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Chapter 5
Riemann Sums and Definite Integrals
Area under a curve
Area under a curve = area between the function curve and the x axis
Area above the x axis is positive
Area below the x axis is negative
Ex: Velocity = -20 mph ∗ time = 1 hour.
Distance traveled = (-20)(1.5) = -30 miles
Ex: Velocity = 30 mph * time = 2 hours.
Distance traveled = 30(2) = 60 miles
V
(mph)
V
30
(mph)
2
1.5
time (hours)
-20
time (hours)
A negative velocity (going in opposite direction)
leads to a negative distance (in opposite
direction)
Dimensional Analysis – Just watch units!
Ex. Distance
V
(m/s)
40
𝑚
40 𝑠 ˙5 𝑠 = 200 𝑚
Distance traveled = vel * time
5
time (sec)
Ex. Cell phone cost
cost
(¢/min)
40
¢
40𝑚𝑖𝑛 ˙ 5 𝑚𝑖𝑛 = 200 ¢ = $2.00
Total cost = cost per time * time
5
time (min)
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Ex.
Gas efficiency
(miles/gal)
40
𝑚𝑖𝑙𝑒𝑠
˙5
𝑔𝑎𝑙
40
𝑔𝑎𝑙 = 200 𝑚𝑖𝑙𝑒𝑠
Distance traveled = gas efficiency * volume of gas
5
vol of gas (gal)
Ex.
Cost per pound
($/lb)
40
$
40 ˙5 𝑙𝑏 = $200
𝑙𝑏
Total cost = cost per lb* weight
5
weight (lb)
Interpretation of Area – Just watch units!
Ex. 5.1 #30
Sales
($millions per
year)
Total sales over the 10 year period in millions of dollars.
10
time (years)
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Finding Area Using Geometry Shapes
Ex.
8
5
2
6
A = rectangle + triangle
= 4*5 + ½ * 4 * 3
= 20 + 6
= 26
A = trapezoid
= ½(b1 + b2) h
=1/2 (5+8) 4
= 26
Ex.
A = ½ circle
9
A = ½ (𝜋𝑟 2 )= ½ (𝜋 ∗ 9)= 𝜋
2
-3
3
LRAM, RRAM, MRAM
Method for approximating area of a shape.
RAM = Rectangular Area Method
LRAM = Left Rectangular Area Method
RRAM = Right Rectangular Area Method
MRAM = Midpoint Rectangular Area Method
Divide up area shape into thin upright rectangles.
𝑏−𝑎
Divide up interval [a,b] into n subintervals of width ∆x = 𝑛
Add up areas of each individual rectangle to get approximate area of whole shape.
Width of each rectangle is ∆x. height of each rectangle is f(x) on either left or right edge of rectangle or
at center (midpoint) of rectangle.
y
3
2.5
2
1.5
1
0.5
x
1
Left
Right
2
3
Midpoint
4
5
6
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Ex. Using a function f(x) = x2+4x [0,3] use n = 6 subintervals
∆x =
𝑏−𝑎 3−0
= 6
𝑛
x
0
0.5
1
1.5
2
2.5
3
= 0.5 each rectangle will have width 0.5
f(x)
0
2.25
5
8.25
12
16.25
21
LRAM6 = (0.5)( f(0) + f(0.5) + f(1) + f(1.5) + f(2) + f(2.5) )
= (0.5) (0 + 2.25 + 5 + 8.25 + 12 + 16.25)
= (0.5) (43.75)
= 21.875
RRAM6 = (0.5) (f(0.5) + f(1) + f(1.5) + f(2) + f(2.5) + f(3) )
= (0.5) ( 2.25 + 5 + 8.25 + 12 + 16.25 + 21)
= (0.5) (64.75)
= 32.375
x
f(x)
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
2.25
2.5
2.75
3
0
1.0625
2.25
3.5625
5
6.5625
8.25
10.0625
12
14.0625
16.25
18.5625
21
MRAM6 = (0.5) ( f(.25) + f(.75) + f(1.25) + f(1.75) + f(2.25) + f(2.75) )
= (0.5) (1.0625 + 3.5625 + 6.5625 + 10.0625 + 14.0625 + 18.5625)
= (0.5) (53.875)
= 26.9375
Note the subscript on LRAM, RRAM, and MRAM denotes the number of subintervals used (n).
