Prove this inequality using integrals [on hold]

-3

Prove:

$$ frac{e-1}{2e} le int_0^1 frac{e^{-x}}{1+x}dx le ln 2$$

Can clearly see that $ 2e ge 1+x ge 1 $ for every $ 0 le x le 1 $ but $ frac{e^{-x}}{1+x} ge ln 2 $ for $ x = 1 $

asked Jan 6 at 10:35

bm1125

63116

put on hold as off-topic by Saad, RRL, Xander Henderson, Did, A. Pongrácz 2 days ago

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-3

Prove:

$$ frac{e-1}{2e} le int_0^1 frac{e^{-x}}{1+x}dx le ln 2$$

Can clearly see that $ 2e ge 1+x ge 1 $ for every $ 0 le x le 1 $ but $ frac{e^{-x}}{1+x} ge ln 2 $ for $ x = 1 $

asked Jan 6 at 10:35

bm1125

63116

put on hold as off-topic by Saad, RRL, Xander Henderson, Did, A. Pongrácz 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, Xander Henderson, Did, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment |

-3

Prove:

$$ frac{e-1}{2e} le int_0^1 frac{e^{-x}}{1+x}dx le ln 2$$

Can clearly see that $ 2e ge 1+x ge 1 $ for every $ 0 le x le 1 $ but $ frac{e^{-x}}{1+x} ge ln 2 $ for $ x = 1 $

asked Jan 6 at 10:35

bm1125

63116

Prove:

$$ frac{e-1}{2e} le int_0^1 frac{e^{-x}}{1+x}dx le ln 2$$

Can clearly see that $ 2e ge 1+x ge 1 $ for every $ 0 le x le 1 $ but $ frac{e^{-x}}{1+x} ge ln 2 $ for $ x = 1 $

calculus

asked Jan 6 at 10:35

bm1125

63116

asked Jan 6 at 10:35

bm1125

63116

asked Jan 6 at 10:35

bm1125

63116

asked Jan 6 at 10:35

bm1125

63116

asked Jan 6 at 10:35

bm1125

63116

put on hold as off-topic by Saad, RRL, Xander Henderson, Did, A. Pongrácz 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, Xander Henderson, Did, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as off-topic by Saad, RRL, Xander Henderson, Did, A. Pongrácz 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, Xander Henderson, Did, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment |

1 Answer
1

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First inequality: in the domain, $frac{1}{1+x} geq frac{1}{2}$. Thus $int_0^1{frac{e^{-x}}{1+x}} geq frac{1}{2}int_0^1{e^{-x}}=frac{e-1}{2e}$.

Second inequality: in the domain, $e^{-x} leq 1$.
Thus $int_0^1{frac{e^{-x}}{1+x}} leq int_0^1{frac{1}{1+x}} = log(2)$.

edited Jan 6 at 11:12

answered Jan 6 at 10:38

Mindlack

2,32217

thanks but still if $ x = 0 $ then $ e^{-x} = 1 $ and $ frac{1}{1+x} = 1 $ and then it is greater than $ ln 2 $
– bm1125
Jan 6 at 11:01

For $xin[0,1]$,$$frac{e^{-x}}2lefrac{e^{-x}}{1+x}lefrac1{1+x}$$
– Shubham Johri
Jan 6 at 11:10

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

First inequality: in the domain, $frac{1}{1+x} geq frac{1}{2}$. Thus $int_0^1{frac{e^{-x}}{1+x}} geq frac{1}{2}int_0^1{e^{-x}}=frac{e-1}{2e}$.

Second inequality: in the domain, $e^{-x} leq 1$.
Thus $int_0^1{frac{e^{-x}}{1+x}} leq int_0^1{frac{1}{1+x}} = log(2)$.

edited Jan 6 at 11:12

answered Jan 6 at 10:38

Mindlack

2,32217

thanks but still if $ x = 0 $ then $ e^{-x} = 1 $ and $ frac{1}{1+x} = 1 $ and then it is greater than $ ln 2 $
– bm1125
Jan 6 at 11:01

For $xin[0,1]$,$$frac{e^{-x}}2lefrac{e^{-x}}{1+x}lefrac1{1+x}$$
– Shubham Johri
Jan 6 at 11:10

add a comment |

First inequality: in the domain, $frac{1}{1+x} geq frac{1}{2}$. Thus $int_0^1{frac{e^{-x}}{1+x}} geq frac{1}{2}int_0^1{e^{-x}}=frac{e-1}{2e}$.

Second inequality: in the domain, $e^{-x} leq 1$.
Thus $int_0^1{frac{e^{-x}}{1+x}} leq int_0^1{frac{1}{1+x}} = log(2)$.

edited Jan 6 at 11:12

answered Jan 6 at 10:38

Mindlack

2,32217

thanks but still if $ x = 0 $ then $ e^{-x} = 1 $ and $ frac{1}{1+x} = 1 $ and then it is greater than $ ln 2 $
– bm1125
Jan 6 at 11:01

For $xin[0,1]$,$$frac{e^{-x}}2lefrac{e^{-x}}{1+x}lefrac1{1+x}$$
– Shubham Johri
Jan 6 at 11:10

add a comment |

First inequality: in the domain, $frac{1}{1+x} geq frac{1}{2}$. Thus $int_0^1{frac{e^{-x}}{1+x}} geq frac{1}{2}int_0^1{e^{-x}}=frac{e-1}{2e}$.

Second inequality: in the domain, $e^{-x} leq 1$.
Thus $int_0^1{frac{e^{-x}}{1+x}} leq int_0^1{frac{1}{1+x}} = log(2)$.

edited Jan 6 at 11:12

answered Jan 6 at 10:38

Mindlack

2,32217

First inequality: in the domain, $frac{1}{1+x} geq frac{1}{2}$. Thus $int_0^1{frac{e^{-x}}{1+x}} geq frac{1}{2}int_0^1{e^{-x}}=frac{e-1}{2e}$.

Second inequality: in the domain, $e^{-x} leq 1$.
Thus $int_0^1{frac{e^{-x}}{1+x}} leq int_0^1{frac{1}{1+x}} = log(2)$.

edited Jan 6 at 11:12

answered Jan 6 at 10:38

Mindlack

2,32217

edited Jan 6 at 11:12

answered Jan 6 at 10:38

Mindlack

2,32217

answered Jan 6 at 10:38

Mindlack

2,32217

answered Jan 6 at 10:38

Mindlack

2,32217

thanks but still if $ x = 0 $ then $ e^{-x} = 1 $ and $ frac{1}{1+x} = 1 $ and then it is greater than $ ln 2 $
– bm1125
Jan 6 at 11:01

For $xin[0,1]$,$$frac{e^{-x}}2lefrac{e^{-x}}{1+x}lefrac1{1+x}$$
– Shubham Johri
Jan 6 at 11:10

add a comment |

thanks but still if $ x = 0 $ then $ e^{-x} = 1 $ and $ frac{1}{1+x} = 1 $ and then it is greater than $ ln 2 $
– bm1125
Jan 6 at 11:01

For $xin[0,1]$,$$frac{e^{-x}}2lefrac{e^{-x}}{1+x}lefrac1{1+x}$$
– Shubham Johri
Jan 6 at 11:10

thanks but still if $ x = 0 $ then $ e^{-x} = 1 $ and $ frac{1}{1+x} = 1 $ and then it is greater than $ ln 2 $
– bm1125
Jan 6 at 11:01

For $xin[0,1]$,$$frac{e^{-x}}2lefrac{e^{-x}}{1+x}lefrac1{1+x}$$
– Shubham Johri
Jan 6 at 11:10

add a comment |

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