# CONSTRUCTING ANTIDERIVATIVES ANALYTICALLY What Is an Antiderivative of f (x) = 0 ? A function whose derivative is zero everywhere on an interval must have a horizontal tangent line at every point of its graph, and the only way this can happen is if the function is constant. Alternatively, if we think of the derivative as a velocity, and if the velocity is always zero, then the object is standing still; the position function is constant. A rigorous proof of this result using the definition of the derivative is surprisingly subtle. (See the Constant Function Theorem.)

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6.2 CONSTRUCTING ANTIDERIVATIVES ANALYTICALLY What Is an Antiderivative of f (x) = 0 ? A function whose derivative is zero everywhere on an interval must have a horizontal tangent line at every point of its graph, and the only way this can happen is if the function is constant. Alternatively, if we think of the derivative as a velocity, and if the velocity is always zero, then the object is standing still; the position function is constant. A rigorous proof of this result using the definition of the derivative is surprisingly subtle. (See the Constant Function Theorem.)

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What Is the Most General Antiderivative of f ? We know that if a function f has an antiderivative F, then it has a family of antiderivatives of the form F (x) + C, where C is any constant. You might wonder if there are any others. To decide, suppose that we have two functions F and G with F′ = f and G′ = f: that is, F and G are both antiderivatives of the same function f. Since F′ = G′ we have (G − F)′ = 0. But this means that we must have G − F = C, so G(x) = F (x) + C, where C is a constant. Thus, any two antiderivatives of the same function differ only by a constant.

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The Indefinite Integral All antiderivatives of f (x) are of the form F (x) + C. We introduce a notation for the general antiderivative that looks like the definite integral without the limits and is called the indefinite integral:

() It is important to understand the difference between

() The first is a number and the second is a family of functions. The word “integration” is frequently used for

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the process of finding the antiderivative as well as of finding the definite integral. The context usually makes clear which is intended.

What Is an Antiderivative of f (x) = k ? If k is a constant, the derivative of kx is k, so we have

() Using the indefinite integral notation, we have If k is constant,

()

Finding Antiderivatives Finding antiderivatives of functions is like taking square roots of numbers: if we pick a number at random, such as 7 or 493, we may have trouble finding its square root without a calculator. But if we happen to pick a number such as 25 or 64, which we know is a perfect square, then we can find its square root exactly. Similarly, if we pick a function which we recognize as a derivative, then we can find its antiderivative easily. For example, to find an antiderivative of f (x) = x, notice that 2x is the derivative of x2; this tells us that x2 is an antiderivative of 2x. If we divide by 2, then we guess that

() To check this statement, take the derivative of x2/2:

() What about an antiderivative of x2? The derivative of x3 is 3×2, so the derivative of x3/3 is 3×2/3 = x2. Thus,

()

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The pattern looks like

() (We assume n ≠ −1, or we would have x0/0, which does not make sense.) It is easy to check this formula by differentiation:

() In indefinite integral notation, we have shown that

() What about when n = −1? In other words, what is an antiderivative of 1/x? Fortunately, we know a function whose derivative is 1/x, namely, the natural logarithm. Thus, since

() we know that

() If x < 0, then ln x is not defined, so it can’t be an antiderivative of 1/x. In this case, we can try ln(−x):

() so

()

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This means ln x is an antiderivative of 1/x if x > 0, and ln(−x) is an antiderivative of 1/x if x < 0. Since |x| = x when x > 0 and |x| = −x when x < 0, we can collapse these two formulas into:

() Therefore

() Since the exponential function is its own derivative, it is also its own antiderivative; thus

() Also, antiderivatives of the sine and cosine are easy to guess. Since

() we get

()

Example 1 Find ∫ (3x + x2) dx. Solution We know that x2/2 is an antiderivative of x and that x3/3 is an antiderivative of x2, so we expect

() You should always check your antiderivatives by differentiation—it’s easy to do. Here

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()

The preceding example illustrates that the sum and constant multiplication rules of differentiation work in reverse:

Theorem 6.1: Properties of Antiderivatives: Sums and Constant Multiples In indefinite integral notation,

1.∫ (f (x) ± g(x)) dx = ∫ f (x) dx ± ∫ g(x) dx 2.∫ cf(x) dx = c ∫ f (x) dx.

In words, 1.An antiderivative of the sum (or difference) of two functions is the sum (or difference) of their antiderivatives. 2.An antiderivative of a constant times a function is the constant times an antiderivative of the function.

These properties are analogous to the properties for definite integrals given in Section 5.4, even though definite integrals are numbers and antiderivatives are functions.