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Ex. Using a table. 5.1 #17
Estimate the distance using LRAM with n=12 subintervals. The velocity is measured every 5 sec.
Dist = Vel * time.
Time (min)
0
5
10
15
20
25
30
35
40
45
50
55
60
Velocity (m/sec)
1.0
1.2
1.7
2.0
1.8
1.6
1.4
1.2
1.0
1.8
1.5
1.2
0
LRAM12 = 10 (1.0 + 1.2 + 1.7 + 2.0 + 1.8 + 1.6 + 1.4 + 1.2 + 1.0 + 1.8 + 1.5 + 1.2) = 5220 m
Note: you can use rectangles of different widths.
LRAM, RRAM, MRAM comparisons
For an increasing function (f ‘ > 0), LRAM < MRAM < RRAM
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LRAM, RRAM for inc/dec function, concave up/concave down
Inc/dec function
Increasing function
(f ‘ >0)
concavity
CCU (f ‘’ > 0)
RAM used
LRAM
Approximation
Underrepresents
the real value
Increasing function
(f ‘ >0)
CCU (f ‘’ > 0)
RRAM
Overrepresents
the real value
Increasing function
(f ‘ >0)
CCD (f ‘’ < 0)
LRAM
Underrepresents
the real value
Increasing function
(f ‘ >0)
CCD (f ‘’ < 0)
RRAM
Overrepresents
the real value
Decreasing function CCU (f ‘’ > 0)
(f ‘ < 0)
LRAM
Overrepresents
the real value
Decreasing function CCU (f ‘’ > 0)
(f ‘ < 0)
RRAM
Underrepresents
the real value
Decreasing function CCD (f ‘’ < 0)
(f ‘ < 0)
LRAM
Overrepresents
the real value
Decreasing function CCD (f ‘’ < 0)
(f ‘ < 0)
RRAM
Underrepresents
the real value
Graph
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Equations to know:
Parabola
y = x2
y = -x2
y = x2+7
y = 4 – x2
Y = 4x – x2
Page |8
Circle x2 + y2 = r2
x2 + y2 = 16
y = √16 − 𝑥 2
y = − √16 − 𝑥 2
Sphere volume
V = 4/3 π r3
Cylinder volume
V = π r2 h
Other graphs of functions you should know:
𝑥
y=2+3
y = |x|
y = |x - 3|
y = - |x|
y = 2 - |x|
y = √4 − 𝑥 2
y = 1 + √4 − 𝑥 2
y=x
Page |9
Volumes of a shape by RAM
Divide up the object into slices (usually all slices are same thickness)
To estimate the volume of sphere or bowl or paraboloid, use cylinder slices with same
thickness (height) but different radii set by outside edge of shape. Add up the
volumes of each individual cylinder piece to get an approximate volume of the whole
shape.
See. 5.1 #22, 24, 25 p 272.
The volume of each cylinder slice is π r2 h.
h is the ∆x that is set depending on the number of subintervals desired.
The radius of each cylinder is set by the outside edge of the shape.
Use an equation for the radius.
The outside circle of the sphere can be modeled with an equation. x2 + y2 = r2
For example, if the sphere has radius 8 (R=8), then the radius of each cylinder
will vary according to r2 + y2 = 82 and r = √64 − 𝑦 2
If you are doing a whole sphere, then y goes from 0 to 16.
If you are doing a half sphere, then y goes from 0 to 8.
If you are doing a half sphere as a bowl that holds water and the water is filled to level 4, then y goes
from 0 to 4.
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Ex. 5.1 #22a. Approximate the volume of water in a hemispheric bowl of radius = 8 m, when the
water is filled to level 4 m. Use n=8 subintervals.
