Is my proof of second fundamental theorem of calculus (without mean value theorem) valid?












2












$begingroup$



Second Fundamental Theorem Of Calculus (The Evaluation Theorem):



For the function $f$ continuous everywhere in the $[A, B]$, and the
function $F$ whose derivative with respect to $x$ equals to $f$:



$$int_A^B f(x)dx = F(B) - F(A)$$






Let $P = {x_i}$ to be a regular partition of the closed interval from $A$ to $B$, where $i = 1, 2, 3, ..., n$. Then lets construct the Riemann Series for the function $f$:



$$sum_{i=1}^n f(x_i)Delta x$$



By the definition, limit of the series will be equal to the integral of $f$ within the same closed interval:



$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = int_A^B f(x)dx$$



Where



$$Delta x = frac{B - A}{n}$$
$$x_i = A + i*Delta x$$



and so that



$$x_0 = A$$
$$x_n = A + n*frac{B - A}{n} = A + B - A = B.$$



Since we also know that



$$n to infty Rightarrow Delta x to 0 $$
$$f(x_i) = F'(x_i)$$
$$f(x_i) = lim_{Delta x to 0} frac{F(x_i) - F(x_i - Delta x)}{Delta x},$$



we can write that



$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} f(x_i)Delta x.$$



And following that we can notate the series without need of a second limit via applying the limit of $Delta x$ to each element of the sum (other than this, it only affects the total count of elements. Also look to the clarificiation at the bottom of the question) as



$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x),$$



which is that



$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x) = lim_{n to infty}sum_{i=1}^n F(x_i) - F(x_{i - 1}).$$



And so what we obtain is equivalent with



$$(F_1 - F_0) + (F_2 - F_1) + (F_3 - F_2) + ... + (F_n - F_{n-1})$$



which is nothing but



$$F_n - F_0 = F(x_n) - f(x_0) = F(B) - F(A).$$



Following that:



$$int_A^B f(x)dx = F(B) - F(A)$$



Q.E.D.



Clarification for the part "without need of a second limit via applying the limit of $Delta x$ to each element of sum":



$$lim_{Delta x to 0} f(x_i)Delta x = lim_{Delta x to 0} (f(x_i)) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x}) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x} * Delta x). $$



So we can continue with



$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x. $$










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  • 1




    $begingroup$
    In spirit, yes. To get it airtight, you need to invoke the MVT somewhere.
    $endgroup$
    – Randall
    Jan 18 at 16:24








  • 1




    $begingroup$
    $F_n = F(x_n)$ and $F_0 = F(x_0)$. Also $x_n = B$ and $x_0 = A$. So notation error i think?
    $endgroup$
    – İbrahim İpek
    Jan 18 at 17:05






  • 1




    $begingroup$
    I mean technically you need the mean value theorem to be able to say $f(x_i)Delta x=F(x_i)-F(x_i-Delta x)$, right?
    $endgroup$
    – Cardioid_Ass_22
    Jan 18 at 17:30








  • 1




    $begingroup$
    @İbrahimİpek: I am well aware that sigma notation represents a sum, thank you. The point I made, though, is that you have a limit (as $Delta xto 0$) of a limit (as $tto 0$), which you are saying can be done as a single limit. I am not saying you cannot, but you have to justify it; in general, nested limits are not equivalent to single limits, and hand waving by itself does not justify it.
    $endgroup$
    – Arturo Magidin
    Jan 18 at 20:47






  • 3




    $begingroup$
    Note that your “clarification” just compounds the abuse of notation, because the $Delta x$ in $lim_{Delta xto 0}f(x_i)$ in the second expression is not the same $Delta x$ in $lim_{Delta xto 0}(F(x_i)-F(x_i-Delta x))/Delta x)$ that appears in the third expression. Your $f(x_i)$ is independent of $Delta x$ in the second expression, but the term in the third expression is not independent of $Delta x$, so they cannot represent the same limit variable. It’s just handwaving to pretend that they are the same.
    $endgroup$
    – Arturo Magidin
    Jan 18 at 20:52


















