Mathematics
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?