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Why is $int_0^1e^e^xdx$ equal to $int_1^ee^uu^-1du$?

1

$begingroup$

Why is this equality $displaystyleint_0^1e^e^xdx=int_1^ee^uu^-1du$?

I don't see the change of variable that was used to pass from one integral to the other?

Could someone please explain?

And how to solve the latter?

Thank you

calculus integration change-of-variable

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asked Apr 1 at 1:17

IsaIsa

353519

$endgroup$

add a comment | 

1

$begingroup$

Why is this equality $displaystyleint_0^1e^e^xdx=int_1^ee^uu^-1du$?

I don't see the change of variable that was used to pass from one integral to the other?

Could someone please explain?

And how to solve the latter?

Thank you

calculus integration change-of-variable

share|cite|improve this question

asked Apr 1 at 1:17

IsaIsa

353519

$endgroup$

add a comment | 

$begingroup$

Why is this equality $displaystyleint_0^1e^e^xdx=int_1^ee^uu^-1du$?

I don't see the change of variable that was used to pass from one integral to the other?

Could someone please explain?

And how to solve the latter?

Thank you

calculus integration change-of-variable

share|cite|improve this question

asked Apr 1 at 1:17

IsaIsa

353519

$endgroup$

share|cite|improve this question

asked Apr 1 at 1:17

IsaIsa

353519

share|cite|improve this question

asked Apr 1 at 1:17

IsaIsa

353519

asked Apr 1 at 1:17

IsaIsa

353519

asked Apr 1 at 1:17

IsaIsa

353519

2 Answers
2

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oldest

votes

1

$begingroup$

$umapsto e^x$. This change of variable can be seen pretty easily from the bounds of integration and the integrand itself. $$beginbmatrixu\mathrm duendbmatrixmapstobeginbmatrixe^x\e^xmathrm dxendbmatrix$$ Notice that $mathrm du=e^xmathrm dx$ is equivalent to saying $mathrm du/u=mathrm dx$. $$int_0^1e^e^xmathrm dx=int_1^ee^ucdotmathrm du/u$$
This is a special integral and has a name, it is the exponential integral denoted $mathrm Ei(x)$. The definite integral can be approximated, and doing so gives $6.316563839027679$.

share|cite|improve this answer

answered Apr 1 at 9:03

Paras KhoslaParas Khosla

3,260627

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much Paras :) .You are younger than me and definitely better than me..
  $endgroup$
  – Isa
  Apr 2 at 22:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your kind words and you're most welcome. :))
  $endgroup$
  – Paras Khosla
  Apr 3 at 3:56
  

  
  
  
  

add a comment | 

3

$begingroup$

The change of variables is $u = e^x$. Notice that $du = e^x dx = u dx $

Therefore,

$$ intlimits_u(0)^u(1) e^u cdot frac1u du = intlimits_1^e e^u u^-1 du $$

share|cite|improve this answer

answered Apr 1 at 1:19

JamesJames

2,636425

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add a comment | 

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1

$begingroup$

$umapsto e^x$. This change of variable can be seen pretty easily from the bounds of integration and the integrand itself. $$beginbmatrixu\mathrm duendbmatrixmapstobeginbmatrixe^x\e^xmathrm dxendbmatrix$$ Notice that $mathrm du=e^xmathrm dx$ is equivalent to saying $mathrm du/u=mathrm dx$. $$int_0^1e^e^xmathrm dx=int_1^ee^ucdotmathrm du/u$$
This is a special integral and has a name, it is the exponential integral denoted $mathrm Ei(x)$. The definite integral can be approximated, and doing so gives $6.316563839027679$.

share|cite|improve this answer

answered Apr 1 at 9:03

Paras KhoslaParas Khosla

3,260627

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much Paras :) .You are younger than me and definitely better than me..
  $endgroup$
  – Isa
  Apr 2 at 22:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your kind words and you're most welcome. :))
  $endgroup$
  – Paras Khosla
  Apr 3 at 3:56
  

  
  
  
  

add a comment | 

1

$begingroup$

$umapsto e^x$. This change of variable can be seen pretty easily from the bounds of integration and the integrand itself. $$beginbmatrixu\mathrm duendbmatrixmapstobeginbmatrixe^x\e^xmathrm dxendbmatrix$$ Notice that $mathrm du=e^xmathrm dx$ is equivalent to saying $mathrm du/u=mathrm dx$. $$int_0^1e^e^xmathrm dx=int_1^ee^ucdotmathrm du/u$$
This is a special integral and has a name, it is the exponential integral denoted $mathrm Ei(x)$. The definite integral can be approximated, and doing so gives $6.316563839027679$.

share|cite|improve this answer

answered Apr 1 at 9:03

Paras KhoslaParas Khosla

3,260627

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much Paras :) .You are younger than me and definitely better than me..
  $endgroup$
  – Isa
  Apr 2 at 22:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your kind words and you're most welcome. :))
  $endgroup$
  – Paras Khosla
  Apr 3 at 3:56
  

  
  
  
  

add a comment | 

$begingroup$

$umapsto e^x$. This change of variable can be seen pretty easily from the bounds of integration and the integrand itself. $$beginbmatrixu\mathrm duendbmatrixmapstobeginbmatrixe^x\e^xmathrm dxendbmatrix$$ Notice that $mathrm du=e^xmathrm dx$ is equivalent to saying $mathrm du/u=mathrm dx$. $$int_0^1e^e^xmathrm dx=int_1^ee^ucdotmathrm du/u$$
This is a special integral and has a name, it is the exponential integral denoted $mathrm Ei(x)$. The definite integral can be approximated, and doing so gives $6.316563839027679$.

share|cite|improve this answer

answered Apr 1 at 9:03

Paras KhoslaParas Khosla

3,260627

$endgroup$

share|cite|improve this answer

answered Apr 1 at 9:03

Paras KhoslaParas Khosla

3,260627

answered Apr 1 at 9:03

Paras KhoslaParas Khosla

3,260627

answered Apr 1 at 9:03

Paras KhoslaParas Khosla

3,260627

3

$begingroup$

The change of variables is $u = e^x$. Notice that $du = e^x dx = u dx $

Therefore,

$$ intlimits_u(0)^u(1) e^u cdot frac1u du = intlimits_1^e e^u u^-1 du $$

share|cite|improve this answer

answered Apr 1 at 1:19

JamesJames

2,636425

$endgroup$

add a comment | 

3

$begingroup$

The change of variables is $u = e^x$. Notice that $du = e^x dx = u dx $

Therefore,

$$ intlimits_u(0)^u(1) e^u cdot frac1u du = intlimits_1^e e^u u^-1 du $$

share|cite|improve this answer

answered Apr 1 at 1:19

JamesJames

2,636425

$endgroup$

add a comment | 

$begingroup$

The change of variables is $u = e^x$. Notice that $du = e^x dx = u dx $

Therefore,

$$ intlimits_u(0)^u(1) e^u cdot frac1u du = intlimits_1^e e^u u^-1 du $$

share|cite|improve this answer

answered Apr 1 at 1:19

JamesJames

2,636425

$endgroup$

share|cite|improve this answer

answered Apr 1 at 1:19

JamesJames

2,636425

answered Apr 1 at 1:19

JamesJames

2,636425

answered Apr 1 at 1:19

JamesJames

2,636425