There are 8 subintervals, or 8 cylinder slices which total a height of 4 m. So ∆x =4/8 = 0.5 m.
The volume of a cylinder slice is π r2 h. h = 0.5 for all cylinders. r will vary.
r = √64 − 𝑦 2
Use y heights of 4 to 8 to get the top of the sphere which would be a bowl inverted and filled to a level
of 4 m.
y
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
r of cylinder slice
r = √64 − 𝑦 2
√48 =6.928
√43.75 =6.614
√39 =6.245
√33.75 = 5.809
√28 =5.292
√21.75 =4.664
√15 =3.873
√7.75 = 2.784
0.0
Volume of 8 cylinder slices
= 0.5 π [ 02 + 2.7842 + 3.8732 + 4.6642 + 5.2922 + 5.8092 + 6.2452 + 6.6142]
= 0.5 π [ 188.9984 ] = 296.878 m3
Check reasonableness: Volume of whole sphere is 4/3 π r3 = 4/3 π 83 = 2144.659. our bowl filled to 4 m
should be less than ¼ of the sphere. ¼ sphere = 536.1647 so ok, it’s reasonable.
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Riemann Sum
Sn=∑𝑛𝑘=1 𝑓(𝐶𝑘 ) ∙
Partial sum of n rectangles with height f(x) and width ∆x.
K is counter, tells how many times to keep adding rectangles to the sum.
Ex. F(x) = x2 on [0,2] use n = 4 subintervals.
𝑏−𝑎 2−0
∆x = = 𝑛 = 4 = 0.5 each ∆xk for all k values will be 0.5. each rectangle has width = 0.5.
Ck
0
.5
1
1.5
2
F(Ck)
0
.25
1
2.25
4
2
S4 = ∑4𝑘=1 𝐶𝑘 ∙ (0.5)
= 0.5 [ f(0) + f(.5) + f(1) + f(1.5) ]
= 0.5 [ 0 + .25 + 1 + 2.25 ] = 1.75
Limit to ∞ of Riemann Sum
Take lim 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑠𝑢𝑚
𝑛→∞
The number of subintervals increases to ∞. The width ∆x shrinks to being infinitesimally small. ∆x-> 0.
Adding up ∞ rectangles of infinitesimally small width. Each rectangle is very very slim but has the height
of f(x) at each x value. Add up all the teeny width rectangles. The approximation becomes perfect. The
approximate sum becomes the exact sum.
Integral Notation
The integral of f from a to b.
The integral from a to b of f(x).
Upper limit of integration
Integral sign
Lower limit of integration
𝑏
∫𝑎 𝑓(𝑥 )𝑑𝑥
Integrand
Variable of Integration
P a g e | 12
Dummy Variable
The variable of integration. The variable of integration can be any variable: x, y, z, m, t, θ
3
3
3
3
3
3
3
3
∫ 𝑥 𝑑𝑥 = ∫ 𝜃 𝑑𝜃 = ∫ 𝑡 𝑑𝑡 = ∫ 𝑚 𝑑𝑚
2
2
2
2
Changing the look from Summation Notation to Integral Notation
𝑛
2
lim ∑ 𝐶𝑘 ∆𝑥𝑘 𝑜𝑛 [0,2]
𝑛→∞
2
↔
∫ 𝑥 2 𝑑𝑥
0
𝑘=1
𝑛
5
2
lim ∑(𝐶𝑘 − 3𝐶𝑘 )∆𝑥𝑘 𝑜𝑛 [−7,5]
𝑛→∞
↔
−7
𝑘=1
𝑛
1
lim ∑
∆𝑥 𝑜𝑛 [2,3]
𝑛→∞
1 − 𝐶𝑘 𝑘
𝑘=1
∫ (𝑥 2 − 3𝑥)𝑑𝑥
3
↔
∫
2
1
𝑑𝑥
1−𝑥
P a g e | 13
Definite Integral - Using geometry shape
Finding area under the curve
Ex.
6
∫2 5 𝑑𝑥
= 5 (6-2) = 5 * 4 = 20
5
2
Ex.