Example 2 Find ∫ (sin x + 3 cos x) dx. Solution We break the antiderivative into two terms:

() Check by differentiating:

()

Using Antiderivatives to Compute Definite Integrals As we saw in Section 5.3, the Fundamental Theorem of Calculus gives us a way of calculating definite integrals. Denoting F (b) − F (a) by , the theorem says that if F′ = f and f is continuous, then

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()

To find , we first find F, and then calculate F (b) − F (a). This method of computing definite

integrals gives an exact answer. However, the method only works in situations where we can find the antiderivative F (x). This is not always easy; for example, none of the functions we have encountered so far is an antiderivative of sin(x2). Example 3

Compute using the Fundamental Theorem.

Solution Since F (x) = x3 is an antiderivative of f (x) = 3×2,

() gives

()

Notice in this example we used the antiderivative x3, but x3 + C works just as well because the constant C cancels out:

()

Example 4

Compute exactly.

Solution We use the Fundamental Theorem. Since F (θ) = tan θ is an antiderivative of f (θ) = 1/cos2 θ, we get

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()

Exercises and Problems for Section 6.2 EXERCISES 1. If p′(x) = q(x), write a statement involving an integral sign giving the relationship between p(x) and q(x).

1. If u′(x) = v(x), write a statement involving an integral sign giving the relationship between u(x) and v(x).

2. Which of (I)-(V) are antiderivatives of f (x) = ex/2?

I.ex/2

II.2ex/2

III.2e(1+x)/2

IV.2ex/2 + 1 V.

1. Which of (I)-(V) are antiderivatives of f (x) = 1/x?

I.ln x II.−1/x2

III.ln x + ln 3 IV.ln(2x) V.ln(x + 1)

1. Which of (I)-(V) are antiderivatives of

()

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I.−2 sin x cos x II.2 cos2 x sin2 x III.sin2 x IV.−cos2 x V.2 sin2 x + cos2 x

In Exercises 6-21, find an antiderivative. 6. f (x) = 5

1. f (t) = 5t

2. f (x) = x2

3. g(t) = t2 + t

10.

11.

12.

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1. h(t) = cos t

14.

15.

1. f (z) = ez

2. g(t) = sin t

3. f (t) = 2t2 + 3t3 + 4t4

19.

20.

21.

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In Exercises 22-33, find the general antiderivative. 22. f (t) = 6t

1. h(x) = x3 − x

2. f (x) = x2 − 4x + 7

25.

1. r(t) = t3 + 5t − 1

2. f (z) = z + ez

3. g(x) = sin x + cos x

4. h(x) = 4×3 − 7

30.

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1. p(t) = 2 + sin t

32.

33.

In Exercises 34-41, find an antiderivative F (x) with F′(x) = f (x) and F (0) = 0. Is there only one possible solution? 34. f (x) = 3

1. f (x) = 2x

2. f (x) = −7x

3. f (x) = 2 + 4x + 5×2

38.

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1. f (x) = x2

40.

1. f (x) = sin x

In Exercises 42-55, find the indefinite integrals. 42. ∫ (5x + 7) dx

43.

1. ∫ (2 + cos t) dt

2. ∫ 7ex dx

3. ∫ (3ex + 2 sin x) dt

4. ∫ (4ex − 3 sin x) dx

48.

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1. ∫ (x + 3)2 dx

50.

51.

1. ∫ (ex + 5) dx

2. ∫ t3(t2 + 1) dt

54.

55.

In Exercises 56-65, evaluate the definite integrals exactly [as in ], using the Fundamental Theorem,

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and numerically : 56.

57.

58.

59.

60.

61.

62.

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63.

64.

65.

In Exercises 66-75, decide if the statement is True or False by differentiating the right-hand side. 66.

67.

68.

69.

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70.

71.

72.

73.

74.

75.

PROBLEMS 76. Use the Fundamental Theorem to find the area under f (x) = x2 between x = 0 and x = 3.

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1. Find the exact area of the region bounded by the x-axis and the graph of y = x3 − x.

78.

Calculate the exact area above the graph of and below the graph of y = cos x. The curves intersect at .

1. Find the exact area of the shaded region in Figure 6.27 between y = 3×2 − 3 and the x-axis.

Figure 6.27

1. (a) Find the exact area between f (x) = x3 − 7×2 + 10x, the x-axis, x = 0, and x = 5.

(b)

Find exactly and interpret this integral in terms of areas.

1. Find the exact area between the curve y = ex − 2 and the x-axis for 0 ≤ x ≤ 2.

2. Find the exact area between the curves y = x2 and y = 2 − x2.

3. Find the exact area between the x-axis and the graph of f (x) = (x − 1)(x − 2)(x − 3).

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1. The area under on the interval 1 ≤ x ≤ b is equal to 6. Find the value of b using the Fundamental Theorem.