2












$begingroup$



Second Fundamental Theorem Of Calculus (The Evaluation Theorem):



For the function $f$ continuous everywhere in the $[A, B]$, and the
function $F$ whose derivative with respect to $x$ equals to $f$:



$$int_A^B f(x)dx = F(B) - F(A)$$






Let $P = {x_i}$ to be a regular partition of the closed interval from $A$ to $B$, where $i = 1, 2, 3, ..., n$. Then lets construct the Riemann Series for the function $f$:



$$sum_{i=1}^n f(x_i)Delta x$$



By the definition, limit of the series will be equal to the integral of $f$ within the same closed interval:



$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = int_A^B f(x)dx$$



Where



$$Delta x = frac{B - A}{n}$$
$$x_i = A + i*Delta x$$



and so that



$$x_0 = A$$
$$x_n = A + n*frac{B - A}{n} = A + B - A = B.$$



Since we also know that



$$n to infty Rightarrow Delta x to 0 $$
$$f(x_i) = F'(x_i)$$
$$f(x_i) = lim_{Delta x to 0} frac{F(x_i) - F(x_i - Delta x)}{Delta x},$$



we can write that



$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} f(x_i)Delta x.$$



And following that we can notate the series without need of a second limit via applying the limit of $Delta x$ to each element of the sum (other than this, it only affects the total count of elements. Also look to the clarificiation at the bottom of the question) as



$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x),$$



which is that



$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x) = lim_{n to infty}sum_{i=1}^n F(x_i) - F(x_{i - 1}).$$



And so what we obtain is equivalent with



$$(F_1 - F_0) + (F_2 - F_1) + (F_3 - F_2) + ... + (F_n - F_{n-1})$$



which is nothing but



$$F_n - F_0 = F(x_n) - f(x_0) = F(B) - F(A).$$



Following that:



$$int_A^B f(x)dx = F(B) - F(A)$$



Q.E.D.



Clarification for the part "without need of a second limit via applying the limit of $Delta x$ to each element of sum":



$$lim_{Delta x to 0} f(x_i)Delta x = lim_{Delta x to 0} (f(x_i)) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x}) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x} * Delta x). $$



So we can continue with



$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x. $$










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  • 1




    $begingroup$
    In spirit, yes. To get it airtight, you need to invoke the MVT somewhere.
    $endgroup$
    – Randall
    Jan 18 at 16:24








  • 1




    $begingroup$
    $F_n = F(x_n)$ and $F_0 = F(x_0)$. Also $x_n = B$ and $x_0 = A$. So notation error i think?
    $endgroup$
    – İbrahim İpek
    Jan 18 at 17:05






  • 1




    $begingroup$
    I mean technically you need the mean value theorem to be able to say $f(x_i)Delta x=F(x_i)-F(x_i-Delta x)$, right?
    $endgroup$
    – Cardioid_Ass_22
    Jan 18 at 17:30








  • 1




    $begingroup$
    @İbrahimİpek: I am well aware that sigma notation represents a sum, thank you. The point I made, though, is that you have a limit (as $Delta xto 0$) of a limit (as $tto 0$), which you are saying can be done as a single limit. I am not saying you cannot, but you have to justify it; in general, nested limits are not equivalent to single limits, and hand waving by itself does not justify it.
    $endgroup$
    – Arturo Magidin
    Jan 18 at 20:47






  • 3




    $begingroup$
    Note that your “clarification” just compounds the abuse of notation, because the $Delta x$ in $lim_{Delta xto 0}f(x_i)$ in the second expression is not the same $Delta x$ in $lim_{Delta xto 0}(F(x_i)-F(x_i-Delta x))/Delta x)$ that appears in the third expression. Your $f(x_i)$ is independent of $Delta x$ in the second expression, but the term in the third expression is not independent of $Delta x$, so they cannot represent the same limit variable. It’s just handwaving to pretend that they are the same.
    $endgroup$
    – Arturo Magidin
    Jan 18 at 20:52
















2












2








2


2



$begingroup$



Second Fundamental Theorem Of Calculus (The Evaluation Theorem):



