calculus a chapter 5 -...
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Calculus AChapter 5
Faten Said Abu-Shoga
December 7, 2017
4.8 Integration
Starting with a function f , we want to find a function F whose derivative is f . If such afunction F exists, it is called an antiderivative of f . We will see in the next chapter thatantiderivatives are the link connecting the two major elements of calculus: derivativesand definite integrals.
Antiderivatives
Definition 1. (Antiderivative) A function F is an antiderivative of f on an interval Iif F ′(x) = f(x) for all x ∈ I.
Example 2. Find an antiderivative of
1. f(x) = 4x3
2. f(x) = cos x
Theorem 3. If F is an antiderivative of f on an interval I, then the most generalantiderivative of f on I is F (x) + C where C is an arbitrary constant.
Antiderivative linearity rules
1. Constant Multiple Rule: kf(x) has antiderivative kF (x) + C, k a constant.
2. Negative Rule: −f(x) has antiderivative −F (x) + C.
3. Sum or Difference Rule: f(x)± g(x) has antiderivative F (X)±G(x) + C.
Definition 4. (Indefinite integral) The collection of all antiderivatives of f is calledthe indefinite integral of f with respect to x, and is denoted by
∫f(x)dx.
The symbol∫
is an integral sign. The function f is the integrand of the integral, and xis the variable of integration.
Remark 5. If F is an antiderivative of f , then∫f(x) dx = F (x) + C
where C is an arbitrary constant called the constant of integration.
Example 6. Find
1.∫
1√xdx
2.∫
cscx cotx dx
3.∫
sec2 x dx
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Integral Formulas
1.∫xn dx = xn+1
n+1+ C, n 6= −1.
2.∫
1√xdx = 2
√x+ C.
3.∫
sinx dx = − cosx+ C.
4.∫
cosx dx = sinx+ C.
5.∫
sec2 x dx = tanx+ C.
6.∫
secx tanx dx = secx+ C.
7.∫
csc2 x dx = − cotx+ C.
8.∫
cscx cotx dx = − cscx+ C.
Example 7. Find
1.∫
(2x3 − 5x+ 7) dx =
2.∫ √
x+ x23 + 1
x− 13dx =
3.∫
t√t−√t
tdt =
4.∫
3 cos 5x− sin 3x dx =
5.∫
cosx tanx dx =
6.∫
cosxsin2 x
dx =
2
Chapter 5 Integrals
5.3 Integrable and Nonintegrable Functions
Definition 8. If∫ b
af(x) dx exists, then we say that f is integrable over [a, b]. Otherwise
we say that f is nonintegrable over [a, b].
Theorem 9. (Continuous functions are integrable) If a function is continuousover [a, b], then f is integrable over [a, b].
Properties of Definite Integrals
Theorem 10. If f and g are integrable over the interval [a, b], then
1.∫ b
af(x) dx = −
∫ a
bf(x) dx.
2.∫ a
af(x) dx = 0.
3.∫ b
akf(x) dx = k
∫ b
af(x) dx.
4.∫ b
a(f(x)± g(x)) dx =
∫ b
af(x) dx±
∫ b
ag(x) dx.
5.∫ b
af(x) dx+
∫ c
bf(x) dx =
∫ c
af(x) dx.
6. (Max-Min Inequality): If f has maximum value max f and minimum value min fon [a, b] , then
min f · (b− a) ≤∫ b
a
f(x) dx ≤ max f · (b− a)
.
7. If f(x) ≥ g(x) on [a, b] then∫ b
af(x) dx ≥
∫ b
ag(x) dx.
8. If f(x) ≥ 0 on [a, b] then∫ b
af(x) dx ≥ 0.
Example 11. Let∫ 4
1f(x) dx = 3 and
∫ 1
−1 f(x) dx = 1 and∫ 4
1h(x) dx = 3, then
1.∫ 1
4f(x) dx =
3
2.∫ 4
1(3f(x)− 4h(x)) dx =
3.∫ 4
−1 3f(x) dx =
Example 12. Find upper and lower bounds for∫ 2
01
x4+1dx =
Example 13. Show that the value of∫ 1
0
√1 + cos x dx is less than or equal to
√2.
