exercise on double integration

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I am not sure if i am not understanding the concept of double integration right, or if i am just making a silly mistake, but here is the problem:

given the function $f(x,y)=24xy, y>0, x>0$ and $x+y<=1 $ fint the $E[X,Y]$

$E[X,Y] = int int 24x^2y^2dxdy$ =

$24 int_0^xx^2 (int_0^1-xy^2dy) dx $ =

$24 int_0^xx^2 [fracy^33]_0^1-xdx $=

$24 int_0^xx^2 frac(1-x)^33dx $=

$8 int_0^xx^2 (1-x)^3dx $=

$8[fracx^3x frac(1-x)^44frac-x^22]_0^1$

= $ 8-frac 512$

However, my professor gave us different solution:

$E[X,Y] = int int 24x^2y^2dxdy$ =

$24 int_0^xx^2 (int_0^1-xy^2dy) dx $ =

$3 int_0^1 x^2 (1-x)^3 dx = $

$ frac215$

What is the problem with my calculation?

Where does the 3 come from in $3 int_0^1 x^2 (1-x)^3 dx $ ?

probability integration definite-integrals expectation

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edited Jul 28 at 9:34

BCLC

6,98421973

asked Jul 28 at 9:19

user1607

608

add a comment | 

up vote
1
down vote

favorite

I am not sure if i am not understanding the concept of double integration right, or if i am just making a silly mistake, but here is the problem:

given the function $f(x,y)=24xy, y>0, x>0$ and $x+y<=1 $ fint the $E[X,Y]$

$E[X,Y] = int int 24x^2y^2dxdy$ =

$24 int_0^xx^2 (int_0^1-xy^2dy) dx $ =

$24 int_0^xx^2 [fracy^33]_0^1-xdx $=

$24 int_0^xx^2 frac(1-x)^33dx $=

$8 int_0^xx^2 (1-x)^3dx $=

$8[fracx^3x frac(1-x)^44frac-x^22]_0^1$

= $ 8-frac 512$

However, my professor gave us different solution:

$E[X,Y] = int int 24x^2y^2dxdy$ =

$24 int_0^xx^2 (int_0^1-xy^2dy) dx $ =

$3 int_0^1 x^2 (1-x)^3 dx = $

$ frac215$

What is the problem with my calculation?

Where does the 3 come from in $3 int_0^1 x^2 (1-x)^3 dx $ ?

probability integration definite-integrals expectation

share|cite|improve this question

edited Jul 28 at 9:34

BCLC

6,98421973

asked Jul 28 at 9:19

user1607

608

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I am not sure if i am not understanding the concept of double integration right, or if i am just making a silly mistake, but here is the problem:

given the function $f(x,y)=24xy, y>0, x>0$ and $x+y<=1 $ fint the $E[X,Y]$

$E[X,Y] = int int 24x^2y^2dxdy$ =

$24 int_0^xx^2 (int_0^1-xy^2dy) dx $ =

$24 int_0^xx^2 [fracy^33]_0^1-xdx $=

$24 int_0^xx^2 frac(1-x)^33dx $=

$8 int_0^xx^2 (1-x)^3dx $=

$8[fracx^3x frac(1-x)^44frac-x^22]_0^1$

= $ 8-frac 512$

However, my professor gave us different solution:

$E[X,Y] = int int 24x^2y^2dxdy$ =

$24 int_0^xx^2 (int_0^1-xy^2dy) dx $ =

$3 int_0^1 x^2 (1-x)^3 dx = $

$ frac215$

What is the problem with my calculation?

Where does the 3 come from in $3 int_0^1 x^2 (1-x)^3 dx $ ?

probability integration definite-integrals expectation

share|cite|improve this question

edited Jul 28 at 9:34

BCLC

6,98421973

asked Jul 28 at 9:19

user1607

608

I am not sure if i am not understanding the concept of double integration right, or if i am just making a silly mistake, but here is the problem:

given the function $f(x,y)=24xy, y>0, x>0$ and $x+y<=1 $ fint the $E[X,Y]$

$E[X,Y] = int int 24x^2y^2dxdy$ =

$24 int_0^xx^2 (int_0^1-xy^2dy) dx $ =

$24 int_0^xx^2 [fracy^33]_0^1-xdx $=

$24 int_0^xx^2 frac(1-x)^33dx $=

$8 int_0^xx^2 (1-x)^3dx $=

$8[fracx^3x frac(1-x)^44frac-x^22]_0^1$

= $ 8-frac 512$

However, my professor gave us different solution:

$E[X,Y] = int int 24x^2y^2dxdy$ =

$24 int_0^xx^2 (int_0^1-xy^2dy) dx $ =

$3 int_0^1 x^2 (1-x)^3 dx = $

$ frac215$

What is the problem with my calculation?