For the function $f$ continuous everywhere in the $[A, B]$, and the
function $F$ whose derivative with respect to $x$ equals to $f$:



$$int_A^B f(x)dx = F(B) - F(A)$$






Let $P = {x_i}$ to be a regular partition of the closed interval from $A$ to $B$, where $i = 1, 2, 3, ..., n$. Then lets construct the Riemann Series for the function $f$:



$$sum_{i=1}^n f(x_i)Delta x$$



By the definition, limit of the series will be equal to the integral of $f$ within the same closed interval:



$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = int_A^B f(x)dx$$



Where



$$Delta x = frac{B - A}{n}$$
$$x_i = A + i*Delta x$$



and so that



$$x_0 = A$$
$$x_n = A + n*frac{B - A}{n} = A + B - A = B.$$



Since we also know that



$$n to infty Rightarrow Delta x to 0 $$
$$f(x_i) = F'(x_i)$$
$$f(x_i) = lim_{Delta x to 0} frac{F(x_i) - F(x_i - Delta x)}{Delta x},$$



we can write that



$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} f(x_i)Delta x.$$



And following that we can notate the series without need of a second limit via applying the limit of $Delta x$ to each element of the sum (other than this, it only affects the total count of elements. Also look to the clarificiation at the bottom of the question) as



$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x),$$



which is that



$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x) = lim_{n to infty}sum_{i=1}^n F(x_i) - F(x_{i - 1}).$$



And so what we obtain is equivalent with



$$(F_1 - F_0) + (F_2 - F_1) + (F_3 - F_2) + ... + (F_n - F_{n-1})$$



which is nothing but



$$F_n - F_0 = F(x_n) - f(x_0) = F(B) - F(A).$$



Following that:



$$int_A^B f(x)dx = F(B) - F(A)$$



Q.E.D.



Clarification for the part "without need of a second limit via applying the limit of $Delta x$ to each element of sum":



$$lim_{Delta x to 0} f(x_i)Delta x = lim_{Delta x to 0} (f(x_i)) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x}) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x} * Delta x). $$



So we can continue with



$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x. $$










share|cite|improve this question











$endgroup$





Second Fundamental Theorem Of Calculus (The Evaluation Theorem):



For the function $f$ continuous everywhere in the $[A, B]$, and the
function $F$ whose derivative with respect to $x$ equals to $f$:



$$int_A^B f(x)dx = F(B) - F(A)$$






Let $P = {x_i}$ to be a regular partition of the closed interval from $A$ to $B$, where $i = 1, 2, 3, ..., n$. Then lets construct the Riemann Series for the function $f$:



$$sum_{i=1}^n f(x_i)Delta x$$



By the definition, limit of the series will be equal to the integral of $f$ within the same closed interval:



$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = int_A^B f(x)dx$$



Where



$$Delta x = frac{B - A}{n}$$
$$x_i = A + i*Delta x$$



and so that



$$x_0 = A$$
$$x_n = A + n*frac{B - A}{n} = A + B - A = B.$$



Since we also know that



$$n to infty Rightarrow Delta x to 0 $$
$$f(x_i) = F'(x_i)$$
$$f(x_i) = lim_{Delta x to 0} frac{F(x_i) - F(x_i - Delta x)}{Delta x},$$



we can write that



$$lim_{nto infty}sum_{i=1}^n f(x_i)Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} f(x_i)Delta x.$$



And following that we can notate the series without need of a second limit via applying the limit of $Delta x$ to each element of the sum (other than this, it only affects the total count of elements. Also look to the clarificiation at the bottom of the question) as



$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x = lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x),$$



which is that



$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} F(x_i) - F(x_i - Delta x) = lim_{n to infty}sum_{i=1}^n F(x_i) - F(x_{i - 1}).$$



And so what we obtain is equivalent with



$$(F_1 - F_0) + (F_2 - F_1) + (F_3 - F_2) + ... + (F_n - F_{n-1})$$



which is nothing but



$$F_n - F_0 = F(x_n) - f(x_0) = F(B) - F(A).$$



Following that:



$$int_A^B f(x)dx = F(B) - F(A)$$



Q.E.D.



