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Question regarding upper and lower integral sums

0

$begingroup$

This is from Binmore’s Mathematical analysis. My question is why does the author say that use of the second definition gives (7) with the negative signs. Second definition is depicted in figure 2. Also how does he say that (6) and (7) are equal? I’m afraid I’m missing something obvious. Is the negative sign due to the fact that the upper sum values go above the graph and we talk about area under the curve?

real-analysis

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asked Mar 31 at 13:56

Daniel EvansDaniel Evans

1846

$endgroup$

add a comment | 

0

$begingroup$

This is from Binmore’s Mathematical analysis. My question is why does the author say that use of the second definition gives (7) with the negative signs. Second definition is depicted in figure 2. Also how does he say that (6) and (7) are equal? I’m afraid I’m missing something obvious. Is the negative sign due to the fact that the upper sum values go above the graph and we talk about area under the curve?

real-analysis

share|cite|improve this question

asked Mar 31 at 13:56

Daniel EvansDaniel Evans

1846

$endgroup$

add a comment | 

$begingroup$

This is from Binmore’s Mathematical analysis. My question is why does the author say that use of the second definition gives (7) with the negative signs. Second definition is depicted in figure 2. Also how does he say that (6) and (7) are equal? I’m afraid I’m missing something obvious. Is the negative sign due to the fact that the upper sum values go above the graph and we talk about area under the curve?

real-analysis

share|cite|improve this question

asked Mar 31 at 13:56

Daniel EvansDaniel Evans

1846

$endgroup$

share|cite|improve this question

asked Mar 31 at 13:56

Daniel EvansDaniel Evans

1846

share|cite|improve this question

asked Mar 31 at 13:56

Daniel EvansDaniel Evans

1846

asked Mar 31 at 13:56

Daniel EvansDaniel Evans

1846

asked Mar 31 at 13:56

Daniel EvansDaniel Evans

1846

1 Answer
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oldest

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0

$begingroup$

It boils down to the property that $sup a =-inf (-a)$ for $a$ in some domain.

The upper integral is defined as $inf U(f,P)$, where $P$ is a partition of the interval.

This is equal to

$inf sum(sup f(x)cdot Delta x) \
= -sup [-sum(sup f(x)cdot Delta x] \
= -sup [sum(inf (-f(x))cdot Delta x] \
= -sup L(-f,P)$,

which is $-int_a^b -f(x)dx$ by the first definition via supremum.

As for the equality between (6) and (7), has the book introduced the fundamental theorem of calculus at this point yet?

share|cite|improve this answer

answered Mar 31 at 15:27

BunburyBunbury

767

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add a comment | 

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0

$begingroup$

It boils down to the property that $sup a =-inf (-a)$ for $a$ in some domain.

The upper integral is defined as $inf U(f,P)$, where $P$ is a partition of the interval.

This is equal to

$inf sum(sup f(x)cdot Delta x) \
= -sup [-sum(sup f(x)cdot Delta x] \
= -sup [sum(inf (-f(x))cdot Delta x] \
= -sup L(-f,P)$,

which is $-int_a^b -f(x)dx$ by the first definition via supremum.

As for the equality between (6) and (7), has the book introduced the fundamental theorem of calculus at this point yet?

share|cite|improve this answer

answered Mar 31 at 15:27

BunburyBunbury

767

$endgroup$

add a comment | 

0

$begingroup$

It boils down to the property that $sup a =-inf (-a)$ for $a$ in some domain.

The upper integral is defined as $inf U(f,P)$, where $P$ is a partition of the interval.

This is equal to

$inf sum(sup f(x)cdot Delta x) \
= -sup [-sum(sup f(x)cdot Delta x] \
= -sup [sum(inf (-f(x))cdot Delta x] \
= -sup L(-f,P)$,

which is $-int_a^b -f(x)dx$ by the first definition via supremum.

As for the equality between (6) and (7), has the book introduced the fundamental theorem of calculus at this point yet?

share|cite|improve this answer

answered Mar 31 at 15:27

BunburyBunbury

767

$endgroup$

add a comment | 

$begingroup$

It boils down to the property that $sup a =-inf (-a)$ for $a$ in some domain.

The upper integral is defined as $inf U(f,P)$, where $P$ is a partition of the interval.

This is equal to

$inf sum(sup f(x)cdot Delta x) \
= -sup [-sum(sup f(x)cdot Delta x] \
= -sup [sum(inf (-f(x))cdot Delta x] \
= -sup L(-f,P)$,

which is $-int_a^b -f(x)dx$ by the first definition via supremum.

As for the equality between (6) and (7), has the book introduced the fundamental theorem of calculus at this point yet?

share|cite|improve this answer

answered Mar 31 at 15:27

BunburyBunbury

767

$endgroup$

share|cite|improve this answer

answered Mar 31 at 15:27

BunburyBunbury

767

answered Mar 31 at 15:27

BunburyBunbury

767

answered Mar 31 at 15:27

BunburyBunbury

767