u (c) 2
u (d) 1 3 e
u
5.4 THE DEFINITION OF AREA AS A LIMIT; SIGMA NOTATION
Our main goal in this section is to use the rectangle method to give a precise mathema- tical definition of the “area under a curve.”
SIGMA NOTATION
To simplify our computations, we will begin by discussing a useful notation for expressing lengthy sums in a compact form. This notation is called sigma notation or summation notation because it uses the uppercase Greek letter $ (sigma) to denote various kinds of sums. To illustrate how this notation works, consider the sum
1 2
in which each term is of the form k 2 , where k is one of the integers from 1 to 5. In sigma notation this sum can be written as (^) ∑ 5
k= 1
k 2
which is read “the summation of k 2 , where k runs from 1 to 5.” The notation tells us to form the sum of the terms that result when we substitute successive integers for k in the expression k^2 , starting with k = 1 and ending with k = 5. More generally, if f(k) is a function of k, and if m and n are integers such that m ≤ n, then (^) ∑n
k=m
f(k) (1)
denotes the sum of the terms that result when we substitute successive integers for k, starting
f ( k)
k = m
n
Ending value of k
This tells us to add
Starting value of k Figure 5.4.1 (^) with k = m and ending with k = n (Figure 5.4.1).
Sullivan Sullivan˙Chapter05 November 30, 2013 10:
362 Chapter 5 •^ The Integral
5.3 The Fundamental Theorem of Calculus
OBJECTIVES When you finish this section, you should be able to:
1 Use Part 1 of the Fundamental Theorem of Calculus (p. 363)
2 Use Part 2 of the Fundamental Theorem of Calculus (p. 365)
3 Interpret an integral using Part 2 of the Fundamental Theorem of Calculus (p. 365)
In this section, we discuss the Fundamental Theorem of Calculus, a method for finding integrals more easily, avoiding the need to find the limit of Riemann sums. The Fundamental Theorem is aptly named because it links the two branches of calculus: differential calculus and integral calculus. As it turns out, the Fundamental Theorem of Calculus has two parts, each of which relates an integral to an antiderivative.
Suppose f is a function that is continuous on a closed interval [ a , b ]. Then the
definite integral
∫ (^) b a f ( x ) d x exists and is equal to a real number. Now if x denotes
any number in [ a , b ], the definite integral
∫ (^) x
∫^ a^ f^ ( t )^ dt^ exists and depends on^ x.^ That is, x a f^ ( t )^ dt^ is a function of^ x , which we name^ I ,^ for “integral.”
I ( x ) =
∫ (^) x
a
f ( t ) dt
The domain of I is the closed interval [ a , b ]. The integral that defines I has a variable upper limit of integration x. The t that appears in the integrand is a dummy variable. Surprisingly, when we differentiate I with respect to x , we get back the original
function f. That is,
∫ (^) x a f^ ( t )^ dt is an antiderivative of^ f.
NEED TO REVIEW? Antiderivatives are discussed in Section 4.8, pp. 328--331.
y! f ( x )
f^ (
x^
"
h
)
a x x " h b
y
h x
f^ (
x )
DF (^) Figure 22
THEOREM Fundamental Theorem of Calculus, Part 1
Let f be a function that is continuous on a closed interval [ a , b ]. The function I defined by
I ( x ) =
∫ (^) x
a
f ( t ) dt
has the properties that it is continuous on [ a , b ] and differentiable on ( a , b ). Moreover,
I ′( x ) =
d
d x
[∫ (^) x
a
f ( t ) dt
]
= f ( x )
for all x in ( a , b ).
The proof of Part 1 of the Fundamental Theorem of Calculus is given in Appendix B.
However, if the integral
∫ (^) x a f^ ( t )^ dt^ represents area, we can interpret the theorem using geometry.
Figure 22 shows the graph of a function f that is nonnegative and continuous on
a closed interval [ a , b ]. Then I ( x ) =
∫ (^) x a f^ ( t )^ dt^ equals the area under the graph of^ f from a to x.
I ( x ) =
∫ (^) x
a
f ( t ) dt = the area under the graph of f from a to x
I ( x + h ) =
∫ (^) x + h
a
f ( t ) dt = the area under the graph of f from a to x + h
I ( x + h ) − I ( x ) = the area under the graph of f from x to x + h
I ( x + h ) − I ( x )
h
the area under the graph of f from x to x + h
h
