Contents 1 Chapter 1 1 2 Chapter 2 1 3 Chapter 3 1 4 Chapter 4 1 5 Applications of Integrals 2 5.1 Area and Definite Integrals............................ 2 5.1.1 Area of a Region Between Two Curves................. 2 5.1.2 Changing Variable to Find Area..................... 8 5.2 Disk Method and Washer Method........................ 12 5.2.1 Disk Method............................... 12 5.2.2 The Washer Method........................... 13 5.3 The Shell Method................................. 18 5.3.1 Deriving the Shell Method Formula................... 18 5.3.2 The Shell Method Formula........................ 19 5.3.3 Example of the Shell Method...................... 19 5.4 Arc length..................................... 26 5.4.1 Definition of Arc Length......................... 26 5.4.2 Surface Area................................ 29 1 Chapter 1 See Math 151 Notes 2 Chapter 2 See Math 151 Notes 3 Chapter 3 See Math 151 Notes 4 Chapter 4 See Chapter 4 Notes Math 152 1
5 Applications of Integrals 5.1 Area and Definite Integrals Let s look a the graph of two functions f(x) and g(x) in the cartesian plane. We can use the antiderivative to find a area between the two curves. 5.1.1 Area of a Region Between Two Curves If f(x) and g(x) are two continuous functions and g(x) f(x) and [a, b], then the area of the region bounded by the functions f(x) and g(x), and the vertical lines x = a and x = b, can be found by the following definite integral: Area = b a [f(x) g(x)]dx This in the case we use vertical rectangles with a base of dx IF we use the case we use horizontal rectangles with a base of dx, we the following formula. Area = b a [f(y) g(y)]dy 2
Now, let s use integration to find the areas of some regions. Example 1 Find the area of a region that is bounded by the following functions g(x) = x and f(x) = 2 x 2 First, find the intersection of the functions. f(x) = g(x) 2 x 2 = x 3
x 2 x 2 = 0 (x 2)(x + 1) = 0 x 2 = 0 or x + 1 0 x = 2 or x = 1 Now, make a sketch of f(x) and g(x) with their intersections. The area of the region can be found by the following antiderivative. Area= 2 1 [(2 x 2 ) x]dx = 2 1 [ x 2 x + 2]dx = [ 1 3 x3 1 2 x2 + 2x] 2 1 = [ 1 3 (1)3 1 2 (1)2 + 2(1)] [ 1 3 ( 2)3 1 2 ( 2)2 + 2( 2)] = [ 1 3 1 2 + 2] [8 3 2 4] = ( 10 7 )6 + 3 = 27 6 = 9 2 Example 2 Find the area of a region that is bounded by the following functions g(x) = x 2 + 4x + 1 and f(x) = x + 1 First, find the intersection of the functions. f(x) = g(x) x + 1 = x 2 + 4x + 1 4
