alternative proof of the fundamental theorem of calculus
An alternative proof for the first part involves the use of a formula derived by the method of exhaustion:
∫baf(t)𝑑t=(b-a)∞∑n=12n-1∑m=1(-1)m+12-nf(a+m(b-a)/2n). |
Given that
F(x)=∫xaf(t)𝑑t, |
and
F′(x)=limΔx→0F(x+Δx)-F(x)Δx=limΔx→01Δx∫x+Δxxf(t)𝑑t, |
the above formula leads to:
F′(x)=limΔx→0(x+Δx-x)Δx∞∑n=12n-1∑m=1(-1)m+12-nf(x+mΔx/2n), |
or
F′(x)=∞∑n=12n-1∑m=1(-1)m+12-nf(x). |
Since it can be shown that
∞∑n=12n-1∑m=1(-1)m+12-n=∞∑n=12-n=1, |
It follows that
F′(x)=f(x). |
The second part of the proof is identical to the parent.
Title | alternative proof of the fundamental theorem of calculus![]() |
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Canonical name | AlternativeProofOfTheFundamentalTheoremOfCalculus |
Date of creation | 2013-03-22 15:55:24 |
Last modified on | 2013-03-22 15:55:24 |
Owner | ruffa (7723) |
Last modified by | ruffa (7723) |
Numerical id | 4 |
Author | ruffa (7723) |
Entry type | Proof |
Classification | msc 26-00 |