Area Under the Graph of a Nonnegative Function
Definition 14. If y = f(x) is nonnegative and integrable over a closed interval [a, b] ,then the area under the curve y = f(x) over [a, b] is the integral of f from a to b,
A =
∫ b
a
f(x) dx.
Some rules
1.∫ ba k dx = k(b− a), where k is constant.
2.∫ ba x dx = b2
2 −a2
2 .
3.∫ ba x
2 dx = b3
3 −a3
3 .
Example 15. Find the area between the curve y = x2 and the x-axis over[−2, 3].
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Example 16. Find the integral by using area∫ 1
−11 +
√1− x2 dx.
Average Value of a Continuous Function Revisited
Definition 17. If f is integrable on [a, b] , then its average value on [a, b], also called its mean, is
av(f) =1
b− a
∫ b
a
f(x) dx.
Example 18. Find the average value of f(x) =√
4− x2 on [−2, 2].
0.1 5.4 The Fundamental Theorem of Calculus
Theorem 19. (The Mean Value Theorem for Definite Integrals)If f is continuous on [a, b], then at some point c in [a, b],
f(c) =1
b− a
∫ b
a
f(x) dx.
Example 20. Show that if f is continuous on [a, b], a 6= b, and if∫ b
a
f(x) dx = 0
then f(x) = 0 at least once in [a, b].
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Fundamental Theorem, Part 1
Theorem 21. If f is continuous on [a, b] , then F (x) =∫ xa f(t) dt is
continuous on [a, b] and differentiable on (a, b) and its derivative is f(x) :
F ′(x) =d
dx
∫ x
a
f(t) dt = f(x)
Another formulas
F ′(x) =d
dx
∫ u(x)
a
f(t) dt = f(u)du
dx
F ′(x) =d
dx
∫ u(x)
v(x)
f(t) dt = f(u)du
dx− f(v)
dv
dx
Example 22. Find F ′(x),
1. F (x) =∫ x2 (t3 + 3) dt.
2. F (x) =∫ 5
x (3t sin t) dt.
3. F (x) =∫ x23x−1 cos t dt.
Fundamental Theorem, Part 2 (The Evaluation Theorem)
Theorem 23. If f is continuous over [a, b] and F is any antiderivative off on [a, b] , then ∫ b
a
f(x) dx = F (x)|ba = F (b)− F (a).
Example 24. Evaluate the following integrals:
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1.∫ 1
0 x2 +√x dx =
2.∫ π0 1 + sin x dx =
3.∫ 8
1(x
13+1)(2−x
23 )
x13
dx =
4.∫ π0 | cosx| dx =
Total area between the curve y = f(x) and the x−axis
Theorem 25. 1. If y = f(x) ≥ 0 on [a, b], then the area between thecurve y = f(x) and the x−axis over [a, b] is
A =
∫ b
a
f(x) dx.
2. If y = f(x) ≤ 0 on [a, b], then the area between the curve y = f(x)and the x−axis over [a, b] is
A = −∫ b
a
f(x) dx.
3. If y = f(x) change signs on [a, b], then the area between the curvey = f(x) and the x−axis over [a, b] is
A =
∫ b
a
|f(x)| dx.
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Example 26. Find the area between the curve y = x2−4x and the x−axisover [0, 5].
5.5 Indefinite Integrals and the Substitution Method
Theorem 27. (The Substitution Rule)If u = g(x) is a differentiable function whose range is an interval I and fis continuous on I, then∫
f(g(x))g′(x) dx =
∫f(u)du.
