June 2nd, 2014, 03:56 PM  #1 
Senior Member Joined: Feb 2014 From: Louisiana Posts: 156 Thanks: 6 Math Focus: algebra and the calculus  Fundamental theorem of calculus
Could somebody explain to me the two parts of the fundamental theorem of calculus? I know that this is fairly rudimentary stuff, but I cannot seem to find out why they are true. Whenever I look for an answer to this, all I find is that they are the case, not why they are the case.

June 2nd, 2014, 08:10 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,876 Thanks: 2240 Math Focus: Mainly analysis and algebra 
The Wikipedia page has proofs of both parts. Does that help?

June 3rd, 2014, 06:30 AM  #3 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 468 Thanks: 40 
I made a video about finding areas underneath curves about six weeks ago. You can find it here: Traditional Way Of Finding Areas Underneath Curves From x=b to x=a  YouTube. It seems as though you want a deeper explanation. $\displaystyle Area\quad PMTN<\delta A<Area\quad SMTQ\\ \\ \therefore \quad y\delta x<\delta A<\left( y+\delta y \right) \delta x\\ \\ \therefore \quad y<\frac { \delta A }{ \delta x } <y+\delta y\\ \\ But...\\ \\ \lim _{ \delta x\rightarrow 0 }{ \left( \frac { \delta A }{ \delta x } \right) } =\frac { dA }{ dx } \\ \\ \lim _{ \delta x\rightarrow 0 }{ \left( \frac { \delta A }{ \delta x } \right) =y } \\ \\ \therefore \quad y=\frac { dA }{ dx } \\ \therefore \quad 1dA=ydx\\ \\ INTEGRATE\quad both\quad sides\quad of\quad equation:\\ \\ \int { 1dA=\int { ydx } } \\ \\ \therefore \quad A=\int { ydx } =F\left( x \right) +C\\ \\ But\quad at\quad x=a,\quad Area\quad or\quad A=O\\ \\ \therefore \quad O=F\left( a \right) +C\\ \\ \therefore \quad C=F\left( a \right) \\ \\ \therefore \quad A=\int _{ a }^{ x }{ ydx=F\left( x \right) } F\left( a \right) \\ \\ But,\quad you\quad want\quad the\quad area\quad from\quad x=b\quad to\quad x=a,\\ so\quad you\quad use:\\ \\ A=\int _{ a }^{ b }{ ydx=F\left( b \right) } F\left( a \right) $ 
June 3rd, 2014, 06:39 AM  #4 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 468 Thanks: 40 
Integration is the polar opposite of differentiation: $\displaystyle If\quad y={ x }^{ 2 }+C,\\ \\ \frac { dy }{ dx } =2x\\ \\ So...\quad 1dy=2xdx\\ \\ INTEGRATE\quad BOTH\quad SIDES\quad OF\quad EQUATION:\\ \\ \int { 1dy=\int { 2xdx } } \\ \\ y=\frac { 2{ x }^{ 2 } }{ 2 } +C={ x }^{ 2 }+C\\ \\ Now\quad back\quad to\quad square\quad one.$ Last edited by perfect_world; June 3rd, 2014 at 06:41 AM. 
June 3rd, 2014, 07:03 AM  #5 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 468 Thanks: 40 
Differentiation Example: $\displaystyle Let's\quad say\quad that:\\ \\ y=f\left( x \right) \quad \therefore \quad \frac { dy }{ dx } =f'\left( x \right) \\ \\ f\left( x \right) ={ x }^{ 2 },\quad \therefore \quad f\left( x+\delta x \right) ={ \left( x+\delta x \right) }^{ 2 }.\\ \\ Rememember\quad that:\quad \lim _{ \delta x\rightarrow 0 }{ \left( \frac { \delta y }{ \delta x } \right) } =\frac { dy }{ dx } .\\ \\ But\quad what\quad is\quad \frac { \delta y }{ \delta x } ?\quad It\quad is\quad m=\frac { { y }_{ 2 }{ y }_{ 1 } }{ { x }_{ 2 }{ x }_{ 1 } } =\frac { f\left( x+\delta x \right) f\left( x \right) }{ \left( x+\delta x \right) x } =\frac { f\left( x+\delta x \right) f\left( x \right) }{ \delta x } .\\ \\ So...\\ \\ f'\left( x \right) =\frac { dy }{ dx } =\lim _{ \delta x\rightarrow 0 }{ \frac { f\left( x+\delta x \right) f\left( x \right) }{ \delta x } } =\lim _{ \delta x\rightarrow 0 }{ \frac { { \left( x+\delta x \right) }^{ 2 }{ x }^{ 2 } }{ \delta x } } =\lim _{ \delta x\rightarrow 0 }{ \frac { { x }^{ 2 }+2x\delta x+{ \left( \delta x \right) }^{ 2 }{ x }^{ 2 } }{ \delta x } } \\ \\ \lim _{ \delta x\rightarrow 0 }{ \frac { \delta x\left( 2x+\delta x \right) }{ \delta x } } =\lim _{ \delta x\rightarrow 0 }{ 2x+\delta x=2x } \\ \\ That\quad is\quad why\quad if\quad y={ x }^{ 2 },\quad \frac { dy }{ dx } =2x.\\ \\ \\ $ Don't know why the change in x approaches zero has been placed to the side of 'lim'. Must be a latex issue. 
June 3rd, 2014, 07:11 AM  #6 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 468 Thanks: 40 
If you need something more advanced than this (I mean, if it has been oversimplified), I'm sure the mathematical experts in this forum will assist you. This is what I understand to be true in simple terms. 
June 5th, 2014, 02:29 PM  #7  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,876 Thanks: 2240 Math Focus: Mainly analysis and algebra  Quote:
\begin{align*} \text{We have} \qquad F(x) &= \int_a^x'{ f(t) dt} \\ \text{so} \qquad F(x+h)  F(x) &= \int_a^{x+h}{ f(t) dt}  \int_a^x {f(t)dt} \\ &= \int_x^{x+h} {f(t) dt} \\ \text{Now note that} \qquad f(t) &= f(x) + \left( f(t)  f(x) \right) \\ \text{giving us} \qquad F(x+h)  F(x) &= \int_x^{x+h} {f(x) dt} + \int_x^{x+h} {f(t)f(x) dt} \\ &= hf(x) + \int_x^{x+h} {f(t)  f(x) dt} \\ \frac{F(x+h)  F(x)}{h} &= f(x) + \frac1h \int_x^{x+h} {f(t)  f(x) dt} \\[12pt] \text{To get the result we seek, we will take the limit as $h \to 0$. We want the last term to go to zero when we do this.} \\[12pt] \text{$f$ is continous at $x$, so for all $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that} \\ \left f(t)  f(x) \right &\lt \epsilon \qquad \text{whenever} \; \left t  x \right \lt \delta \\ \text{thus} \qquad \left \int_x^{x+h} {f(t)l  f(x) dt} \right &\le \int_x^{x+h} { \left f(t)  f(x) \right dt} \lt \int_x^{x+h} {\epsilon dt } = h\epsilon & \text{when $h \lt \delta$} \\ \text{Thus} \qquad \left \frac1h \int_x^{x+h} {f(t)  f(x) dt} \right &\lt \epsilon & \text{and tends to zero with $h$} \\ \lim_{h \ to 0}{ \frac{F(x+h)  F(x)}{h} } = F^\prime(x) &= f(x) \\ \end{align*}  

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