Math 180
5.3 – The Fundamental Theorem of
Calculus
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In 5.1, we saw how to visualize an estimation of
distance traveled using velocity samples at 5second intervals.
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In 5.1, we saw how to visualize an estimation of
distance traveled using velocity samples at 5second intervals.
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In 5.1, we saw how to visualize an estimation of
distance traveled using velocity samples at 5second intervals.
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In 5.1, we saw how to visualize an estimation of
distance traveled using velocity samples at 5second intervals.
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In 5.1, we saw how to visualize an estimation of
distance traveled using velocity samples at 5second intervals.
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To get a more accurate estimation, we could
sample every second.
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To get a more accurate estimation, we could
sample every second.
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To get a more accurate estimation, we could
sample every second.
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To get a more accurate estimation, we could
sample every second.
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What if we could sample every instant?
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What if we could sample every instant?
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What if we could sample every instant?
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What if we could sample every instant?
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The Fundamental Theorem of Calculus, Part 2
(also called The Evaluation Theorem)
If 𝑓 is continuous on 𝑎, 𝑏 and 𝐹 is any
antiderivative of 𝑓 on 𝑎, 𝑏 , then
𝒃
𝒇 𝒙 𝒅𝒙 = 𝑭 𝒃 − 𝑭(𝒂)
𝒂
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Ex 1.
Evaluate.
𝜋
cos 𝑥 𝑑𝑥
0
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Notation:
𝐹 𝑥
𝑏
𝑎
=𝐹 𝑥
𝑏
𝑎
=𝐹 𝑏 −𝐹 𝑎
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Ex 2.
Evaluate.
0
𝜋
−4
sec 𝑥 tan 𝑥 𝑑𝑥
4
1
−2
−3
3
4
𝑥 − 2 𝑑𝑥
2
𝑥
𝑑𝑥
𝑥+1
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Let’s go back to the velocity function, 𝑣 𝑡 .
Consider the area under 𝑣 𝑡 from 𝑡 = 0 to 𝑡 =
𝑥 (where 𝑥 is a variable here). This area
represents the change in position over that time
interval. Note that the area is a function of 𝑥.
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The Fundamental Theorem of Calculus, Part 1
If 𝑓 is continuous on 𝑎, 𝑏 , then
𝑥
𝑓
𝑎
𝐹 𝑥 =
𝑡 𝑑𝑡 is continuous on 𝑎, 𝑏 and
differentiable on 𝑎, 𝑏 and its derivative is
𝑓 𝑥 :
𝒅
′
𝑭 𝒙 =
𝒅𝒙
𝒙
𝒇 𝒕 𝒅𝒕 = 𝒇 𝒙
𝒂
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Ex 3.
Use the Fundamental Theorem to find
𝑥
𝑑𝑦
.
𝑑𝑥
𝑡 3 + 1 𝑑𝑡
𝑦=
𝑎
5
𝑦=
3𝑡 sin 𝑡 𝑑𝑡
𝑥
𝑥2
𝑦=
cos 𝑡 𝑑𝑡
1
4
1
𝑦=
𝑑𝑡
𝑡
1+3𝑥 2 2 + 𝑒
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By the way, here’s another way to get intuition
about the FTOC, Part 1…
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What is the rate at which the area is changing?
That is, what is 𝐹′(𝑥)? Let’s estimate it:
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𝐹 𝑥+ℎ −𝐹 𝑥 ≈
actual change in area
𝑓 𝑥 ⋅ℎ
estimated
change in area
𝐹 𝑥+ℎ −𝐹 𝑥
≈𝑓 𝑥
ℎ
𝐹 𝑥+ℎ −𝐹 𝑥
lim
=𝑓 𝑥
ℎ→0
ℎ
𝐹′ 𝑥 = 𝑓 𝑥
So,
𝒅
𝒅𝒙
𝒙
𝒇
𝒂
𝒕 𝒅𝒕 = 𝒇 𝒙 .
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𝐹 𝑥+ℎ −𝐹 𝑥 ≈
actual change in area
𝑓 𝑥 ⋅ℎ
estimated
change in area
𝐹 𝑥+ℎ −𝐹 𝑥
≈𝑓 𝑥
ℎ
𝐹 𝑥+ℎ −𝐹 𝑥
lim
=𝑓 𝑥
ℎ→0
ℎ
𝐹′ 𝑥 = 𝑓 𝑥
So,
𝒅
𝒅𝒙
𝒙
𝒇
𝒂
𝒕 𝒅𝒕 = 𝒇 𝒙 .
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𝐹 𝑥+ℎ −𝐹 𝑥 ≈
actual change in area
𝑓 𝑥 ⋅ℎ
estimated
change in area
𝐹 𝑥+ℎ −𝐹 𝑥
≈𝑓 𝑥
ℎ
𝐹 𝑥+ℎ −𝐹 𝑥
lim
=𝑓 𝑥
ℎ→0
ℎ
𝐹′ 𝑥 = 𝑓 𝑥
So,
𝒅
𝒅𝒙
𝒙
𝒇
𝒂
𝒕 𝒅𝒕 = 𝒇 𝒙 .
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𝐹 𝑥+ℎ −𝐹 𝑥 ≈
actual change in area
𝑓 𝑥 ⋅ℎ
estimated
change in area
𝐹 𝑥+ℎ −𝐹 𝑥
≈𝑓 𝑥
ℎ
𝐹 𝑥+ℎ −𝐹 𝑥
lim
=𝑓 𝑥
ℎ→0
ℎ
𝐹′ 𝑥 = 𝑓 𝑥
So,
𝒅
𝒅𝒙
𝒙
𝒇
𝒂
𝒕 𝒅𝒕 = 𝒇 𝒙 .
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𝐹 𝑥+ℎ −𝐹 𝑥 ≈
actual change in area
𝑓 𝑥 ⋅ℎ
estimated
change in area
𝐹 𝑥+ℎ −𝐹 𝑥
≈𝑓 𝑥
ℎ
𝐹 𝑥+ℎ −𝐹 𝑥
lim
=𝑓 𝑥
ℎ→0
ℎ
𝐹′ 𝑥 = 𝑓 𝑥
So,
𝒅
𝒅𝒙
𝒙
𝒇
𝒂
𝒕 𝒅𝒕 = 𝒇 𝒙 .
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𝒅
𝒅𝒙
𝒙
𝒇 𝒕 𝒅𝒕 = 𝒇 𝒙
𝒂
In other words, the rate at which the area is
changing is equal to the function value of 𝑓 at 𝑥.
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