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FundamentalTheoremofCalculus(Part2)
Created by
Barbara Forrest and Brian Forrest
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FundamentalTheoremofCalculus(Part2)
Example: Evaluate Z
2
3
t dt.
Observation: If 0
Z x 3
G(x) = t dt,
0
then Z
2
3
G(2) = t dt.
WeknowfromtheFTC1that 0
G′(x) = x3
3
so Gis an antiderivative of x . Hence there exists a constant C such
that Z
x 4
3 x
G(x) = t dt = +C.
0 4
Question: How does this help us?
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FundamentalTheoremofCalculus(Part2)
Note: Wehavejustseenthat
G(x) = Z xt3dt = x4 +C
0 4
for some constant C ∈ R.
However, we also know that
Z 0 3 04
0 = t dt = G(0) = +C=C
0 4
so Z x 3 x4
G(x) = t dt = .
0 4
Z 2 4
Finally, 3 2
t dt = G(2) = =4.
0 4
Question: Did we really need to find C?
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FundamentalTheoremofCalculus(Part2)
KeyObservation:
Let F and G be any two antiderivatives of the same function f. Then
G(x) = F(x)+C.
Let a,b ∈ R. Then
G(b)−G(a) = (F(b)+C)−(F(a)+C)
= F(b)−F(a).
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