x 2 3x = 0 x(x 3) = 0 x = 0 or x 3 0 x = 0 or x = 3 Now, make a sketch of f(x) and g(x) with their intersections. The area of the region can be found by the following antiderivative. Area= 3 0 [( x2 + 4x + 1) (x + 1)]dx = 3 0 [ x2 + 3x]dx = [ 1 3 x3 + 3 2 x2 ] 3 0 = [ 1 3 (3)3 + 3 2 (3)2 ] [ 1 3 (0)3 + 3 2 (0)2 ] = [ 9 + 27 2 ] [0 + 0] = 18 2 + 27 2 = 9 2 Example 3 Find the area of a region that is bounded by the following functions g(x) = x 3 and f(x) = x First, find the intersection of the functions. f(x) = g(x) x = x 3 5
x 3 x = 0 x(x 2 1) = 0 x(x + 1)(x 1) = 0 x = 0 or x + 1 0 or x 1 = 0 x = 0 or x = 1 0r x = 1 Now, make a sketch of f(x) and g(x) with their intersections. The area of the region can be found by the following antiderivative. Area= 0 1 [(x3 x]dx + 1 0 [(x x3 ]dx = [ 1 3 x3 1 2 x2 ] 0 1 + [1 2 x2 1 3 x3 ] 1 0 = [ 1 4 (0)4 1 2 (0)2 ] [ 1 4 ( 1)4 1 2 ( 1)2 ] [ 1 2 (1)2 1 4 (1)4 ] [ 1 2 (0)2 1 4 (0)4 ] = [0 + 0] [ 1 4 + 1 2 ] + [1 2 1 4 ] + [0 + 0] = [ 1 4 ] + 1 4 = 1 2 6
Example 4 Find the area of a region that is bounded by the following functions g(x) = cos(x) and f(x) = sin(x) First, find the intersection of the functions. f(x) = g(x) sin(x) = cos(x) x = π 4, 5π 4 Points of Intersection( π 4, 2 2 ) and (π 4, 2 2 ) Now, make a sketch of f(x) and g(x) with their intersections. The area of the region can be found by the following antiderivative. Area= 5π 4 π [sin(x) cos(x)]dx 4 = [ cos(x) sin(x)] 5π 4 π 4 = [ cos( 5π 4 ) sin(5π 4 )] [ cos(π 4 ) sin(π 4 )] = [ ( 2 2 ) ( 2 2 )] [ 2 2 2 2 ] = 2 ( 2) = 2 2 7
5.1.2 Changing Variable to Find Area Example 5 Find the area of a region that is bounded by the following functions g(x) = x 2 and f(x) = x First, find the intersection of the functions. f(x) = g(x) x = x 2 ( x) 2 = (x 2 ) 2 x = x 4 x 4 x = 0 x(x 3 1) = 0 x(x 1)(x 2 + x + 1) = 0 x = 0 or x 1 = 0 x = 0 or x = 1 Now, make a sketch of f(x) and g(x) with their intersections. The area of the region can be found by the following antiderivative. Area= 0 1 [ x x 2 ]dx = 0 1 [x 1 2 x 2 ]dx = [ 2 3 x 3 2 1 3 x3 ] 1 0 = [ 2 3 (1) 3 2 1 3 (1)3 ] [ 2 3 (0) 3 2 1 3 (0)3 ] = 2 3 1 3 + 0 = 1 3 8
The area of the region can be found by the following antiderivative. Area= 0 1 [ y y 2 ]dx = 0 1 [y 1 2 y 2 ]dx = [ 2 3 y 3 2 1 3 y3 ] 1 0 = [ 2 3 (1) 3 2 1 3 (1)3 ] [ 2 3 (0) 3 2 1 3 (0)3 ] = 2 3 1 3 + 0 = 1 3 Example 6 Find the area of a region that is bounded by the following functions g(y) = 3 y 2 and f(y) = y + 1 First, find the intersection of the functions. f(y) = g(y) y + 1 = 3 y 2 y 2 + y 2 = 0 (y + 2)(y 1) = 0 y + 2 = 0 or y 1 = 0 y = 2 or y = 1 9
Now, make a sketch of f(y) and g(y) with their intersections. The area of the region can be found by the following antiderivative. Area= 2 1 [(3 y 2 ) (y + 1)]dy = 1 2 [3 y2 y 1]dy = 1 2 [ y2 y + 2]dy = [ 1 3 y3 1 2 y2 + 2y] 1 2 = [ 1 3 (1)3 1 2 (1)2 +2(1)] [ 1 3 ( 2)3 1 2 ( 2)2 +2( 2)] = [ 1 3 1 2 +2] [ 8 3 2 4] = 7 6 ( 10 3 ) = 9 2 Now, find the area using dx and vertical rectangles. 10
