How change the order and integrate this function $e^{x^{2}}$?












0












$begingroup$


I need to change the order of integration and evaluate the following:



$int_{0}^{1} int_{sqrt{y}}^{1} e^{x^2}dxdy$



$x=sqrt{y}, x=1$



Would this work: $int_{0}^{1} int_{0}^{x^2} e^{x^3}dydx$ ?






integration multivariable-calculus







share|cite|improve this question











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  • 2




    $begingroup$
    Your notation is ambigious. Please write integrals at $int_a^b dots ,color{red}{mathrm dx}$ etc. so that we know what is the integration variable.
    $endgroup$
    – Christoph
    Dec 19 '18 at 16:52










  • $begingroup$
    Sorry about that; I added them in.
    $endgroup$
    – Jaigus
    Dec 19 '18 at 16:53










  • $begingroup$
    I believe you got the right idea, except that $e^{x^2}$ should not have become $e^{x^3}$
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 16:55










  • $begingroup$
    Draw a picture! You must draw a picture when you first start working on these problems to build an intuition.
    $endgroup$
    – Squirtle
    Dec 19 '18 at 16:58
















0












$begingroup$


I need to change the order of integration and evaluate the following:



$int_{0}^{1} int_{sqrt{y}}^{1} e^{x^2}dxdy$



$x=sqrt{y}, x=1$



Would this work: $int_{0}^{1} int_{0}^{x^2} e^{x^3}dydx$ ?






integration multivariable-calculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Your notation is ambigious. Please write integrals at $int_a^b dots ,color{red}{mathrm dx}$ etc. so that we know what is the integration variable.
    $endgroup$
    – Christoph
    Dec 19 '18 at 16:52










  • $begingroup$
    Sorry about that; I added them in.
    $endgroup$
    – Jaigus
    Dec 19 '18 at 16:53










  • $begingroup$
    I believe you got the right idea, except that $e^{x^2}$ should not have become $e^{x^3}$
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 16:55










  • $begingroup$
    Draw a picture! You must draw a picture when you first start working on these problems to build an intuition.
    $endgroup$
    – Squirtle
    Dec 19 '18 at 16:58














0












0








0





$begingroup$


I need to change the order of integration and evaluate the following:



$int_{0}^{1} int_{sqrt{y}}^{1} e^{x^2}dxdy$



$x=sqrt{y}, x=1$



Would this work: $int_{0}^{1} int_{0}^{x^2} e^{x^3}dydx$ ?






integration multivariable-calculus







share|cite|improve this question











$endgroup$




I need to change the order of integration and evaluate the following:



$int_{0}^{1} int_{sqrt{y}}^{1} e^{x^2}dxdy$



$x=sqrt{y}, x=1$



Would this work: $int_{0}^{1} int_{0}^{x^2} e^{x^3}dydx$ ?






integration multivariable-calculus




integration multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 16:53







Jaigus

















asked Dec 19 '18 at 16:46









JaigusJaigus

2218




2218








  • 2




    $begingroup$
    Your notation is ambigious. Please write integrals at $int_a^b dots ,color{red}{mathrm dx}$ etc. so that we know what is the integration variable.
    $endgroup$
    – Christoph
    Dec 19 '18 at 16:52










  • $begingroup$
    Sorry about that; I added them in.
    $endgroup$
    – Jaigus
    Dec 19 '18 at 16:53










  • $begingroup$
    I believe you got the right idea, except that $e^{x^2}$ should not have become $e^{x^3}$
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 16:55










  • $begingroup$
    Draw a picture! You must draw a picture when you first start working on these problems to build an intuition.
    $endgroup$
    – Squirtle
    Dec 19 '18 at 16:58














  • 2




    $begingroup$
    Your notation is ambigious. Please write integrals at $int_a^b dots ,color{red}{mathrm dx}$ etc. so that we know what is the integration variable.
    $endgroup$
    – Christoph
    Dec 19 '18 at 16:52










