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Value of the double integral of $int_0^1int_1^3 (4x^3-27x^2y^2)dydx$

On the convergence of an improper integralIntegral$int_1^infty log log left(xright)fracdx1-x+x^2=frac2pisqrt 3left(frac56log (2pi)-log Gamma frac16right)$Find the second derivative of a double integralIntegral with Limit and ParametersHow to get the asymptotic form of this oscilatting integral?On the convergence of $-int_0^1left(int_1^infty left(frac1zeta(x/y)-1right)dxright)dy$How to prove the following ineqiulity : $expleft(int_0^t f(s)ds right) le 1+ int_0^t e^max(1,s)f(s)ds$Solid volume using double integralJustify an approximation of the first few decimals of $int_1^infty(int_0^1fracdx-1+x^3+y^2)dy$Double integral over triangle, what bounds should be chosen?

1

$begingroup$

$$int_0^1int_1^3left(4x^3-27x^2y^2right)dy,dx$$

$$int_0^1left(4x^3y-9x^2y^3right)Big|_1^3,dx$$

$$int_0^1left(12x^3-243x^2-4x^3+9x^2right)dx$$

$$int_0^1left(8x^3-234x^2right)dx$$

$$left(2x^4-78x^3right)Big|_0^1$$

Of course this gives a negative answer and is supposedly wrong, but I don't see any problem with my work. I know for a fact that when you take the derivative with respect to $y$, you will get $4x^3y-9x^2y^3$, and I know for a fact that when you plug in $3$ you will get $3^3$ which is $27$ and multiply that by $9$ and you get $243$! These supposed arithmetic errors are killing me! I don't even see any in this problem.

integration

share|cite|improve this question

edited Mar 16 at 21:39

Robert Howard

2,2383935

asked Dec 3 '16 at 20:30

K. GibsonK. Gibson

54621027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the region and the function?
  $endgroup$
  – AlgorithmsX
  Dec 3 '16 at 20:34
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You're solving the integral correctly, W|A confirms it as well. If that's not the right answer, perhaps the problem is in setting up the integral.
  $endgroup$
  – 2012ssohn
  Dec 3 '16 at 20:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To stop banging your head against your wall, you might want to compute the double integral the other way round, that is, integrating first with respect to $x$ then with respect to $y$. That is, start with $int_0^1(4x^3-27x^2y^2)dx=left.x^4-9x^3y^2right|_0^1=1-9y^2$ hence your double integral is $int_1^3(1-9y^2)dy=left.y-3y^3right|_1^3=3-3cdot3^3-1+3=-76$. This does not prove your first computation is correct but this definitely suggests that it might be so...
  $endgroup$
  – Did
  Dec 3 '16 at 20:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are we supposed to be integrating over the rectangle bound by the $y$-axis and the lines $x=1$, $y=1$, and $y=3$? If not, then I'm afraid you've set up the integral incorrectly.
  $endgroup$
  – teadawg1337
  Dec 3 '16 at 20:58
  

  
  
  
  

add a comment | 

1

$begingroup$

$$int_0^1int_1^3left(4x^3-27x^2y^2right)dy,dx$$

$$int_0^1left(4x^3y-9x^2y^3right)Big|_1^3,dx$$

$$int_0^1left(12x^3-243x^2-4x^3+9x^2right)dx$$

$$int_0^1left(8x^3-234x^2right)dx$$

$$left(2x^4-78x^3right)Big|_0^1$$

Of course this gives a negative answer and is supposedly wrong, but I don't see any problem with my work. I know for a fact that when you take the derivative with respect to $y$, you will get $4x^3y-9x^2y^3$, and I know for a fact that when you plug in $3$ you will get $3^3$ which is $27$ and multiply that by $9$ and you get $243$! These supposed arithmetic errors are killing me! I don't even see any in this problem.

integration

share|cite|improve this question

edited Mar 16 at 21:39

Robert Howard

2,2383935

asked Dec 3 '16 at 20:30

K. GibsonK. Gibson

54621027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the region and the function?
  $endgroup$
  – AlgorithmsX
  Dec 3 '16 at 20:34
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You're solving the integral correctly, W|A confirms it as well. If that's not the right answer, perhaps the problem is in setting up the integral.
  $endgroup$
  – 2012ssohn
  Dec 3 '16 at 20:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To stop banging your head against your wall, you might want to compute the double integral the other way round, that is, integrating first with respect to $x$ then with respect to $y$. That is, start with $int_0^1(4x^3-27x^2y^2)dx=left.x^4-9x^3y^2right|_0^1=1-9y^2$ hence your double integral is $int_1^3(1-9y^2)dy=left.y-3y^3right|_1^3=3-3cdot3^3-1+3=-76$. This does not prove your first computation is correct but this definitely suggests that it might be so...
  $endgroup$
  – Did
  Dec 3 '16 at 20:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are we supposed to be integrating over the rectangle bound by the $y$-axis and the lines $x=1$, $y=1$, and $y=3$? If not, then I'm afraid you've set up the integral incorrectly.
  $endgroup$
  – teadawg1337
  Dec 3 '16 at 20:58
  

  
  
  
  

add a comment | 

$begingroup$

$$int_0^1int_1^3left(4x^3-27x^2y^2right)dy,dx$$

$$int_0^1left(4x^3y-9x^2y^3right)Big|_1^3,dx$$

$$int_0^1left(12x^3-243x^2-4x^3+9x^2right)dx$$

$$int_0^1left(8x^3-234x^2right)dx$$

$$left(2x^4-78x^3right)Big|_0^1$$

Of course this gives a negative answer and is supposedly wrong, but I don't see any problem with my work. I know for a fact that when you take the derivative with respect to $y$, you will get $4x^3y-9x^2y^3$, and I know for a fact that when you plug in $3$ you will get $3^3$ which is $27$ and multiply that by $9$ and you get $243$! These supposed arithmetic errors are killing me! I don't even see any in this problem.

integration

share|cite|improve this question

edited Mar 16 at 21:39

Robert Howard

2,2383935

asked Dec 3 '16 at 20:30

K. GibsonK. Gibson

54621027

$endgroup$

share|cite|improve this question

edited Mar 16 at 21:39

Robert Howard

2,2383935

asked Dec 3 '16 at 20:30

K. GibsonK. Gibson

54621027

share|cite|improve this question

edited Mar 16 at 21:39

Robert Howard

2,2383935

asked Dec 3 '16 at 20:30

K. GibsonK. Gibson

54621027

edited Mar 16 at 21:39

Robert Howard

2,2383935

edited Mar 16 at 21:39

Robert Howard

2,2383935

asked Dec 3 '16 at 20:30

K. GibsonK. Gibson

54621027

asked Dec 3 '16 at 20:30

K. GibsonK. Gibson

54621027