Proof area of {$(x,y) R^2 | 0 le y < e^x$ and $0le x< h$} is $e^h-1$
$begingroup$
I'm supposed to prove that the area of {$(x,y) R^2 | 0 le y < e^x$ and $0le x< h$} is $e^h-1$
I was going to try to make it a function and calculate it using a Riemanns sum.
That led me to
$F(x) = e^2 = y$
Assuming n rectangles with the width $h/n$ and height $e^frac{hi}{n}$
That got me to the sum and now I'm stuck at
$$frac hn sum_{i=0}^n e^frac{hi}{n} $$
How should I proceed?
The proof is supposed to use simple sets, e.i. not supposed to use an integral.
proof-writing power-series supremum-and-infimum riemann-sum
asked Dec 6 '18 at 11:05
user9516437user9516437
11
$endgroup$
add a comment |
$begingroup$
I'm supposed to prove that the area of {$(x,y) R^2 | 0 le y < e^x$ and $0le x< h$} is $e^h-1$
I was going to try to make it a function and calculate it using a Riemanns sum.
That led me to
$F(x) = e^2 = y$
Assuming n rectangles with the width $h/n$ and height $e^frac{hi}{n}$
That got me to the sum and now I'm stuck at
$$frac hn sum_{i=0}^n e^frac{hi}{n} $$
How should I proceed?
The proof is supposed to use simple sets, e.i. not supposed to use an integral.
proof-writing power-series supremum-and-infimum riemann-sum
asked Dec 6 '18 at 11:05
user9516437user9516437
11
$endgroup$
1
$begingroup$
look up the fundamental theorem of calculus
$endgroup$
– wilkersmon
Dec 6 '18 at 11:08
$begingroup$
It is a geomtrical serie, the sum is equal to $$h/nsumlimits_{i = 0}^{n - 1} {{e^{ih/n}}} = left( {h/n} right)frac{{1 - {e^{(h + frac{h}{n})}}}}{{1 - {e^{h/n}}}}$$
$endgroup$
– Gustave
Dec 6 '18 at 11:35
add a comment |
$begingroup$
I'm supposed to prove that the area of {$(x,y) R^2 | 0 le y < e^x$ and $0le x< h$} is $e^h-1$
I was going to try to make it a function and calculate it using a Riemanns sum.
That led me to
$F(x) = e^2 = y$
Assuming n rectangles with the width $h/n$ and height $e^frac{hi}{n}$
That got me to the sum and now I'm stuck at
$$frac hn sum_{i=0}^n e^frac{hi}{n} $$
How should I proceed?
The proof is supposed to use simple sets, e.i. not supposed to use an integral.
proof-writing power-series supremum-and-infimum riemann-sum
asked Dec 6 '18 at 11:05
user9516437user9516437
11
$endgroup$
I'm supposed to prove that the area of {$(x,y) R^2 | 0 le y < e^x$ and $0le x< h$} is $e^h-1$
I was going to try to make it a function and calculate it using a Riemanns sum.
That led me to
$F(x) = e^2 = y$
Assuming n rectangles with the width $h/n$ and height $e^frac{hi}{n}$
That got me to the sum and now I'm stuck at
$$frac hn sum_{i=0}^n e^frac{hi}{n} $$
How should I proceed?
The proof is supposed to use simple sets, e.i. not supposed to use an integral.
proof-writing power-series supremum-and-infimum riemann-sum
proof-writing power-series supremum-and-infimum riemann-sum
asked Dec 6 '18 at 11:05
user9516437user9516437
11
asked Dec 6 '18 at 11:05
user9516437user9516437
11
edited Dec 6 '18 at 11:30
user9516437
asked Dec 6 '18 at 11:05
user9516437user9516437
11
asked Dec 6 '18 at 11:05
user9516437user9516437
11
asked Dec 6 '18 at 11:05
user9516437user9516437
11
11
1
$begingroup$
look up the fundamental theorem of calculus
$endgroup$
– wilkersmon
Dec 6 '18 at 11:08
$begingroup$
It is a geomtrical serie, the sum is equal to $$h/nsumlimits_{i = 0}^{n - 1} {{e^{ih/n}}} = left( {h/n} right)frac{{1 - {e^{(h + frac{h}{n})}}}}{{1 - {e^{h/n}}}}$$
$endgroup$
– Gustave
Dec 6 '18 at 11:35
add a comment |
1
$begingroup$
look up the fundamental theorem of calculus
$endgroup$
– wilkersmon
Dec 6 '18 at 11:08
$begingroup$