Where does the 3 come from in $3 int_0^1 x^2 (1-x)^3 dx $ ?

probability integration definite-integrals expectation

share|cite|improve this question

edited Jul 28 at 9:34

BCLC

6,98421973

asked Jul 28 at 9:19

user1607

608

share|cite|improve this question

share|cite|improve this question

edited Jul 28 at 9:34

BCLC

6,98421973

edited Jul 28 at 9:34

BCLC

6,98421973

edited Jul 28 at 9:34

BCLC

6,98421973

6,98421973

asked Jul 28 at 9:19

user1607

608

asked Jul 28 at 9:19

user1607

608

asked Jul 28 at 9:19

user1607

608

608

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

I believe you either misread '8' as '3', or there is a typo on your prof's part.

beginalign
E &= int_x=0^x=1int_y=0^y=1-x 24x^2y^2 , dy , dx \
&= 24int_x=0^x=1x^2int_y=0^y=1-x y^2 , dy ,dx \
&= 24int_x=0^x=1x^2left[fracy^33right]_y=0^y=1-x ,dx \
&= 8int_0^1x^2(1-x)^3 ,dx \
&= 8int_0^1x^2(1-x)^3 ,dx \
&= 8int_0^1x^2(1-3x+3x^2-x^3) ,dx \
&= 8int_0^1x^2-3x^3+3x^4-x^5 ,dx \
&= 8left[fracx^33-frac3x^44+frac3x^55-fracx^66right]_0^1\
&= 8left(frac 13-frac34+frac35-frac16right)\
&= 8cdotfrac160 \
&=frac215
endalign

share|cite|improve this answer

edited Jul 28 at 10:00

answered Jul 28 at 9:39

Karn Watcharasupat

3,7992426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  must be a typo. Can you also provide how you solved $int_0^1x^2(1-x)^3dx$ ? I am not getting $frac160 but frac512$
  – user1607
  Jul 28 at 9:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user1607 sure give me a moment
  – Karn Watcharasupat
  Jul 28 at 9:55
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

I believe you either misread '8' as '3', or there is a typo on your prof's part.

beginalign
E &= int_x=0^x=1int_y=0^y=1-x 24x^2y^2 , dy , dx \
&= 24int_x=0^x=1x^2int_y=0^y=1-x y^2 , dy ,dx \
&= 24int_x=0^x=1x^2left[fracy^33right]_y=0^y=1-x ,dx \
&= 8int_0^1x^2(1-x)^3 ,dx \
&= 8int_0^1x^2(1-x)^3 ,dx \
&= 8int_0^1x^2(1-3x+3x^2-x^3) ,dx \
&= 8int_0^1x^2-3x^3+3x^4-x^5 ,dx \
&= 8left[fracx^33-frac3x^44+frac3x^55-fracx^66right]_0^1\
&= 8left(frac 13-frac34+frac35-frac16right)\
&= 8cdotfrac160 \
&=frac215
endalign

share|cite|improve this answer

edited Jul 28 at 10:00

answered Jul 28 at 9:39

Karn Watcharasupat

3,7992426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  must be a typo. Can you also provide how you solved $int_0^1x^2(1-x)^3dx$ ? I am not getting $frac160 but frac512$
  – user1607
  Jul 28 at 9:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user1607 sure give me a moment
  – Karn Watcharasupat
  Jul 28 at 9:55
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

I believe you either misread '8' as '3', or there is a typo on your prof's part.