Clarification for the part "without need of a second limit via applying the limit of $Delta x$ to each element of sum":



$$lim_{Delta x to 0} f(x_i)Delta x = lim_{Delta x to 0} (f(x_i)) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x}) * lim_{Delta x to 0} (Delta x) = lim_{Delta x to 0} (frac{F(x_i) - F(x_i - Delta x)}{Delta x} * Delta x). $$



So we can continue with



$$lim_{Delta x to 0}sum_{i=1}^{frac{B - A}{Delta x}} frac{F(x_i) - F(x_i - Delta x)}{Delta x}Delta x. $$







calculus






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edited Jan 18 at 20:39







İbrahim İpek

















asked Jan 18 at 16:22









İbrahim İpekİbrahim İpek

197




197








  • 1




    $begingroup$
    In spirit, yes. To get it airtight, you need to invoke the MVT somewhere.
    $endgroup$
    – Randall
    Jan 18 at 16:24








  • 1




    $begingroup$
    $F_n = F(x_n)$ and $F_0 = F(x_0)$. Also $x_n = B$ and $x_0 = A$. So notation error i think?
    $endgroup$
    – İbrahim İpek
    Jan 18 at 17:05






  • 1




    $begingroup$
    I mean technically you need the mean value theorem to be able to say $f(x_i)Delta x=F(x_i)-F(x_i-Delta x)$, right?
    $endgroup$
    – Cardioid_Ass_22
    Jan 18 at 17:30








  • 1




    $begingroup$
    @İbrahimİpek: I am well aware that sigma notation represents a sum, thank you. The point I made, though, is that you have a limit (as $Delta xto 0$) of a limit (as $tto 0$), which you are saying can be done as a single limit. I am not saying you cannot, but you have to justify it; in general, nested limits are not equivalent to single limits, and hand waving by itself does not justify it.
    $endgroup$
    – Arturo Magidin
    Jan 18 at 20:47






  • 3




    $begingroup$
    Note that your “clarification” just compounds the abuse of notation, because the $Delta x$ in $lim_{Delta xto 0}f(x_i)$ in the second expression is not the same $Delta x$ in $lim_{Delta xto 0}(F(x_i)-F(x_i-Delta x))/Delta x)$ that appears in the third expression. Your $f(x_i)$ is independent of $Delta x$ in the second expression, but the term in the third expression is not independent of $Delta x$, so they cannot represent the same limit variable. It’s just handwaving to pretend that they are the same.
    $endgroup$
    – Arturo Magidin
    Jan 18 at 20:52
















  • 1




    $begingroup$
    In spirit, yes. To get it airtight, you need to invoke the MVT somewhere.
    $endgroup$
    – Randall
    Jan 18 at 16:24








  • 1




    $begingroup$
    $F_n = F(x_n)$ and $F_0 = F(x_0)$. Also $x_n = B$ and $x_0 = A$. So notation error i think?
    $endgroup$
    – İbrahim İpek
    Jan 18 at 17:05






  • 1




    $begingroup$
    I mean technically you need the mean value theorem to be able to say $f(x_i)Delta x=F(x_i)-F(x_i-Delta x)$, right?
    $endgroup$
    – Cardioid_Ass_22
    Jan 18 at 17:30








  • 1




    $begingroup$
    @İbrahimİpek: I am well aware that sigma notation represents a sum, thank you. The point I made, though, is that you have a limit (as $Delta xto 0$) of a limit (as $tto 0$), which you are saying can be done as a single limit. I am not saying you cannot, but you have to justify it; in general, nested limits are not equivalent to single limits, and hand waving by itself does not justify it.
    $endgroup$
    – Arturo Magidin
    Jan 18 at 20:47






  • 3




    $begingroup$
    Note that your “clarification” just compounds the abuse of notation, because the $Delta x$ in $lim_{Delta xto 0}f(x_i)$ in the second expression is not the same $Delta x$ in $lim_{Delta xto 0}(F(x_i)-F(x_i-Delta x))/Delta x)$ that appears in the third expression. Your $f(x_i)$ is independent of $Delta x$ in the second expression, but the term in the third expression is not independent of $Delta x$, so they cannot represent the same limit variable. It’s just handwaving to pretend that they are the same.
    $endgroup$
    – Arturo Magidin
    Jan 18 at 20:52










