Proof area of {$(x,y) R^2 | 0 le y < e^x$ and $0le x< h$} is $e^h-1$












0












$begingroup$


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







share|cite|improve this question











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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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0


0



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 11:30







user9516437

















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














  • 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










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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



$frac hn sum_{i=0}^n e^frac{hi}{n} rightarrow e^h-1$






share|cite|improve this 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



    $frac hn sum_{i=0}^n e^frac{hi}{n} rightarrow e^h-1$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $sum_{i=0}^n e^frac{hi}{n}$ 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



      $frac hn sum_{i=0}^n e^frac{hi}{n} rightarrow e^h-1$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $sum_{i=0}^n e^frac{hi}{n}$ 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



        $frac hn sum_{i=0}^n e^frac{hi}{n} rightarrow e^h-1$






        share|cite|improve this answer









        $endgroup$



        $sum_{i=0}^n e^frac{hi}{n}$ 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



        $frac hn sum_{i=0}^n e^frac{hi}{n} rightarrow e^h-1$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 11:34









        gandalf61gandalf61

        8,579725




        8,579725






























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