2. Use the Fundamental Theorem to find the value of b if the area under the graph of f (x) = 8x between x = 1 and x = b is equal to 192. Assume b > 1.

3. Find the exact positive value of c which makes the area under the graph of y = c(1 − x2) and above the x-axis equal to 1.

4. Sketch the parabola and the curve y = sin x, showing their points of intersection. Find the exact area between the two graphs.

5. Find the exact average value of on the interval 0 ≤ x ≤ 9. Illustrate your answer on a graph of

.

1. (a) What is the average value of f (t) = sin t over ? Why is this a reasonable answer?

(b) Find the average of f (t) = sin t over .

90.

Let where Q(3) = 12. Given that , find Q(8).

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1. Water is pumped into a cylindrical tank, standing vertically, at a decreasing rate given at time t minutes by

() The tank has radius 5 ft and is empty when t = 0. Find the depth of water in the tank at t = 4.

1. A car moves along a straight line with velocity, in feet/second, given by

() (a) Describe the car’s motion in words. (When is it moving forward, backward, and so on?)

(b) The car’s position is measured from its starting point. When is it farthest forward? Backward?

(c) Find s, the car’s position measured from its starting point, as a function of time.

1. In drilling an oil well, the total cost, C, consists of fixed costs (independent of the depth of the well) and marginal costs, which depend on depth; drilling becomes more expensive, per meter, deeper into the earth. Suppose the fixed costs are 1,000,000 riyals (the riyal is the unit of currency of Saudi Arabia), and the marginal costs are

() where x is the depth in meters. Find the total cost of drilling a well x meters deep.

1. A helicopter rotor slows down at a constant rate from 350 revs/min to 260 revs/min in 1.5 minutes. (a) Find the angular acceleration (i.e. change in rev/min) during this time interval. What are the units of this acceleration?

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(b) Assuming the angular acceleration remains constant, how long does it take for the rotor to stop? (Measure time from the moment when speed was 350 revs/min.)

(c) How many revolutions does the rotor make between the time the angular speed was 350 revs/min and stopping?

95.

Use the fact that (xx)′ = xx(1 + ln x) to evaluate exactly: .

1. Assuming that ∫g(x) dx = G(x) + C, where G(4) = 9, G(6) = 4, and G(9) = 6, evaluate the definite integral: (a)

(b)

(c)

For Problems 97-99, let ∫g(x) dx = G(x) + C. Which of (I)-(III), if any, is equal to the given integral? 97. ∫g(2x) dx

I.0.5G(0.5x) + C II.0.5G(2x) + C III.2G(0.5x) + C

98.

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∫cos(G(x)) g(x) dx I.sin(G(x)) g(x) + C II.sin(G(x)) G(x) + C III.sin(G(x)) + C

1. ∫ xg(x) dx

I.G(x2) + C II.xG(x) + C III.

Strengthen Your Understanding In Problems 100-101, explain what is wrong with the statement.

100.

101.

For all n, .

In Problems 102-103, give an example of: 102. Two different functions F (x) and G(x) that have the same derivative.

1. A function f (x) whose antiderivative F (x) has a graph which is a line with negative slope.

Are the statements in Problems 104-112 true or false? Give an explanation for your answer. 104. An antiderivative of is 2(x + 1)3/2.

105.

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An antiderivative of 3×2 is .

1. An antiderivative of 1/x is ln |x| + ln 2.

2. An antiderivative of is .

3. ∫ f (x) dx = (1/x) ∫ xf(x) dx.

4. If F (x) is an antiderivative of f (x) and G(x) = F (x) + 2, then G(x) is an antiderivative of f (x).

5. If F (x) and G(x) are two antiderivatives of f (x) for −∞ < x < ∞ and F (5) > G(5), then F (10) > G(10).

6. If F (x) is an antiderivative of f (x) and G(x) is an antiderivative of g(x), then F (x) ⋅ G(x) is an antiderivative of f (x) ⋅ g(x).

7. If F (x) and G(x) are both antiderivatives of f (x) on an interval, then F (x) − G(x) is a constant function.

Additional Problems In Problems 113-114, evaluate the integral using f (x) 4x−3.

AP113.

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AP114.

AP115. If An is the area between the curves y = x and y = xn, show that as n → ∞ and explain this result graphically.

AP116. Consider the area between the curve y = ex − 2 and the x-axis, between x = 0 and x = c for c > 0. Find the value of c making the area above the axis equal to the area below the axis.

AP117. The origin and the point (a, a) are at opposite corners of a square. Calculate the ratio of the areas of the two parts into which the curve divides the square.

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