  • $begingroup$
    Sorry about that; I added them in.
    $endgroup$
    – Jaigus
    Dec 19 '18 at 16:53










  • $begingroup$
    I believe you got the right idea, except that $e^{x^2}$ should not have become $e^{x^3}$
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 16:55










  • $begingroup$
    Draw a picture! You must draw a picture when you first start working on these problems to build an intuition.
    $endgroup$
    – Squirtle
    Dec 19 '18 at 16:58








2




2




$begingroup$
Your notation is ambigious. Please write integrals at $int_a^b dots ,color{red}{mathrm dx}$ etc. so that we know what is the integration variable.
$endgroup$
– Christoph
Dec 19 '18 at 16:52




$begingroup$
Your notation is ambigious. Please write integrals at $int_a^b dots ,color{red}{mathrm dx}$ etc. so that we know what is the integration variable.
$endgroup$
– Christoph
Dec 19 '18 at 16:52












$begingroup$
Sorry about that; I added them in.
$endgroup$
– Jaigus
Dec 19 '18 at 16:53




$begingroup$
Sorry about that; I added them in.
$endgroup$
– Jaigus
Dec 19 '18 at 16:53












$begingroup$
I believe you got the right idea, except that $e^{x^2}$ should not have become $e^{x^3}$
$endgroup$
– SmileyCraft
Dec 19 '18 at 16:55




$begingroup$
I believe you got the right idea, except that $e^{x^2}$ should not have become $e^{x^3}$
$endgroup$
– SmileyCraft
Dec 19 '18 at 16:55












$begingroup$
Draw a picture! You must draw a picture when you first start working on these problems to build an intuition.
$endgroup$
– Squirtle
Dec 19 '18 at 16:58




$begingroup$
Draw a picture! You must draw a picture when you first start working on these problems to build an intuition.
$endgroup$
– Squirtle
Dec 19 '18 at 16:58










1 Answer
1






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2












$begingroup$

The original domain is $0 le y le 1, sqrt{y} le x le 1$



It is equivalent to $0 le x le 1, 0 le y le x^2$.



Hence the part about changing the variable is correct, however, note that the function that you are integrating should remain the same.



begin{align}
int_0^1 x^2 exp(x^2) , dx &= int_0^1 frac{x}2 (2x) exp(x^2) , dx \
&= frac{x}2 exp(x^2) |_0^1 - int_0^1 frac12 exp(x^2) , dx \
&= frac{e}2 - frac12 int_0^1 exp(x^2) , dx \
&= frac{e}{2} - frac{sqrt{pi}}{4}erfi(1)
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, so it should be $int_{0}^{1} int_{0}^{x^2} e^{x^2}dydx$. But when integrating this, I am stuck at $int_{0}^{1} e^{x^2}x^2dx$ because substitution creates more variables, and by parts doesn't seem to be getting me anywhere.
    $endgroup$
    – Jaigus
    Dec 19 '18 at 17:01













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

The original domain is $0 le y le 1, sqrt{y} le x le 1$



It is equivalent to $0 le x le 1, 0 le y le x^2$.



Hence the part about changing the variable is correct, however, note that the function that you are integrating should remain the same.



begin{align}
int_0^1 x^2 exp(x^2) , dx &= int_0^1 frac{x}2 (2x) exp(x^2) , dx \
&= frac{x}2 exp(x^2) |_0^1 - int_0^1 frac12 exp(x^2) , dx \
&= frac{e}2 - frac12 int_0^1 exp(x^2) , dx \
&= frac{e}{2} - frac{sqrt{pi}}{4}erfi(1)
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, so it should be $int_{0}^{1} int_{0}^{x^2} e^{x^2}dydx$. But when integrating this, I am stuck at $int_{0}^{1} e^{x^2}x^2dx$ because substitution creates more variables, and by parts doesn't seem to be getting me anywhere.
    $endgroup$
    – Jaigus
    Dec 19 '18 at 17:01


















2












$begingroup$

The original domain is $0 le y le 1, sqrt{y} le x le 1$



It is equivalent to $0 le x le 1, 0 le y le x^2$.



