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

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anonymous · Answer

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.

anonymous · Answer

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

anonymous · Answer

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