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June 2nd, 2014, 03:56 PM   #1
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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.
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June 2nd, 2014, 08:10 PM   #2
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The Wikipedia page has proofs of both parts. Does that help?
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June 3rd, 2014, 06:30 AM   #3
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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) $
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June 3rd, 2014, 06:39 AM   #4
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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.
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June 3rd, 2014, 07:03 AM   #5
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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.
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June 3rd, 2014, 07:11 AM   #6
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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.
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June 5th, 2014, 02:29 PM   #7
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Quote:
Originally Posted by v8archie View Post
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 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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