Area= 1 2 [(x 1) ( 3 x)]dx + 2 3 [ 3 x ( 3 x)]dx = 1 2 [(x 1) + (3 x) 2]dx 1 + 2 3 [2(3 x) 1 2]dx = [ 1 2 x2 + x 2 3 (3 x) 2] 3 2 1 [4 3 (3 x) 3 2] 3 2 = [ 1 2 (2)2 2 2 3 (3 2) 3 2] [ 1 2 ( 1)2 ( 1) 2 3 (3 ( 1)) 3 2] 4 3 (3 3) 3 2 [ 4 3 (3 2) 3 2 = [2 2 2 3 ] [1 2 + 1 16 3 ] 0 + 4 3 = 2 3 + 23 6 + 4 3 = 27 6 = 9 2 11
5.2 Disk Method and Washer Method 5.2.1 Disk Method If a region in a plane is revolved around an axis of revolution, the resulting figure will resemble a disk. The volume of the disk can be represented by the following formula. V = πr 2 x The Volume of Solid = lim n π The Disk Method Formulas n [ ] 2 x b R(x) = π [R(x)] 2 dx i=1 Horizontal Axis of Revolution Volume: V = π b a [R(x)]2 dx Vertical Axis of Revolution Volume: V = π b a [R(y)]2 dy Example 1 Find the volume of the solid formed by revolving the region by f(x) = 2 x 2 and g(x) = 1 about the line y = 1. a R(x) = f(x) g(x) = 2 x 2 1 = 1 x 2 Now, find the points of intersection. g(x) = f(x) 2 x 2 = 1 x 2 = 1 x 2 1 = 0 (x 1)(x + 1) = 0 x = 1 or x = 1 12
V = π 1 1 [R(x)]2 dx = π 1 1 (1 x2 ) 2 dx = π 1 1 (1 x2 )(1 x 2 )dx = π 1 1 (1 2x2 + x 4 )dx = π [x 2 3 x3 + 1 5 x5 ] 1 1 = π[1 2 3 (1)3 + 1 5 (1)5 ] π[ 1 2 3 ( 1)3 + 1 5 ( 1)5 ] = π[1 2 3 + 1 5 ] π[ 1 + 2 3 1 5 ] = π[ 8 15 ] π[ 8 15 ] = 16π 15 5.2.2 The Washer Method V = π b a ( [R(x)] 2 [r(x)] 2) Example 2 Find the volume of the solid formed by revolving the region bounded the graphs of f(x) = x and g(x) = x 2 about the x-axis. Now, find the intersections of f(x) and g(x) x = x 2 ( x) 2 = (x 2 ) 2 x = x 4 13
x 4 x = 0 x(x 3 1) = 0 x(x 1)(x 2 + x + 1) = 0 x = 0 or x = 1 (0, 0) or (1, 1) V = π 1 0 V = π 0 1 ( [R(x)] 2 [r(x)] 2) dx ( ( x) 2 (x 2 ) 2) dx ( ) x x 4 dx V = π 0 1 V = π [ 1 2 x2 1 5 x5 ] 1 0 V = π[ 1 2 (1)2 1 5 (1)5 ] π[ 1 2 (0)2 1 5 (0)5 ] V = π[ 1 2 1 3π 5 ] π[0 0] = 10 Example 3 Find the volume of the solid formed by revolving the region bounded the graphs of f(x) = sin(x) and g(x) = 0 about the x-axis. Now, find the intersections of f(x) and g(x) sin(x) = 0 ( sin(x)) 2 = (0) 2 sin(x) = 0 x = 0 or x = π 14
(0, 0) or (π, 0) Now, use the Disk Method to find the volume of the solid. R(x) = x 0 = x V = π 1 0 [R(x)]2 dx V = π 1 0 [ sin(x)] 2 dx V = π 1 0 sin(x)dx V = π [ cos(x)] π 0 V = π[ cos(π) ( cos(0))] V = π[ ( 1) + 1] = 2π Example 4 Find the volume of the solid generated by the region bounded the graphs of y = 2x 2, y = 0, and x = 2 using the following axis of revolution. a) the x-axis Using the Disk Method make a sketch of the region bound by the curves given in the problem. 15