6
6
∫2 (2𝑥 + 3) 𝑑𝑥
= rectangle + triangle
=7 (6-2) + ½ (15-7)(6-2)
=7*4+ ½ (8)(4) = 28+16=44
15
7
53
2
Ex.
6
3
∫0 𝑥 𝑑𝑥
= ½ (3) (3) = 9/2
3
3
Ex.
12
∫10 3𝑥 𝑑𝑥
= rectangle + triangle
= 30*2 + ½ (2)(6) = 60 + 6 = 66
36
30
10
12
P a g e | 14
Numeric Integration on Calculator fnInt
fnInt( f, x, a, b)
Math 9:fnInt(
fnInt (5, x, 2, 6) = 20
fnInt (2x+3,x,2,6) = 44
You can put function into y1 and refer to it inside fnInt.
fnInt( y1, x, a, b)
VARS Y-VARS 1:Funtion 1:Y1
Definite Integral - Using antiderivative
Fundamental Theorem of Calculus Part 2
Let derivative of F(x) be f(x)
𝑑
𝐹(𝑥) = 𝑓(𝑥)
𝑑𝑥
𝑏
∫𝑎 𝑓(𝑥) 𝑑𝑥 = F(b) - F(a)
Then
6
Ex. ∫2 5 𝑑𝑥
∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥)
= F(6) – F(2) = 5(6) – 5(2) = 20
f(x) = 5
F(x) = 5x
This works out to be:
For constant function, f(x) = constant, then
𝑏
∫ 𝐶 𝑑𝑥 = 𝐶(𝑏 − 𝑎)
𝑎
P a g e | 15
Finding the antiderivative
1.
f(x) = constant
Ex.
f(x) = C
f(x) = 5
F(x) = Cx
F(x) = 5x
2. Reverse power rule. Add 1 to exponent. Divide by new exponent.
f(x) = 2x
F(x) =
2𝑥 2
2
= x2
f(x) = 5x3 + 10x2 + 3x + 5
F(x) =
5𝑥 4
4
+
10𝑥 3
3
+
3𝑥 2
2
+ 5x
Reverse power rule for antiderivatives:
If
f (x) = axb then,
𝐹(𝑥) =
6
Ex. ∫2 (2𝑥 + 3) 𝑑𝑥
𝑎 𝑥 𝑏+1
𝑏+1
= F(6) – F(2) = [ (6)2+3*6 ] – [ (2)2 + 3*2 ] = [36+18] – [4+5] = 54 -10 = 44
f(x) = 2x+3
F(x) = x2+3x
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Definite Integral - Using antiderivative – more practice:
1
Ex. 5.2 p. 282 #7 ∫−2 5 𝑑𝑥
7
Ex. 5.2 p. 282 #8 ∫3 −20 𝑑𝑥
√18
= F(1) – F(-2) = 5(1) – 5(-2) = 5+10 = 15
f(x) = 5
F(x) = 5x
fnInt(5,x,-2,1) = 15
= F(7) – F(3) =[ -20(7)] –[ -20(3)] = -140 + 60 = -80
f(x) = -20
F(x) = -20x
fnInt(-20,x,3,7) = -80
Ex. 5.2 p. 282 #12 ∫√2 √2 𝑑𝑟
= F(√18) – F(√2) =[ √2(√18)] –[ √2(√2)] = √36 − 2 = 6-2 = 4
f(x) = √2
F(x) = √2 x
fnInt(√2 ,x, √2, √18) = 4
−1 𝜋
Ex. 5.2 p. 282 #10 ∫−4
2
𝑑𝜃
𝜋
𝜋
= F(-1) – F(-4) =[2 (-1)] –[ 2 (-4)] =−
f(θ) =
𝜋
2
𝜋
F(x) = 2 θ
𝜋
2
- (-2π) =
3𝜋
2
≈ 4.712
𝜋
fnInt(2 ,x,-4,-1) = 4.712
π
Ex. ∫0 sin 𝑥 𝑑𝑥 =F(π) – F(0) = [−cos(π)] − [− cos(0)] = [− (−1)] – [−1] = 1 – (−1) = 2
f(x) = sin 𝑥
F(x) =− cos 𝑥 x
fnInt(sin x,x,0,π) = 2
MUST BE IN RADIANS!!!!