It is a geomtrical serie, the sum is equal to $$h/nsumlimits_{i = 0}^{n - 1} {{e^{ih/n}}} = left( {h/n} right)frac{{1 - {e^{(h + frac{h}{n})}}}}{{1 - {e^{h/n}}}}$$
$endgroup$
– Gustave
Dec 6 '18 at 11:35
1
1
$begingroup$
look up the fundamental theorem of calculus
$endgroup$
– wilkersmon
Dec 6 '18 at 11:08
$begingroup$
look up the fundamental theorem of calculus
$endgroup$
– wilkersmon
Dec 6 '18 at 11:08
$begingroup$
It is a geomtrical serie, the sum is equal to $$h/nsumlimits_{i = 0}^{n - 1} {{e^{ih/n}}} = left( {h/n} right)frac{{1 - {e^{(h + frac{h}{n})}}}}{{1 - {e^{h/n}}}}$$
$endgroup$
– Gustave
Dec 6 '18 at 11:35
$begingroup$
It is a geomtrical serie, the sum is equal to $$h/nsumlimits_{i = 0}^{n - 1} {{e^{ih/n}}} = left( {h/n} right)frac{{1 - {e^{(h + frac{h}{n})}}}}{{1 - {e^{h/n}}}}$$
$endgroup$
– Gustave
Dec 6 '18 at 11:35
add a comment |
1 Answer
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$begingroup$
$sum_{i=0}^n e^frac{hi}{n}$ is a geometric series with ratio $e^{frac hn}$ so
$sum_{i=0}^n e^frac{hi}{n} = frac{e^{frac{h(n+1)}{n}}-1}{e^{frac hn}-1}$
As $n rightarrow infty$, we have $e^{frac{h(n+1)}{n}}-1 rightarrow e^h-1$ and $e^{frac hn}-1 rightarrow frac hn$ so
$frac hn sum_{i=0}^n e^frac{hi}{n} rightarrow e^h-1$
answered Dec 6 '18 at 11:34
gandalf61gandalf61
8,579725
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$begingroup$
$sum_{i=0}^n e^frac{hi}{n}$ is a geometric series with ratio $e^{frac hn}$ so
$sum_{i=0}^n e^frac{hi}{n} = frac{e^{frac{h(n+1)}{n}}-1}{e^{frac hn}-1}$
As $n rightarrow infty$, we have $e^{frac{h(n+1)}{n}}-1 rightarrow e^h-1$ and $e^{frac hn}-1 rightarrow frac hn$ so
$frac hn sum_{i=0}^n e^frac{hi}{n} rightarrow e^h-1$
answered Dec 6 '18 at 11:34
gandalf61gandalf61
8,579725
$endgroup$
add a comment |
$begingroup$
$sum_{i=0}^n e^frac{hi}{n}$ is a geometric series with ratio $e^{frac hn}$ so
$sum_{i=0}^n e^frac{hi}{n} = frac{e^{frac{h(n+1)}{n}}-1}{e^{frac hn}-1}$
As $n rightarrow infty$, we have $e^{frac{h(n+1)}{n}}-1 rightarrow e^h-1$ and $e^{frac hn}-1 rightarrow frac hn$ so
$frac hn sum_{i=0}^n e^frac{hi}{n} rightarrow e^h-1$
answered Dec 6 '18 at 11:34
gandalf61gandalf61
8,579725
$endgroup$
add a comment |
$begingroup$
$sum_{i=0}^n e^frac{hi}{n}$ is a geometric series with ratio $e^{frac hn}$ so
$sum_{i=0}^n e^frac{hi}{n} = frac{e^{frac{h(n+1)}{n}}-1}{e^{frac hn}-1}$
As $n rightarrow infty$, we have $e^{frac{h(n+1)}{n}}-1 rightarrow e^h-1$ and $e^{frac hn}-1 rightarrow frac hn$ so
$frac hn sum_{i=0}^n e^frac{hi}{n} rightarrow e^h-1$
answered Dec 6 '18 at 11:34
gandalf61gandalf61
8,579725
$endgroup$
$sum_{i=0}^n e^frac{hi}{n}$ is a geometric series with ratio $e^{frac hn}$ so
$sum_{i=0}^n e^frac{hi}{n} = frac{e^{frac{h(n+1)}{n}}-1}{e^{frac hn}-1}$
As $n rightarrow infty$, we have $e^{frac{h(n+1)}{n}}-1 rightarrow e^h-1$ and $e^{frac hn}-1 rightarrow frac hn$ so
$frac hn sum_{i=0}^n e^frac{hi}{n} rightarrow e^h-1$
answered Dec 6 '18 at 11:34
gandalf61gandalf61
8,579725
answered Dec 6 '18 at 11:34
gandalf61gandalf61
8,579725
answered Dec 6 '18 at 11:34
gandalf61gandalf61
8,579725
answered Dec 6 '18 at 11:34
gandalf61gandalf61
8,579725
8,579725
add a comment |
add a comment |
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1
$begingroup$
look up the fundamental theorem of calculus
$endgroup$
– wilkersmon
Dec 6 '18 at 11:08
$begingroup$
It is a geomtrical serie, the sum is equal to $$h/nsumlimits_{i = 0}^{n - 1} {{e^{ih/n}}} = left( {h/n} right)frac{{1 - {e^{(h + frac{h}{n})}}}}{{1 - {e^{h/n}}}}$$
$endgroup$
– Gustave
Dec 6 '18 at 11:35