Example 28. Evaluate the following integrals:
1.∫
2x(x2 + 1)−4 dx
2.∫ √
x sin2(x32 − 1) dx
3.∫ √
sinx cos3 x dx
8
4.∫ √
x4
x3−1 dx
5.∫
3x5√x3 + 1 dx
6.∫
x3√x2+1
dx
7.∫ √
x−1x5 dx
8.∫
1√3x+1
dx
9.∫
tan7(x2)sec2(x2) dx
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5.6 Substitution and Area between Curves
Theorem 29. (Substitution in definite integrals) If g is continuouson the interval [a, b] and f is continuous on the range of g, then∫ b
a
f(g(x))g′(x) dx =
∫ g(b)
g(a)
f(u) du
Example 30. Evaluate the following integrals:
1.∫ π
2
0 (2 + tan(x2)) sec2(x2) dx
2.∫ π
2
0cosx
(2+sinx)3 dx
3.∫ π2
4
π2
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cos√x√
x sin√xdx
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4. Use the integral∫ 2
1 f(z) dz = 3 to find the value of the integral∫ 1
12
1
x2f(
1
x) dx
5.∫ π
3
0tanx√2 secx
dx
Definite Integrals of Symmetric Functions
Theorem 31. Let f be continuous on the symmetric interval [−a, a].
1. If f is even, then∫ a−a f(x) dx = 2
∫ a0 f(x) dx.
2. If f is odd, then∫ a−a f(x) dx = 0.
Example 32. Evaluate∫ π
2
−π2
2 sin θ1+cos2 θ dθ
Area Between Curves
Definition 33. If f and g are continuous with f(x) ≥ g(x) throughout[a, b], then the area of the region between the curves y = f(x) and y = g(x)
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from a to b is
A =
∫ b
a
[f(x)− g(x)] dx
Example 34. Find the area of the region enclosed by y = x4, y = 16.
Example 35. Find the area of the region enclosed by y = 4 − x2, y =2− x, x = −2, and x = 2.
Example 36. Find the area of the region in the first quadrant that isbounded above by y = 2
√x and below by the x−axis and the line , y = x−3.
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Example 37. Find the area of the region enclosed by y = sin πx2 and y = x.
Integration with Respect to y
The area of the region between the curves x = f(y) and x = g(y) fromy = a to y = b is
A =
∫ b
a
|f(y)− g(y)| dy
Example 38. Find the area of the region enclosed by x = y2, x = y + 2.
Example 39. Find the area of the region enclosed by the lines x = 0, y =3, and the curve x = 2y2.
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Example 40. Find the area of the region enclosed by the curves x+y2 = 3and 4x+ y2 = 0.
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Chapter 6Applications of Definite Integrals
6.1 Volumes Using Cross-Sections
Solids of Revolution: The Disk Method
The solid generated by rotating (or revolving) a plane region about an axisin its plane is called a solid of revolution. To find the volume of a solid, weneed only observe that the cross-sectional area A(x) is the area of a diskof radius R(x), the distance of the planar region’s boundary from the axisof revolution. The area is then
A(x) = π(radius)2 = π[R(x)]2.
So the definition of volume in this case gives
Volume by Disks for Rotation About the x-axis
V =
∫ b
a
A(x) dx =
∫ b
a
π[R(x)]2 dx.
This method for calculating the volume of a solid of revolution is oftencalled the disk method Volume by Disks for Rotation About the x-axis
Example 41. The region between the curve y =√x, 0 ≤ x ≤ 4, and the
x-axis is revolved about the x-axis to generate a solid. Find its volume.
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Example 42. Find the volume of the solid generated by revolving the re-gion bounded by y =
√x and the lines y = 1, x = 4 about the line y = 1.
To find the volume of a solid generated by revolving a region betweenthe y-axis and a curve x = R(y), c ≤ y ≤ d, about the y-axis, we use thesame method with x replaced by y. In this case, the area of the circularcross-section is
A(y) = π(radius)2 = π[R(y)]2.
So the definition of volume in this case gives
Volume by Disks for Rotation About the y-axis
V =
∫ d
c
A(y) dy =
∫ d
c
π[R(y)]2 dy.
Example 43. Find the volume of the solid generated by revolving the re-gion bounded by the curve x = y
2 , 1 ≤ y ≤ 4, and the y-axis about they-axis.
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Example 44. Find the volume of the solid generated by revolving the re-gion between the parabola x = y2 + 1 and the line x = 3 about the linex = 3.