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Endpoint Playfulness
Rule 1: switch endpoints and switch signs on area
𝑏
𝑎
∫𝑎 𝑓(𝑥) 𝑑𝑥 = − ∫𝑏 𝑓(𝑥) 𝑑𝑥
Rule 2: same endpoint, zero area
𝑎
∫𝑎 𝑓(𝑥)𝑑𝑥 = 0
Rule 3: constant multiplier can move outside integral
𝑏
𝑏
∫𝑎 𝐶𝑓(𝑥) 𝑑𝑥 = 𝐶 ∫𝑎 𝑓(𝑥) 𝑑𝑥
𝑏
𝑏
∫𝑎 3𝑓(𝑥) 𝑑𝑥 = 3 ∫𝑎 𝑓(𝑥) 𝑑𝑥
Rule 4: can add or subtract parts
𝑏
𝑏
𝑏
𝑏
𝑏
𝑏
∫𝑎 [𝑓(𝑥) + 𝑔(𝑥)]𝑑𝑥 = ∫𝑎 𝑓(𝑥)𝑑𝑥 + ∫𝑎 𝑔(𝑥) 𝑑𝑥
∫𝑎 [𝑓(𝑥) − 𝑔(𝑥)]𝑑𝑥 = ∫𝑎 𝑓(𝑥)𝑑𝑥 - ∫𝑎 𝑔(𝑥) 𝑑𝑥
Rule 5: add two adjacent areas and get whole area
𝑏
𝑐
𝑐
2
5
5
∫𝑎 𝑓(𝑥)𝑑𝑥 + ∫𝑏 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(𝑥) 𝑑𝑥
∫0 𝑓(𝑥)𝑑𝑥 + ∫2 𝑓(𝑥)𝑑𝑥 = ∫0 𝑓(𝑥) 𝑑𝑥
Ex. P. 290 5.3 #1
2
Given:
∫1 𝑓(𝑥)𝑑𝑥 = −4
5
5
∫1 𝑓(𝑥)𝑑𝑥 = 6 ∫1 𝑔(𝑥) 𝑑𝑥 = 8
2
1.a. ∫2 𝑔(𝑥)𝑑𝑥 = 0
Rule 2
1
5
2
2
1.b. ∫5 𝑔(𝑥)𝑑𝑥 = − ∫1 𝑔(𝑥) 𝑑𝑥 = - 8
Rule 1
1.c. ∫1 3𝑓(𝑥)𝑑𝑥 = 3 ∫1 𝑓(𝑥)𝑑𝑥 = 3(-4) = -12
5
5
Rule 3
2
1.d. ∫2 𝑓(𝑥)𝑑𝑥 = ∫1 𝑓(𝑥)𝑑𝑥 - ∫1 𝑓(𝑥)𝑑𝑥 = 6 - -4 = 10
5
5
Rule 5
5
1.e. ∫1 [𝑓(𝑥) − 𝑔(𝑥)]𝑑𝑥 = ∫1 𝑓(𝑥)𝑑𝑥 − ∫1 𝑔(𝑥)𝑑𝑥 = 6 - 8 = -2
5
5
5
5
Rule 4
5
1.f. ∫1 [4𝑓(𝑥) − 𝑔(𝑥)]𝑑𝑥 = ∫1 4𝑓(𝑥)𝑑𝑥 − ∫1 𝑔(𝑥)𝑑𝑥 =4 ∫1 𝑓(𝑥)𝑑𝑥 − ∫1 𝑔(𝑥)𝑑𝑥
= 4*6 - 8 = 24 – 8 = 16
Rule 3,4
P a g e | 18
Average value of a function
Finding Derivative of an Integral
Areas by Trapezoids. Trapezoidal Rule