My Math Forum Fundamental theorem of calculus

 Calculus Calculus Math Forum

 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. $\displaystyle Area\quad PMTN<\delta A  June 3rd, 2014, 06:39 AM #4 Senior Member Joined: Jul 2013 From: United Kingdom Posts: 468 Thanks: 40 Integration is the polar opposite of differentiation:$\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.\\ \\ \\ $Don't know why the change in x approaches zero has been placed to the side of 'lim'. Must be a latex issue.  June 3rd, 2014, 07:11 AM #6 Senior Member Joined: Jul 2013 From: United Kingdom Posts: 468 Thanks: 40 If you need something more advanced than this (I mean, if it has been oversimplified), I'm sure the mathematical experts in this forum will assist you. This is what I understand to be true in simple terms. June 5th, 2014, 02:29 PM #7 Math Team Joined: Dec 2013 From: Colombia Posts: 6,876 Thanks: 2240 Math Focus: Mainly analysis and algebra Quote:  Originally Posted by v8archie The Wikipedia page has proofs of both parts. Does that help? I have noticed that the proof on the Wikipedia page requires that$f$be continuous in a neighbourhood of$x$. I here present a proof that requires only that$f$is continuous at$x. \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 ash \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*}

 Tags calculus, fundamental, theorem

### as delta x and delta y tends to 0 what happens to delta y/delta x

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post MadSoulz Real Analysis 2 April 15th, 2014 03:19 PM layd33foxx Calculus 3 December 12th, 2011 07:32 PM riotsandravess Calculus 3 November 25th, 2010 12:44 PM Aurica Calculus 1 June 10th, 2009 05:39 PM mrguitar Calculus 3 December 9th, 2007 01:22 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top