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Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral5.1 The Area ProblemGoal: To find the
area under the graph of f(x)
and above the x-axis between x = a and x = b
How:
Problem:
Use rectangles to approximate the area
curved sides
Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral( ) [ ]2 on the interval 1,5f x x=
using 4 rectangles
and the endpoint
of each interval to get
the height of each rectangle
right
( ) ( ) ( ) ( )Area 2 1 3 1 4 1 5 1f f f f≈ ⋅ + ⋅ + ⋅ + ⋅
Area 4 9 16 25≈ + + +
2Area 54 units≈
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Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral( ) [ ]2 on the interval 1,5f x x=
using 4 rectangles
and the endpoint
of each interval to get
the height of each rectangle
left
Area 1 4 9 16≈ + + +
2Area 30 units≈
( ) ( ) ( ) ( )Area 1 1 2 1 3 1 4 1f f f f≈ ⋅ + ⋅ + ⋅ + ⋅
Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral( ) [ ]2 on the interval 1,5f x x=
using 4 rectangles
and the
of each interval to get
the height of each rectangle
midpoint
9 25 49 81Area
4 4 4 4≈ + + +
2164Area units4
≈
( ) ( ) ( ) ( )3 5 7 92 2 2 2Area 1 1 1 1f f f f≈ ⋅ + ⋅ + ⋅ +
⋅
2Area 41 units≈
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Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral( ) [ ]2 on the interval 1,5f x x=
using 8 rectangles
and the endpoint
of each interval to get
the height of each rectangle
right
Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral( ) [ ]2 on the interval 1,5f x x=
using 8 rectangles
and the endpoint
of each interval to get
the height of each rectangle
right
1 9 25 49 81Area 4 9 16 25
2 4 4 4 4
≈ + + + + + + +
21 164Area 54 units2 4
≈ +
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )3 5 7 91 1 1 1 1 1 1 12 2 2 2 2 2
2 2 2 2 2 2Area 2 3 4 5f f f f f f f f≈ ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ +
⋅
295Area units2
≈
2Area 47.5 units≈
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Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral
Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral
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11/13/2013
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Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral
Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral
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11/13/2013
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Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral
as the number
of rectangles
increases, accuracy
increases
Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral
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Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral5.3 The Definite Integral
Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral
( )b
a
f x dx∫
Definite Integral
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Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral
* and i i i
x x x a i x= = + ∆
( )*1
limn
in
i
f x x→∞
=
∆∑
b ax
n
−∆ =
a a x+ ∆ 2a x+ ∆ 3a x+ ∆
�
a i x+ ∆
0x 1x 2x 3x ix
using right endpoints we can simnplify the Riemann sum
( ) ( ) ( ) ( )1 2lim i nn
x f x f x f x f x→∞
∆ + + + + + � �
( )*1
limn
in
i
f x x→∞
=
∆∑ ( )1
limn
b a
nn
i
b af a i
n
−
→∞=
−= + ⋅∑
5.2 Riemann Sum
Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral
Appendix E : Sigma Notation page A37
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Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral( ) [ ]2 on the interval 1,5f x x=
Find the exact area using the Riemann sum ( ) ( )1
lim
b n
b a
nn
ia
b af x dx f a i
n
−
→∞=
−= + ⋅∑∫
( )5
2 4
11
4lim 1
n
nn
i
x dx f in→∞ =
= + ⋅∑∫ ( )2
4
1
4lim 1
n
nn
i
in→∞ =
= + ⋅∑ ( )2
2
8 16
1
4lim 1
n
i in nn
in→∞ == + +∑
2
2
8 16
1 1 1
4lim 1
n n n
i in nn
i i in→∞ = = =
= + +
∑ ∑ ∑
11
2
21
4 8li 1
16m
n
i
n
n
n
i ini
ni
n ==→ =∞
= + +
∑∑ ∑
( ) ( )( )2
4 8 1 16li
2
21 1
6m
n n n nn
n nn n n
→∞
+ = + ⋅ + ⋅
+
+ ( ) ( )( )2 3
1 1 2 132 64lim 4
2 6n
n n n n n
n n→∞
+ + + = + ⋅ + ⋅
32lim 4n→∞
= +
16
2n
n⋅
( )1
2
n + 64+
32
3n 2
n⋅
( ) ( )22 3 1
1 2 1
6
n n
n n+ +
+ +3
( )22
32 2 3 116 16lim 4
3n
n nn
n n→∞
+ ++ = + +
2 2
644 16
3
16 3 2l
2 3
3im
n n n n→∞
= + + + +
+
0 0 0
644 16
3= + +
60 64
3
+=
124
3=
1= 41
3Too much work. We find an easier way in section 5.3
Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral
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Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral
Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral
Not all functions are integrable
the definite integral measures net area
Area under the x-axis is considered negative
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Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral
( )1. 0a
a
f x dx =∫
Properties :
( ) ( )2. a b
b a
f x dx f x dx= −∫ ∫
( ) ( ) ( ) ( )3. b b b
a a a
f x g x dx f x dx g x dx± = ± ∫ ∫ ∫
( ) ( )4. b b
a a
cf x dx c f x dx=∫ ∫
( )5. b
a
cdx c b a= −∫
( ) ( ) ( )6. If , then b c b
a a c
a c b f x dx f x dx f x dx< < = +∫ ∫ ∫
Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite Integral
2
2 33
−3 −4
−1
( )2
0
. i f x dx =∫ 4
( )5
0
. ii f x dx =∫ ( ) ( )2 5
0 2
f x dx f x dx+∫ ∫ 4 6= + = 10
( )7
5
. iii f x dx =∫ −3
( )9
0
. iv f x dx =∫ ( ) ( )5 9
0 5
f x dx f x dx+∫ ∫ ( )10 8= + − = 2
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Math 103 – Rimmer
5.1 The Area Problem
5.3 The Definite IntegralEvaluate the integral by interpreting
it in terms of areas.
( )5
2
0
1 25 x dx+ −∫21 25y x= + −
only the right upper
quarter circle
( )22
1 25x y+ − =
21 54
rπ= +
=25π
+ 54
12
0
6x dx−∫ 6y x= −
shifted 6
units to the right
y x= ( )( )1
6 62
( )( )1
6 62
18 18
= 36
( )center 0,1 , radius 5=
5
