Math 71 – 1.1

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Transcript Math 71 – 1.1

Math 140
5.3 – 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, let’s try to find the area under
𝑓 π‘₯ = π‘₯ 3 βˆ’ 9π‘₯ 2 + 25π‘₯ βˆ’ 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 , …, π‘₯𝑛
and each rectangle has width Ξ”π‘₯,
then the areas of the rectangles are
𝑓 π‘₯1 Ξ”π‘₯, 𝑓 π‘₯2 Ξ”π‘₯, …, 𝑓 π‘₯𝑛 Ξ”π‘₯.
So, the sum of the areas of the rectangles (called a
Riemann sum) is:
𝑓 π‘₯1 Ξ”π‘₯ + 𝑓 π‘₯2 Ξ”π‘₯ + β‹― + 𝑓 π‘₯𝑛 Ξ”π‘₯
= 𝑓 π‘₯1 + 𝑓 π‘₯2 + β‹― + 𝑓 π‘₯𝑛 Ξ”π‘₯
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So, the sum of the areas of the rectangles (called a
Riemann sum) is:
𝑓 π‘₯1 Ξ”π‘₯ + 𝑓 π‘₯2 Ξ”π‘₯ + β‹― + 𝑓 π‘₯𝑛 Ξ”π‘₯
= 𝑓 π‘₯1 + 𝑓 π‘₯2 + β‹― + 𝑓 π‘₯𝑛 Ξ”π‘₯
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So, the sum of the areas of the rectangles (called a
Riemann sum) is:
𝑓 π‘₯1 Ξ”π‘₯ + 𝑓 π‘₯2 Ξ”π‘₯ + β‹― + 𝑓 π‘₯𝑛 Ξ”π‘₯
= 𝑓 π‘₯1 + 𝑓 π‘₯2 + β‹― + 𝑓 π‘₯𝑛 Ξ”π‘₯
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As the number of rectangles (𝑛) increases, the
above sum gets closer to the true area.
That is,
Area under curve =
π’π’Šπ’Ž 𝒇 π’™πŸ + 𝒇 π’™πŸ + β‹― + 𝒇 𝒙𝒏
𝐧→+∞
πœŸπ’™
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𝑏
π‘Ž
𝑓 π‘₯ 𝑑π‘₯ = lim 𝑓 π‘₯1 + 𝑓 π‘₯2 + β‹― + 𝑓 π‘₯𝑛
nβ†’+∞
Ξ”π‘₯
The above integral is called a definite integral, and π‘Ž
and 𝑏 are called the lower and upper limits of
integration.
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The Fundamental Theorem of Calculus
If 𝑓(π‘₯) is continuous on the interval π‘Ž ≀ π‘₯ ≀ 𝑏,
then
𝒃
𝒇 𝒙 𝒅𝒙 = 𝑭 𝒃 βˆ’ 𝑭(𝒂)
𝒂
where 𝐹(π‘₯) is any antiderivative of 𝑓(π‘₯) on π‘Ž ≀
π‘₯ ≀ 𝑏.
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Ex 1.
Evaluate the given definite integrals using the
fundamental theorem of calculus.
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2π‘₯ 2 𝑑π‘₯
βˆ’1
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Ex 1.
Evaluate the given definite integrals using the
fundamental theorem of calculus.
1
2π‘₯ 2 𝑑π‘₯
βˆ’1
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Ex 1 (cont).
Evaluate the given definite integrals using the
fundamental theorem of calculus.
1
(𝑒 2π‘₯ βˆ’ π‘₯) 𝑑π‘₯
0
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Ex 1 (cont).
Evaluate the given definite integrals using the
fundamental theorem of calculus.
1
(𝑒 2π‘₯ βˆ’ π‘₯) 𝑑π‘₯
0
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Note: Definite integrals can evaluate to
negative numbers. For example,
2
1
(1
βˆ’
π‘₯)
𝑑π‘₯
=
βˆ’
1
2
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Just like with indefinite integrals, terms can be integrated
separately, and constants can be pulled out.
𝑏
𝑏
𝑓 π‘₯ ± 𝑔(π‘₯) 𝑑π‘₯ =
π‘Ž
𝑏
𝑏
π‘˜π‘“ π‘₯ 𝑑π‘₯ = π‘˜
π‘Ž
𝑏
𝑓 π‘₯ 𝑑π‘₯ ±
π‘Ž
𝑔(π‘₯) 𝑑π‘₯
π‘Ž
𝑓 π‘₯ 𝑑π‘₯
π‘Ž
Also,
π‘Ž
𝑓 π‘₯ 𝑑π‘₯ = 0
π‘Ž
π‘Ž
𝑏
𝑓 π‘₯ 𝑑π‘₯ = βˆ’
𝑏
𝑓 π‘₯ 𝑑π‘₯
π‘Ž
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𝑏
𝑐
𝑓 π‘₯ 𝑑π‘₯ =
π‘Ž
𝑏
𝑓 π‘₯ 𝑑π‘₯ +
π‘Ž
𝑓 π‘₯ 𝑑π‘₯
𝑐
(here, 𝑐 is any real number)
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𝑏
𝑐
𝑓 π‘₯ 𝑑π‘₯ =
π‘Ž
𝑏
𝑓 π‘₯ 𝑑π‘₯ +
π‘Ž
𝑓 π‘₯ 𝑑π‘₯
𝑐
(here, 𝑐 is any real number)
=
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𝑏
𝑐
𝑓 π‘₯ 𝑑π‘₯ =
π‘Ž
𝑏
𝑓 π‘₯ 𝑑π‘₯ +
π‘Ž
𝑓 π‘₯ 𝑑π‘₯
𝑐
(here, 𝑐 is any real number)
=
+
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Ex 2.
Let 𝑓(π‘₯) and 𝑔(π‘₯) be functions that are continuous on the interval
βˆ’ 3 ≀ π‘₯ ≀ 5 and that satisfy
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5
𝑓 π‘₯ 𝑑π‘₯ = 2
5
𝑔 π‘₯ 𝑑π‘₯ = 7
βˆ’3
βˆ’3
𝑓 π‘₯ 𝑑π‘₯ = βˆ’8
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Use this information along with the rules definite integrals to evaluate
the following.
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2𝑓 π‘₯ βˆ’ 3𝑔 π‘₯
𝑑π‘₯
βˆ’3
1
𝑓 π‘₯ 𝑑π‘₯
βˆ’3
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Ex 3.
Evaluate:
1
8π‘₯ π‘₯ 2 + 1
3
𝑑π‘₯
0
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Ex 3.
Evaluate:
1
8π‘₯ π‘₯ 2 + 1
3
𝑑π‘₯
0
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Net Change
If we have 𝑄′(π‘₯), then we can calculate the net
change in 𝑄(π‘₯) as π‘₯ goes from π‘Ž to 𝑏 with an
integral:
𝑏
𝑄 𝑏 βˆ’π‘„ π‘Ž =
𝑄 β€² π‘₯ 𝑑π‘₯
π‘Ž
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Net Change
Ex 4.
Suppose marginal cost is 𝐢 β€² π‘₯ = 3 π‘₯ βˆ’ 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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