beginalign
E &= int_x=0^x=1int_y=0^y=1-x 24x^2y^2 , dy , dx \
&= 24int_x=0^x=1x^2int_y=0^y=1-x y^2 , dy ,dx \
&= 24int_x=0^x=1x^2left[fracy^33right]_y=0^y=1-x ,dx \
&= 8int_0^1x^2(1-x)^3 ,dx \
&= 8int_0^1x^2(1-x)^3 ,dx \
&= 8int_0^1x^2(1-3x+3x^2-x^3) ,dx \
&= 8int_0^1x^2-3x^3+3x^4-x^5 ,dx \
&= 8left[fracx^33-frac3x^44+frac3x^55-fracx^66right]_0^1\
&= 8left(frac 13-frac34+frac35-frac16right)\
&= 8cdotfrac160 \
&=frac215
endalign

share|cite|improve this answer

edited Jul 28 at 10:00

answered Jul 28 at 9:39

Karn Watcharasupat

3,7992426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  must be a typo. Can you also provide how you solved $int_0^1x^2(1-x)^3dx$ ? I am not getting $frac160 but frac512$
  – user1607
  Jul 28 at 9:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user1607 sure give me a moment
  – Karn Watcharasupat
  Jul 28 at 9:55
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

I believe you either misread '8' as '3', or there is a typo on your prof's part.

beginalign
E &= int_x=0^x=1int_y=0^y=1-x 24x^2y^2 , dy , dx \
&= 24int_x=0^x=1x^2int_y=0^y=1-x y^2 , dy ,dx \
&= 24int_x=0^x=1x^2left[fracy^33right]_y=0^y=1-x ,dx \
&= 8int_0^1x^2(1-x)^3 ,dx \
&= 8int_0^1x^2(1-x)^3 ,dx \
&= 8int_0^1x^2(1-3x+3x^2-x^3) ,dx \
&= 8int_0^1x^2-3x^3+3x^4-x^5 ,dx \
&= 8left[fracx^33-frac3x^44+frac3x^55-fracx^66right]_0^1\
&= 8left(frac 13-frac34+frac35-frac16right)\
&= 8cdotfrac160 \
&=frac215
endalign

share|cite|improve this answer

edited Jul 28 at 10:00

answered Jul 28 at 9:39

Karn Watcharasupat

3,7992426

I believe you either misread '8' as '3', or there is a typo on your prof's part.

beginalign
E &= int_x=0^x=1int_y=0^y=1-x 24x^2y^2 , dy , dx \
&= 24int_x=0^x=1x^2int_y=0^y=1-x y^2 , dy ,dx \
&= 24int_x=0^x=1x^2left[fracy^33right]_y=0^y=1-x ,dx \
&= 8int_0^1x^2(1-x)^3 ,dx \
&= 8int_0^1x^2(1-x)^3 ,dx \
&= 8int_0^1x^2(1-3x+3x^2-x^3) ,dx \
&= 8int_0^1x^2-3x^3+3x^4-x^5 ,dx \
&= 8left[fracx^33-frac3x^44+frac3x^55-fracx^66right]_0^1\
&= 8left(frac 13-frac34+frac35-frac16right)\
&= 8cdotfrac160 \
&=frac215
endalign

share|cite|improve this answer

edited Jul 28 at 10:00

answered Jul 28 at 9:39

Karn Watcharasupat

3,7992426

share|cite|improve this answer

share|cite|improve this answer

edited Jul 28 at 10:00

edited Jul 28 at 10:00

edited Jul 28 at 10:00

answered Jul 28 at 9:39

Karn Watcharasupat

3,7992426

answered Jul 28 at 9:39

Karn Watcharasupat

3,7992426

answered Jul 28 at 9:39

Karn Watcharasupat

3,7992426

3,7992426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  must be a typo. Can you also provide how you solved $int_0^1x^2(1-x)^3dx$ ? I am not getting $frac160 but frac512$
  – user1607
  Jul 28 at 9:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user1607 sure give me a moment
  – Karn Watcharasupat
  Jul 28 at 9:55
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  must be a typo. Can you also provide how you solved $int_0^1x^2(1-x)^3dx$ ? I am not getting $frac160 but frac512$
  – user1607
  Jul 28 at 9:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user1607 sure give me a moment
  – Karn Watcharasupat
  Jul 28 at 9:55
  

  
  
  
  

must be a typo. Can you also provide how you solved $int_0^1x^2(1-x)^3dx$ ? I am not getting $frac160 but frac512$
– user1607
Jul 28 at 9:54

must be a typo. Can you also provide how you solved $int_0^1x^2(1-x)^3dx$ ? I am not getting $frac160 but frac512$
– user1607
Jul 28 at 9:54

@user1607 sure give me a moment
– Karn Watcharasupat
Jul 28 at 9:55

@user1607 sure give me a moment
– Karn Watcharasupat
Jul 28 at 9:55

add a comment | 

 

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