Now, use the Disk Method to find the volume of the solid. R(x) = 2x 2 0 = 2x 2 V = π 2 0 [R(x)]2 dx V = π 2 0 [2x2 ] 2 dx V = π 2 0 4x4 dx V = π [ 4 5 x5 ] 2 0 V = π[ 4 5 (2)5 ] π[ 4 5 (0)5 ] V = π 128 5 0 V = 128 5 π a) the y-axis Using the Washer Method make a sketch of the region bound by the curves given in the problem. y = 2x 2 y 2 = x2 y 2 = x 2 x = y 2 16
Use the Washer Method to find the volume of the solid. First, find the inner and outer radii. r(x) = y 2 0 = y 2 R(x) = 2 0 = 2 V = π 8 0 [[R(y)]2 [r(y)] 2 ]dx V = π 8 0 [[2]2 [ y V = π 8 0 [4 y 2 ]dx 2 ]2 ]dx V = π [4y y2 4 ] 8 0 V = π[4(8) 82 02 4 ] π[4(0) V = π[32 16] π[0 0] V = 16π 0 V = 16π 4 ] 17
5.3 The Shell Method 5.3.1 Deriving the Shell Method Formula When we use the Shell Method, we rotate a rectangle in a circular path which will result in a cylinder shaped object. To compute the volume of this hollow cylinder, we will slice the cylinder an open the object up so that it is now the shape of a rectangular solid. (See illustration below:) Now, let s compute the volume of this rectangular solid where l = 2πr, h = h(x), and w = dx V = l w h = 2πr h(x) dx Now, convert use integration to compute the volume. V = b a 2πρ(x)h(x)dx = 2π b a ρ(x)h(x)dx 18
5.3.2 The Shell Method Formula Shell Method: Horizontal Axis of Revolution. V = 2π b a ρ(y)h(y)dy Shell Method: Vertical Axis of Revolution. V = 2π b a ρ(x)h(x)dx 5.3.3 Example of the Shell Method Example 1 Find the volume of solid formed by a solid formed by revolving the region bounded by the graphs of y = x 2, y = 0, x = 1 (revolved about y) ρ(x) = x 0 = x h(x) = x 2 + 1 0 = x 2 + 1 V = 2π b a ρ(x)h(x)dx V = 2π 1 0 x(x2 + 1)dx V = 2π b a (x3 + x)dx 19
V = 2π [ x4 4 + x2 2 ] 1 0 V = 2π[ 14 4 + 12 2 ] 2π[04 V = 2π[ 1 4 + 1 2 ] 2π[0 + 0] V = 2π[ 3 4 ] 0 V = 3π 2 Example 2 4 + 02 2 ] Find the volume of solid formed by a solid formed by revolving the region bounded by the graphs of y = 1 x, y = 0, x = 1 (revolved about y) ρ(x) = x 0 = x h(x) = 1 x 0 = 1 x V = 2π b a ρ(x)h(x)dx V = 2π 1 0 x(1 x)dx V = 2π b a (x x2 )dx V = 2π [ x2 2 x3 3 ] 1 V = 2π[ 12 2 13 3 ] 2π[02 V = 2π[ 1 2 1 3 ] 2π[0 + 0] V = 2π[ 1 6 ] 0 = π 3 0 2 + 03 3 ] 20
Example 3 Find the volume of solid formed by a solid formed by revolving the region bounded by the graphs of y = x 2 + 4, y = 8, x = 0 (revolved about y) ρ(x) = x 0 = x h(x) = 8 (x 2 + 4) = 4 x 2 V = 2π b a ρ(x)h(x)dx = 2π 1 0 x(4 x2 )dx V = 2π b a (4x x3 )dx V = 2π [2x 2 x4 4 ] 2 V = 2π[2 2 2 24 4 ] 2π[2 02 + 04 V = 2π[2 4 4] 2π[0 + 0] V = 2π[8 4] 0 V = 8π 0 4 ] 21