Solids of Revolution: The Washer Method
If the region we revolve to generate a solid does not border on or cross theaxis of revolution, the solid has a hole in it. The cross-sections perpendic-ular to the axis of revolution are washers instead of disks. The dimensionsof a typical washer are
Outer radius: R(x)
Inner radius: r(x)
The washer’s area is
A(x) = π[[R(x)]2 − π[r(x)]2 = π([R(x)]2 − [r(x)]2).
Consequently, the definition of volume in this case gives
Volume by Washers for Rotation About the x-axis
V =
∫ b
a
A(x) dx =
∫ b
a
π([R(x)]2 − [r(x)]2) dx.
This method for calculating the volume of a solid of revolution is calledthe washer method.
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Example 45. The region bounded by the curve y = x2 + 1 and the liney = −x+3 is revolved about the x-axis to generate a solid. Find the volumeof the solid.
To find the volume of a solid formed by revolving a region about they-axis, we use the same procedure as before, but integrate with respect toy instead of x. In this situation the line segment sweeping out a typicalwasher is perpendicular to the y-axis (the axis of revolution), and the outerand inner radii of the washer are functions of y.
Volume by Washers for Rotation About the x-axis
V =
∫ d
c
A(y) dy =
∫ d
c
π([R(y)]2 − [r(y)]2) dy.
Example 46. The region bounded by the parabola y = x2 and the liney = 2x in the first quadrant is revolved about the y-axis to generate asolid. Find the volume of the solid.
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Example 47. Find the volume of the solid generated by revolving the re-gion bounded by the parabola y = x2 and the line y = 1 about
1. the line y = 1
2. the line x = 1
3. the line y = 2
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6.3 Arc Length
Length of a Curve y = f(x)
Definition 48. If f ′ is continuous on [a, b] , then the length (arc length) ofthe curve y = f(x) from the point A = (a, f(a)) to the point B = (b, f(b))is the value of the integral
L =
∫ b
a
√1 + [f ′(x)]2 dx =
∫ b
a
√1 + [
dy
dx]2 dx
Example 49. Find the length of the curves
1. y = 13(x2 + 1)
32 from x = 0 to x = 3.
2. y = x3
12 + 1x where 1 ≤ x ≤ 4.
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3. Find a curve with a positive derivative through the point (1, 0) whoselength integral is
L =
∫ 2
1
√1 +
1
x4dx
Length of a Curve x = f(y)
Definition 50. If f ′ is continuous on [c, d] , then the length (arc length) ofthe curve x = f(y) from the point A = (f(a), a) to the point B = (f(b), b)is the value of the integral
L =
∫ d
c
√1 + [f ′(y)]2 dy =
∫ d
c
√1 + [
dx
dy]2 dy
Example 51. Find the length of the curves
1. x = y4
4 + 18y2 from y = 1 to y = 2.
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2. x = y3
6 + 12y from y = 2 to y = 3.
6.4 Areas of Surfaces of Revolution
Revolution About the x-Axis
Definition 52. If the function f(x) ≥ 0 is continuously differentiable on[a, b], the area of the surface generated by revolving the graph of y = f(x)about the x-axis is
S =
∫ b
a
2πf(x)√
1 + [f ′(x)]2 dx =
∫ b
a
2πy
√1 + (
dy
dx)2 dx
Example 53. Find the area of the surface generated by revolving the curvey = 2
√x, 1 ≤ x ≤ 2, about the x-axis.
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Example 54. Find the area of the surface generated by revolving the curvey =
∫ x0
√t2 − 1, 1 ≤ x ≤
√5, about the x-axis.
Revolution About the y-Axis
Definition 55. If the function x = g(y) ≥ 0 is continuously differentiableon [c, d], the area of the surface generated by revolving the graph of x =g(y) about the y-axis is
S =
∫ d
c
2πg(y)√
1 + [g′(y)]2 dy =
∫ d
c
2πy
√1 + (
dx
dy)2 dy
Example 56. The line segment x = 1 − y, 0 ≤ y ≤ 1, is revolved aboutthe y-axis to generate the cone. Find its lateral surface area.
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Example 57. Find the area of the surface generated by revolving the curvex =
∫ y0 tan t dt, 0 ≤ y ≤ π
3 , about the y-axis.
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