The Definite Integral Part 2
The Definite Integral
Part 2
Recall: Definite Integral
If the function π π₯ is defined on π, π , then the definite integral of π from π to π is
οΏ½ π π₯π
πππ₯ = lim
π β0οΏ½π ππ β βπ₯π
π
π=1
provided the limit exists.
Theorems 1. If π π₯ is continuous on π, π , then π π₯ is integrable on
π, π .
2. If π π₯ has only finitely many points of discontinuity on π, π and if there is a positive number π such that βπ β€ π π₯ β€ π for all π₯ in π, π (that is, π π₯ is bounded on π, π ), then π π₯ is integrable on π, π .
3. If π π₯ is continuous and assumes both positive and negative values on π, π , then
οΏ½ π π₯π
πππ₯ = area above π₯βaxis β area below π₯βaxis
Basic Properties of Integrals a) Zero Width Interval
οΏ½ π π₯π
πππ₯ = 0
b) Constant Multiple
οΏ½ π β π π₯π
πππ₯ = ποΏ½ π π₯
π
πππ₯
c) Sum/Difference
οΏ½ π π₯ Β± π π₯π
πππ₯ = οΏ½ π π₯
π
πππ₯ Β± οΏ½ π π₯
π
πππ₯
d) Additivity
οΏ½ π π₯π
πππ₯ + οΏ½ π π₯
π
πππ₯ = οΏ½ π π₯
π
πππ₯
Basic Properties of Integrals (continued)
e) Max-Min Inequality: If π π₯ has maximum value max π and minimum value min π on π, π , then
min π β π β π β€ οΏ½ π π₯π
πππ₯ β€ max π β π β π
f) Domination
β π π₯ β₯ π π₯ on π, π βΉ β« π π₯ππ ππ₯ β₯ β« π π₯π
π ππ₯
β π π₯ β₯ 0 on π, π βΉ β« π π₯ππ ππ₯ β₯ 0
g) Order
οΏ½ π π₯π
πππ₯ = βοΏ½ π π₯
π
πππ₯
Basic Definite Integrals
β’ β« π₯ππ ππ₯ = π2
2β π2
2, π < π
β’ β« πππ ππ₯ = π π β π , π any constant
β’ β« π₯2ππ ππ₯ = π3
3β π3
3, π < π
Example 1
Evaluate:
a) β« 347 ππ₯
b) β« 7π₯150 ππ₯
c) β« 3π₯ + 114 ππ₯
d) β« 5π₯221 2β ππ₯
Example 1 (continued)
Solution (a):
οΏ½ 34
7ππ₯
= βοΏ½ 37
4ππ₯
= β 3 7 β 4 = β9
Example 1 (continued)
Solution (b):
οΏ½ 7π₯15
0ππ₯
= 7οΏ½ π₯15
0ππ₯
= 7152
2β
02
2=
15752
Example 1 (continued) Solution (c):
οΏ½ 3π₯ + 11
4ππ₯
= βοΏ½ 3π₯ + 14
1ππ₯
= β οΏ½ 3π₯4
1ππ₯ + οΏ½ 1
4
1ππ₯
= β 3οΏ½ π₯4
1ππ₯ + οΏ½ 1
4
1ππ₯
= β 342
2 β12
2 + 1 4 β 1 = β512
Example 1 (continued)
Solution (d):
οΏ½ 5π₯22
1 2βππ₯
= 5οΏ½ π₯22
1 2βππ₯
= 523
3β
12οΏ½
3
3=
1058
Average Value of a Continuous Function
If π π₯ is integrable on π, π , then its average value on π,π , also called its mean, is
av π =1
π β ποΏ½ π π₯π
πππ₯
Example 2
Graph π π₯ = βπ₯2 and find its average value over β1,4 .
Solution:
av π =1
π β ποΏ½ π π₯π
πππ₯
We have: π π₯ = βπ₯2 π, π = β1,4
Example 2 (continued)
av π =1
π β ποΏ½ π π₯π
πππ₯
=1
4 β β1οΏ½ βπ₯24
β1ππ₯
= β15οΏ½ π₯24
β1ππ₯
= β15
43
3β
β1 3
3
= β133
Eureka!
The exclamation 'Eureka!' is famously attributed to the ancient Greek scholar Archimedes. He reportedly proclaimed "Eureka!" when he stepped into a bath and noticed that the water level roseβhe suddenly understood that the volume of water displaced must be equal to the volume of the part of his body he had submerged. He then realized that the volume of irregular objects could be measured with precision, a previously intractable problem. He is said to have been so eager to share his discovery that he leapt out of his bathtub and ran through the streets of Syracuse naked.