Example 4 Find the volume of solid formed by a solid formed by revolving the region bounded by the graphs of y = x 2, y = 0, x = 2 (revolved about y) ρ(x) = x 0 = x h(x) = x 2 0 = x 2 V = 2π b a ρ(x)h(x)dx = 2π 1 0 x(x2 )dx V = 2π 2 0 (x3 )dx V = 2π [ x4 4 ] 2 0 V = 2π[ 24 4 ] 2π[04 4 ] V = 2π[ 16 4 ] 2π[0] V = 2π[4] 0 V = 8π 22
Example 5 Find the volume of solid formed by a solid formed by revolving the region bounded by the graphs of y = x 2 and y = 4x x 2 (revolved about x=2) First, find the intersection of the two functions. x 2 = 4x x 2 0 = 4x x 2 x 2 0 = 4x 2x 2 0 = (2 x)2x 2 x = 0 or 2x = 0 x = 2 or x = 0 ρ(x) = 4 x 0 = 4 x h(x) = 4x x 2 x 2 = 4x 2x 2 V = 2π b a ρ(x)h(x)dx V = 2π 2 0 (2 x)(4x 2x2 )dx V = 2π 2 0 [8x 4x2 4x 2 + x 3 ]dx V = 2π 2 0 [8x 8x2 + x 3 ]dx V = 2π [4x 2 8x3 3 + x4 2 ] 2 0 23
V = 2π[4(2) 2 8(2)3 3 + 24 2 ] 2π[8(0)2 4(0) 3 + 04 V = 2π[16 64 3 + 16 2 ] 2π[0] V = 2π[ 16 3 + 8] 0 V = 2π[ 8 3 ] V = 16π 3 2 ] Example 6 Find the volume of solid formed by a solid formed by revolving the region bounded by the graphs of y = x, x = 4, and y = 0(revolved about x=6) ρ(x) = 6 x 0 = 6 x h(x) = x 0 = x V = 2π b a ρ(x)h(x)dx V = 2π 4 0 (6 x) xdx V = 2π 4 0 [6 x x x]dx V = 2π 4 0 [6x 1 2 x 3 2]dx V = 2π [6 2 3 x 3 2 2 5 x 2] 5 4 0 V = 2π [4x 3 2 2 5 x 5 2] 4 0 V = 2π[4(4) 3 2 2 5 (4) 5 2] 2π[4(0) 3 2 2 5 (0) 5 2] 24
V = 2π[32 64 5 ] 2π[0] V = 2π[ 116 5 ] 0 V = 132π 5 25
5.4 Arc length The distance between two points is given by the formula: d = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 A ref1ciable curve is one that has finite length. Next, we will compute the length of this curve using the distance formula and a summation Arc Length: s = n (xi x i 1 ) 2 + (y i y i 1 ) 2 i=1 n s = ( xi ) 2 + ( y i ) 2 i=1 n s = (( x) 2 + ( y i ) x 2 x 2 i=1 i n s = (1 + ( y i ) x 2 ) x 2 i=1 i n s = (1 + ( y i ) x 2 ) x i i=1 i Now, let s observe the value of s as n n s = lim (1 + ( y i ) n x 2 ) x i i i=1 Note that y i x is the slope or derivative of the curve, thus f (c i ) = y i i x i n s = lim (1 + (f (c i )) 2 ) x i n i=1 s = a b (1 + (f (c i )) 2 )dx 5.4.1 Definition of Arc Length Let the function given by y = f(x) represent a smooth curve on the interval [a, b]. The arc length of between the values a and b is given by: s = a b 1 + (f (x)) 2 dx 26
Example 1 Find the arc length of the function over the given interval. y = 2x 3 2 + 1: [0, 4] First, find the derivative of y y = 2x 3 2 + 1 y = 3 2 2x 3 2 1 y = 3x 1 2 Now, find the arc length s = a b 1 + (f (x)) 2 dx s = 0 9 1 + (3x 1 2) 2 dx s = 0 9 1 + 9xdx Let u = 1 + 9x du = 9dx 1 9du = dx 4 4 s = 1 9 0 udu s = 1 9 0 u 1 2du s = 1 9 [2 3 u 3 2] 4 0 s = 2 27 [(1 + 9x) 2] 3 4 0 s = 2 27 [1 + 9(4)] 3 2 2 27 [1 + 9(0)] 3 2 s = 2 27 [37] 3 2 2 27 [1] 3 2 v 16.7.074 16.03 Example 2 Find the arc length of the function over the given interval. y = ln(cos(x)): [0, π 4 ] First, find the derivative of y 27
