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OpenStudy (anonymous):

can anyone please help me with the proof of fundamental theorem of calculus. I would appreciate if you may attach a file

OpenStudy (anonymous):

I will write here a proof of the First Fundamental Theorem of Calculus as seen in Spivak's Calculus. Theorem: Let \(f\) be integrable on \([a,b]\), and define \(F\) on \([a,b]\) by \[ F(x)=\int_a^xf. \]If \(f\) is continuous at \(c\) in \([a,b]\), then \(F\) is differentiable at \(c\), and \[ F'(c)=f(c). \](If \(c=a\) or \(b\), then \(F'(c)\) is understood to mean the right- or left-hand derivative of \(F\).) Proof: We will assume that \(c\) is in \((a,b)\); the easy modifications for \(c=a\) or \(b\) may be supplied by the reader. By definition, \[ F'(c)=\lim_{h\to0}\frac{F(c+h)-F(c)}{h}. \]Suppose first that \(h>0\). Then\[ F(c+h)-F(c)=\int_c^{c+h}f. \]Define \(m_h\) and \(M_h\) as follows:\[ m_h=\inf\{f(x):c\le x\le c+h\},\\ M_h=\sup\{f(x):c\le x\le c+h\}. \]It follows from Theorem 13-7 (this theorem says that if \(m\le f(x)\le M\) for all \(x\) in \([a,b]\), then \(m(b-a)\le\int_a^bf\le M(b-a)\)) that\[ m_h\cdot h \le \int_c^{c+h}f\le M_h\cdot h. \]Therefore\[ m_h\le\frac{F(c+h)-F(c)}{h}\le M_h. \]If \(h<0\), only a few details of the argument have to be changed. Essentially, the \(c\) and \(c+h\) terms switch places and some negatives must be dealt with, but the proof follows the same pattern.

OpenStudy (anonymous):

Thanx, I'm writing a maths exam tomorrow so I just needed some clarification on this

OpenStudy (anonymous):

No worries. Did that all make sense? Any steps you need explained at all?

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