1




1




$begingroup$
In spirit, yes. To get it airtight, you need to invoke the MVT somewhere.
$endgroup$
– Randall
Jan 18 at 16:24






$begingroup$
In spirit, yes. To get it airtight, you need to invoke the MVT somewhere.
$endgroup$
– Randall
Jan 18 at 16:24






1




1




$begingroup$
$F_n = F(x_n)$ and $F_0 = F(x_0)$. Also $x_n = B$ and $x_0 = A$. So notation error i think?
$endgroup$
– İbrahim İpek
Jan 18 at 17:05




$begingroup$
$F_n = F(x_n)$ and $F_0 = F(x_0)$. Also $x_n = B$ and $x_0 = A$. So notation error i think?
$endgroup$
– İbrahim İpek
Jan 18 at 17:05




1




1




$begingroup$
I mean technically you need the mean value theorem to be able to say $f(x_i)Delta x=F(x_i)-F(x_i-Delta x)$, right?
$endgroup$
– Cardioid_Ass_22
Jan 18 at 17:30






$begingroup$
I mean technically you need the mean value theorem to be able to say $f(x_i)Delta x=F(x_i)-F(x_i-Delta x)$, right?
$endgroup$
– Cardioid_Ass_22
Jan 18 at 17:30






1




1




$begingroup$
@İbrahimİpek: I am well aware that sigma notation represents a sum, thank you. The point I made, though, is that you have a limit (as $Delta xto 0$) of a limit (as $tto 0$), which you are saying can be done as a single limit. I am not saying you cannot, but you have to justify it; in general, nested limits are not equivalent to single limits, and hand waving by itself does not justify it.
$endgroup$
– Arturo Magidin
Jan 18 at 20:47




$begingroup$
@İbrahimİpek: I am well aware that sigma notation represents a sum, thank you. The point I made, though, is that you have a limit (as $Delta xto 0$) of a limit (as $tto 0$), which you are saying can be done as a single limit. I am not saying you cannot, but you have to justify it; in general, nested limits are not equivalent to single limits, and hand waving by itself does not justify it.
$endgroup$
– Arturo Magidin
Jan 18 at 20:47




3




3




$begingroup$
Note that your “clarification” just compounds the abuse of notation, because the $Delta x$ in $lim_{Delta xto 0}f(x_i)$ in the second expression is not the same $Delta x$ in $lim_{Delta xto 0}(F(x_i)-F(x_i-Delta x))/Delta x)$ that appears in the third expression. Your $f(x_i)$ is independent of $Delta x$ in the second expression, but the term in the third expression is not independent of $Delta x$, so they cannot represent the same limit variable. It’s just handwaving to pretend that they are the same.
$endgroup$
– Arturo Magidin
Jan 18 at 20:52






$begingroup$
Note that your “clarification” just compounds the abuse of notation, because the $Delta x$ in $lim_{Delta xto 0}f(x_i)$ in the second expression is not the same $Delta x$ in $lim_{Delta xto 0}(F(x_i)-F(x_i-Delta x))/Delta x)$ that appears in the third expression. Your $f(x_i)$ is independent of $Delta x$ in the second expression, but the term in the third expression is not independent of $Delta x$, so they cannot represent the same limit variable. It’s just handwaving to pretend that they are the same.
$endgroup$
– Arturo Magidin
Jan 18 at 20:52












2 Answers
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Your proof would be correct, if you can first prove something like the following:




Theorem? For any increasing integer sequence $b(n)to infty$ and double-index sequence $A(k,n)$ such that [insert nice assumptions here...],
$$ lim_{ntoinfty} sum_{k=1}^{b(n)}frac{A(k,n)}n=lim_{ntoinfty} sum_{k=1}^{b(n)} frac{lim_{mtoinfty}A(k,m)}n $$