y = ln(cos(x)) y = sin(x) cos(x) y = sin(x) cos(x) y = tan(x) Now, find the arc length s = a b 1 + (f (x)) 2 dx s = π 4 0 1 + ( tan(x))2 dx s = π 4 0 1 + tan2 (x)dx s = π 4 0 sec2 (x)dx s = π 4 0 sec(x)dx s = [ln[sec(x) + tan(x)] π 4 0 s = ln[sec( π 4 ) + tan(π 4 )] ln[sec(0) + tan(0)] s = ln[ 2 + 1] ln[1] s = ln[ 2 + 1] Example 3 Find the arc length of the function over the given interval. y = ln(sin(x)): [ π 3, 2π 3 ] First, find the derivative of y y = ln(sin(x)) y = cos(x) sin(x) y = cot(x) Now, find the arc length s = a b 1 + (f (x)) 2 dx s = 2π 3 π 1 + (cot(x))2 dx 3 s = 2π 3 π 1 + cot2 (x)dx 3 s = 2π 3 π csc2 (x)dx 3 28
s = 2π 3 π csc(x)dx 3 s = [ ln[csc(x) + cot(x)] 2π 3 π 3 s = ln[csc( 2π 3 ) + cot(2π 3 )] [ ln[sec(π 3 ) + tan(π 3 ]] s = ln[ 3 3 + 2 3 3 ] + ln[ 3 3 + 2 3 3 ] s = ln[ 3 3 ] + ln[3 3 3 ] s = ln[ 3 3 ] + ln[ 3] s 1.16 5.4.2 Surface Area Definition of Surface Area Let y = f(x) have a continuous derivative on the interval [a, b]. The area of the surface of revolution formed by revolving the graph of f about a horizontal or vertical axis is given by the following formula (where r(x) is the distance between f(x) and the axis of revolution) S = 2π b a r(x) 1 + [f (x)] 2 dx If x = g(y) is a function of y and r(y) is the distance between g(y) and the axis of revolution then: S = 2π b a r(y) 1 + [g (y)] 2 dy 29
Example 4 Set up and solve the definite integral for the surface area generated by revolving the curve point about the y-axis. f(x) = 2x 2 : [0, 2] First, find the derivative of f(x) f(x) = 2x 2 f (x) = 4x r(x) = x 0 = x S = 2π b a r(x) 1 + [f (x)] 2 dx S = 2π 2 0 x 1 + [4x] 2 dx S = 2π 2 0 x 1 + 16x 2 dx Let u = 1 + 16x 2 du = 32xdx 1 32du = xdx 2 S = 2π 1 32 0 udu S = π 2 16 0 udu S = π 2 16 0 u 1 2du S = π 16 [2 3 u 3 2] 2 0 S = π 24 [1 + 16x2 ] 3 2 2 0 S = π 24 [1 + 16(2)2 ] 3 2 π 24 [1 + 16(0)2 ] 3 2 30
S = π 24 [65] 3 2 π 24 [1] 3 2 S 68.6.1 S 68.5 Example 5 Set up and solve the definite integral for the surface area generated by revolving the curve point about the x-axis. f(x) = x 3 : [0, 1] First, find the derivative of f(x) f(x) = x 3 f (x) = 3x 2 r(x) = x 3 0 = x 3 S = 2π b a r(x) 1 + [f (x)] 2 dx S = 2π 1 0 x 1 + [3x 2 ] 2 dx S = 2π 1 0 x 1 + 9x 4 dx Let u = 1 + 9x 4 du = 36x 3 dx 1 36du = xdx 1 S = 2π 1 36 0 udu S = π 1 18 0 udu S = π 1 18 0 u 1 2du 31
S = π 18 [2 3 u 3 2] 2 0 S = π 27 [1 + 9x4 ] 3 2 2 0 S = π 27 [1 + 9(1)4 ] 3 2 π 27 [1 + 9(0)4 ] 3 2 S = π 27 [10] 3 2 π 27 [1] 3 2 S 3.68.12 S 3.56 32