Hence the part about changing the variable is correct, however, note that the function that you are integrating should remain the same.



begin{align}
int_0^1 x^2 exp(x^2) , dx &= int_0^1 frac{x}2 (2x) exp(x^2) , dx \
&= frac{x}2 exp(x^2) |_0^1 - int_0^1 frac12 exp(x^2) , dx \
&= frac{e}2 - frac12 int_0^1 exp(x^2) , dx \
&= frac{e}{2} - frac{sqrt{pi}}{4}erfi(1)
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, so it should be $int_{0}^{1} int_{0}^{x^2} e^{x^2}dydx$. But when integrating this, I am stuck at $int_{0}^{1} e^{x^2}x^2dx$ because substitution creates more variables, and by parts doesn't seem to be getting me anywhere.
    $endgroup$
    – Jaigus
    Dec 19 '18 at 17:01
















2












2








2





$begingroup$

The original domain is $0 le y le 1, sqrt{y} le x le 1$



It is equivalent to $0 le x le 1, 0 le y le x^2$.



Hence the part about changing the variable is correct, however, note that the function that you are integrating should remain the same.



begin{align}
int_0^1 x^2 exp(x^2) , dx &= int_0^1 frac{x}2 (2x) exp(x^2) , dx \
&= frac{x}2 exp(x^2) |_0^1 - int_0^1 frac12 exp(x^2) , dx \
&= frac{e}2 - frac12 int_0^1 exp(x^2) , dx \
&= frac{e}{2} - frac{sqrt{pi}}{4}erfi(1)
end{align}






share|cite|improve this answer











$endgroup$



The original domain is $0 le y le 1, sqrt{y} le x le 1$



It is equivalent to $0 le x le 1, 0 le y le x^2$.



Hence the part about changing the variable is correct, however, note that the function that you are integrating should remain the same.



begin{align}
int_0^1 x^2 exp(x^2) , dx &= int_0^1 frac{x}2 (2x) exp(x^2) , dx \
&= frac{x}2 exp(x^2) |_0^1 - int_0^1 frac12 exp(x^2) , dx \
&= frac{e}2 - frac12 int_0^1 exp(x^2) , dx \
&= frac{e}{2} - frac{sqrt{pi}}{4}erfi(1)
end{align}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 19 '18 at 17:25

























answered Dec 19 '18 at 16:54









Siong Thye GohSiong Thye Goh

100k1466117




100k1466117












  • $begingroup$
    Thanks, so it should be $int_{0}^{1} int_{0}^{x^2} e^{x^2}dydx$. But when integrating this, I am stuck at $int_{0}^{1} e^{x^2}x^2dx$ because substitution creates more variables, and by parts doesn't seem to be getting me anywhere.
    $endgroup$
    – Jaigus
    Dec 19 '18 at 17:01




















  • $begingroup$
    Thanks, so it should be $int_{0}^{1} int_{0}^{x^2} e^{x^2}dydx$. But when integrating this, I am stuck at $int_{0}^{1} e^{x^2}x^2dx$ because substitution creates more variables, and by parts doesn't seem to be getting me anywhere.
    $endgroup$
    – Jaigus
    Dec 19 '18 at 17:01


















$begingroup$
Thanks, so it should be $int_{0}^{1} int_{0}^{x^2} e^{x^2}dydx$. But when integrating this, I am stuck at $int_{0}^{1} e^{x^2}x^2dx$ because substitution creates more variables, and by parts doesn't seem to be getting me anywhere.
$endgroup$
– Jaigus
Dec 19 '18 at 17:01






$begingroup$
Thanks, so it should be $int_{0}^{1} int_{0}^{x^2} e^{x^2}dydx$. But when integrating this, I am stuck at $int_{0}^{1} e^{x^2}x^2dx$ because substitution creates more variables, and by parts doesn't seem to be getting me anywhere.
$endgroup$
– Jaigus
Dec 19 '18 at 17:01




















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