For you, $A(k,m)$ would be something like $frac{f(x_k + 1/m) - f(x_k)}{1/m}$.
As an indication that this isn't easy, note that it isn't true without additional assumptions:



$$ frac12 = lim_{ntoinfty} sum_{k=1}^n frac{k/n}n overset{???}= lim_{ntoinfty} sum_{k=1}^n frac{lim_{mtoinfty} k/m}n = lim_{ntoinfty} sum_{k=0}^n frac{0}n = 0dots $$






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    I once read a similar approach in one of the textbooks meant for high school students of age 16-17 years. As noted in comments the proof is not really a proof but more of a hand-waving.



    Since the function $f$ is continuous it is best to notice that in this case the first and second theorems of calculus are essentially the same. But even then one can't avoid mean value theorem. Here is how you can proceed.




    First Fundamental Theorem of Calculus for Continuous Functions: Let the function $f:[a, b] tomathbb{R} $ be continuous on $[a, b] $. Then the function $G:[a, b] tomathbb {R} $ defined by $$G(x) =int_{a} ^{x} f(t) , dt$$ is differentiable on $[a, b] $ with $G'(x) =f(x) , forall xin[a, b] $.




    Now we prove the second fundamental theorem as mentioned in your question using the first theorem mentioned above. Consider the function $h(x) =F(x) - G(x) $ then $h$ is continuous and differentiable on $[a, b] $ with $h'(x) =f(x)-f(x)=0$. By mean value theorem $h$ is constant on $[a, b] $ and let's say $h(x) =k$ for all all $xin[a, b] $ so that $F(x) =G(x) +k$ for all $xin[a, b] $.



    By definition of $G$ we have $G(a) =0$ and $$int_{a} ^{b} f(x) , dx=G(b) =G(b) - G(a) =G(b) +k-(G(a) +k) =F(b) - F(a) $$



    It is only when the function $f$ is discontinuous the two fundamental theorems of calculus are different and have independent proofs.






    share|cite|improve this answer









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      $begingroup$

      Your proof would be correct, if you can first prove something like the following:




      Theorem? For any increasing integer sequence $b(n)to infty$ and double-index sequence $A(k,n)$ such that [insert nice assumptions here...],
      $$ lim_{ntoinfty} sum_{k=1}^{b(n)}frac{A(k,n)}n=lim_{ntoinfty} sum_{k=1}^{b(n)} frac{lim_{mtoinfty}A(k,m)}n $$




      For you, $A(k,m)$ would be something like $frac{f(x_k + 1/m) - f(x_k)}{1/m}$.
      As an indication that this isn't easy, note that it isn't true without additional assumptions:



      $$ frac12 = lim_{ntoinfty} sum_{k=1}^n frac{k/n}n overset{???}= lim_{ntoinfty} sum_{k=1}^n frac{lim_{mtoinfty} k/m}n = lim_{ntoinfty} sum_{k=0}^n frac{0}n = 0dots $$






      share|cite|improve this answer











      $endgroup$


















        7












        $begingroup$

        Your proof would be correct, if you can first prove something like the following:




        Theorem? For any increasing integer sequence $b(n)to infty$ and double-index sequence $A(k,n)$ such that [insert nice assumptions here...],
        $$ lim_{ntoinfty} sum_{k=1}^{b(n)}frac{A(k,n)}n=lim_{ntoinfty} sum_{k=1}^{b(n)} frac{lim_{mtoinfty}A(k,m)}n $$




        For you, $A(k,m)$ would be something like $frac{f(x_k + 1/m) - f(x_k)}{1/m}$.
        As an indication that this isn't easy, note that it isn't true without additional assumptions:



        $$ frac12 = lim_{ntoinfty} sum_{k=1}^n frac{k/n}n overset{???}= lim_{ntoinfty} sum_{k=1}^n frac{lim_{mtoinfty} k/m}n = lim_{ntoinfty} sum_{k=0}^n frac{0}n = 0dots $$






        share|cite|improve this answer











        $endgroup$
















          7












          7








          7





          $begingroup$

          Your proof would be correct, if you can first prove something like the following:




