Math 71 – 1.1
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Transcript Math 71 – 1.1
Math 140
5.3 β The Definite Integral and the
Fundamental Theorem of Calculus
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Q: How can we find the area under any curve?
For example, letβs try to find the area under
π π₯ = π₯ 3 β 9π₯ 2 + 25π₯ β 19
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What is
the area
of this?
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1 rectangle
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2 rectangles
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4 rectangles
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50 rectangles
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If the left sides of our rectangles are at π₯1 , π₯2 , β¦, π₯π
and each rectangle has width Ξπ₯,
then the areas of the rectangles are
π π₯1 Ξπ₯, π π₯2 Ξπ₯, β¦, π π₯π Ξπ₯.
So, the sum of the areas of the rectangles (called a
Riemann sum) is:
π π₯1 Ξπ₯ + π π₯2 Ξπ₯ + β― + π π₯π Ξπ₯
= π π₯1 + π π₯2 + β― + π π₯π Ξπ₯
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So, the sum of the areas of the rectangles (called a
Riemann sum) is:
π π₯1 Ξπ₯ + π π₯2 Ξπ₯ + β― + π π₯π Ξπ₯
= π π₯1 + π π₯2 + β― + π π₯π Ξπ₯
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So, the sum of the areas of the rectangles (called a
Riemann sum) is:
π π₯1 Ξπ₯ + π π₯2 Ξπ₯ + β― + π π₯π Ξπ₯
= π π₯1 + π π₯2 + β― + π π₯π Ξπ₯
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As the number of rectangles (π) increases, the
above sum gets closer to the true area.
That is,
Area under curve =
πππ π ππ + π ππ + β― + π ππ
π§β+β
ππ
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π
π
π π₯ ππ₯ = lim π π₯1 + π π₯2 + β― + π π₯π
nβ+β
Ξπ₯
The above integral is called a definite integral, and π
and π are called the lower and upper limits of
integration.
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The Fundamental Theorem of Calculus
If π(π₯) is continuous on the interval π β€ π₯ β€ π,
then
π
π π π
π = π π β π(π)
π
where πΉ(π₯) is any antiderivative of π(π₯) on π β€
π₯ β€ π.
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Ex 1.
Evaluate the given definite integrals using the
fundamental theorem of calculus.
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2π₯ 2 ππ₯
β1
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Ex 1.
Evaluate the given definite integrals using the
fundamental theorem of calculus.
1
2π₯ 2 ππ₯
β1
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Ex 1 (cont).
Evaluate the given definite integrals using the
fundamental theorem of calculus.
1
(π 2π₯ β π₯) ππ₯
0
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Ex 1 (cont).
Evaluate the given definite integrals using the
fundamental theorem of calculus.
1
(π 2π₯ β π₯) ππ₯
0
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Note: Definite integrals can evaluate to
negative numbers. For example,
2
1
(1
β
π₯)
ππ₯
=
β
1
2
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Just like with indefinite integrals, terms can be integrated
separately, and constants can be pulled out.
π
π
π π₯ ± π(π₯) ππ₯ =
π
π
π
ππ π₯ ππ₯ = π
π
π
π π₯ ππ₯ ±
π
π(π₯) ππ₯
π
π π₯ ππ₯
π
Also,
π
π π₯ ππ₯ = 0
π
π
π
π π₯ ππ₯ = β
π
π π₯ ππ₯
π
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π
π
π π₯ ππ₯ =
π
π
π π₯ ππ₯ +
π
π π₯ ππ₯
π
(here, π is any real number)
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π
π
π π₯ ππ₯ =
π
π
π π₯ ππ₯ +
π
π π₯ ππ₯
π
(here, π is any real number)
=
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π
π
π π₯ ππ₯ =
π
π
π π₯ ππ₯ +
π
π π₯ ππ₯
π
(here, π is any real number)
=
+
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Ex 2.
Let π(π₯) and π(π₯) be functions that are continuous on the interval
β 3 β€ π₯ β€ 5 and that satisfy
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5
π π₯ ππ₯ = 2
5
π π₯ ππ₯ = 7
β3
β3
π π₯ ππ₯ = β8
1
Use this information along with the rules definite integrals to evaluate
the following.
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2π π₯ β 3π π₯
ππ₯
β3
1
π π₯ ππ₯
β3
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Ex 3.
Evaluate:
1
8π₯ π₯ 2 + 1
3
ππ₯
0
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Ex 3.
Evaluate:
1
8π₯ π₯ 2 + 1
3
ππ₯
0
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Net Change
If we have πβ²(π₯), then we can calculate the net
change in π(π₯) as π₯ goes from π to π with an
integral:
π
π π βπ π =
π β² π₯ ππ₯
π
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Net Change
Ex 4.
Suppose marginal cost is πΆ β² π₯ = 3 π₯ β 4 2
dollars per unit when the level of production is π₯
units. By how much will the total manufacturing
cost increase if the level of production is raised
from 6 units to 10 units?
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