          Theorem? For any increasing integer sequence $b(n)to infty$ and double-index sequence $A(k,n)$ such that [insert nice assumptions here...],
          $$ lim_{ntoinfty} sum_{k=1}^{b(n)}frac{A(k,n)}n=lim_{ntoinfty} sum_{k=1}^{b(n)} frac{lim_{mtoinfty}A(k,m)}n $$




          For you, $A(k,m)$ would be something like $frac{f(x_k + 1/m) - f(x_k)}{1/m}$.
          As an indication that this isn't easy, note that it isn't true without additional assumptions:



          $$ frac12 = lim_{ntoinfty} sum_{k=1}^n frac{k/n}n overset{???}= lim_{ntoinfty} sum_{k=1}^n frac{lim_{mtoinfty} k/m}n = lim_{ntoinfty} sum_{k=0}^n frac{0}n = 0dots $$






          share|cite|improve this answer











          $endgroup$



          Your proof would be correct, if you can first prove something like the following:




          Theorem? For any increasing integer sequence $b(n)to infty$ and double-index sequence $A(k,n)$ such that [insert nice assumptions here...],
          $$ lim_{ntoinfty} sum_{k=1}^{b(n)}frac{A(k,n)}n=lim_{ntoinfty} sum_{k=1}^{b(n)} frac{lim_{mtoinfty}A(k,m)}n $$




          For you, $A(k,m)$ would be something like $frac{f(x_k + 1/m) - f(x_k)}{1/m}$.
          As an indication that this isn't easy, note that it isn't true without additional assumptions:



          $$ frac12 = lim_{ntoinfty} sum_{k=1}^n frac{k/n}n overset{???}= lim_{ntoinfty} sum_{k=1}^n frac{lim_{mtoinfty} k/m}n = lim_{ntoinfty} sum_{k=0}^n frac{0}n = 0dots $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 18 at 21:50

























          answered Jan 18 at 21:44









          Calvin KhorCalvin Khor

          12.1k21438




          12.1k21438























              3












              $begingroup$

              I once read a similar approach in one of the textbooks meant for high school students of age 16-17 years. As noted in comments the proof is not really a proof but more of a hand-waving.



              Since the function $f$ is continuous it is best to notice that in this case the first and second theorems of calculus are essentially the same. But even then one can't avoid mean value theorem. Here is how you can proceed.




              First Fundamental Theorem of Calculus for Continuous Functions: Let the function $f:[a, b] tomathbb{R} $ be continuous on $[a, b] $. Then the function $G:[a, b] tomathbb {R} $ defined by $$G(x) =int_{a} ^{x} f(t) , dt$$ is differentiable on $[a, b] $ with $G'(x) =f(x) , forall xin[a, b] $.




              Now we prove the second fundamental theorem as mentioned in your question using the first theorem mentioned above. Consider the function $h(x) =F(x) - G(x) $ then $h$ is continuous and differentiable on $[a, b] $ with $h'(x) =f(x)-f(x)=0$. By mean value theorem $h$ is constant on $[a, b] $ and let's say $h(x) =k$ for all all $xin[a, b] $ so that $F(x) =G(x) +k$ for all $xin[a, b] $.



              By definition of $G$ we have $G(a) =0$ and $$int_{a} ^{b} f(x) , dx=G(b) =G(b) - G(a) =G(b) +k-(G(a) +k) =F(b) - F(a) $$



              It is only when the function $f$ is discontinuous the two fundamental theorems of calculus are different and have independent proofs.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                I once read a similar approach in one of the textbooks meant for high school students of age 16-17 years. As noted in comments the proof is not really a proof but more of a hand-waving.



                Since the function $f$ is continuous it is best to notice that in this case the first and second theorems of calculus are essentially the same. But even then one can't avoid mean value theorem. Here is how you can proceed.




                First Fundamental Theorem of Calculus for Continuous Functions: Let the function $f:[a, b] tomathbb{R} $ be continuous on $[a, b] $. Then the function $G:[a, b] tomathbb {R} $ defined by $$G(x) =int_{a} ^{x} f(t) , dt$$ is differentiable on $[a, b] $ with $G'(x) =f(x) , forall xin[a, b] $.




                Now we prove the second fundamental theorem as mentioned in your question using the first theorem mentioned above. Consider the function $h(x) =F(x) - G(x) $ then $h$ is continuous and differentiable on $[a, b] $ with $h'(x) =f(x)-f(x)=0$. By mean value theorem $h$ is constant on $[a, b] $ and let's say $h(x) =k$ for all all $xin[a, b] $ so that $F(x) =G(x) +k$ for all $xin[a, b] $.



                By definition of $G$ we have $G(a) =0$ and $$int_{a} ^{b} f(x) , dx=G(b) =G(b) - G(a) =G(b) +k-(G(a) +k) =F(b) - F(a) $$



                It is only when the function $f$ is discontinuous the two fundamental theorems of calculus are different and have independent proofs.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  I once read a similar approach in one of the textbooks meant for high school students of age 16-17 years. As noted in comments the proof is not really a proof but more of a hand-waving.



                  Since the function $f$ is continuous it is best to notice that in this case the first and second theorems of calculus are essentially the same. But even then one can't avoid mean value theorem. Here is how you can proceed.




                  First Fundamental Theorem of Calculus for Continuous Functions: Let the function $f:[a, b] tomathbb{R} $ be continuous on $[a, b] $. Then the function $G:[a, b] tomathbb {R} $ defined by $$G(x) =int_{a} ^{x} f(t) , dt$$ is differentiable on $[a, b] $ with $G'(x) =f(x) , forall xin[a, b] $.




                  Now we prove the second fundamental theorem as mentioned in your question using the first theorem mentioned above. Consider the function $h(x) =F(x) - G(x) $ then $h$ is continuous and differentiable on $[a, b] $ with $h'(x) =f(x)-f(x)=0$. By mean value theorem $h$ is constant on $[a, b] $ and let's say $h(x) =k$ for all all $xin[a, b] $ so that $F(x) =G(x) +k$ for all $xin[a, b] $.



                  By definition of $G$ we have $G(a) =0$ and $$int_{a} ^{b} f(x) , dx=G(b) =G(b) - G(a) =G(b) +k-(G(a) +k) =F(b) - F(a) $$



                  It is only when the function $f$ is discontinuous the two fundamental theorems of calculus are different and have independent proofs.






                  share|cite|improve this answer









                  $endgroup$



                  I once read a similar approach in one of the textbooks meant for high school students of age 16-17 years. As noted in comments the proof is not really a proof but more of a hand-waving.



                  Since the function $f$ is continuous it is best to notice that in this case the first and second theorems of calculus are essentially the same. But even then one can't avoid mean value theorem. Here is how you can proceed.




                  First Fundamental Theorem of Calculus for Continuous Functions: Let the function $f:[a, b] tomathbb{R} $ be continuous on $[a, b] $. Then the function $G:[a, b] tomathbb {R} $ defined by $$G(x) =int_{a} ^{x} f(t) , dt$$ is differentiable on $[a, b] $ with $G'(x) =f(x) , forall xin[a, b] $.




                  Now we prove the second fundamental theorem as mentioned in your question using the first theorem mentioned above. Consider the function $h(x) =F(x) - G(x) $ then $h$ is continuous and differentiable on $[a, b] $ with $h'(x) =f(x)-f(x)=0$. By mean value theorem $h$ is constant on $[a, b] $ and let's say $h(x) =k$ for all all $xin[a, b] $ so that $F(x) =G(x) +k$ for all $xin[a, b] $.



                  By definition of $G$ we have $G(a) =0$ and $$int_{a} ^{b} f(x) , dx=G(b) =G(b) - G(a) =G(b) +k-(G(a) +k) =F(b) - F(a) $$



                  It is only when the function $f$ is discontinuous the two fundamental theorems of calculus are different and have independent proofs.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 19 at 3:22









                  Paramanand SinghParamanand Singh

                  50.4k556166




                  50.4k556166






























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