Integration anomaly? [closed]

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I'm wanting to integrate the following function: $[(a^2-b^2)x^-1.5-x^-0.5] dx$ with the upper limit $(a+b)^2$ and the lower limit $(a-b)^2$.
The result is $ -2 (a^2-b^2+x)/x^0.5$.

My problem arises when I plug in the limits and get zero as the result when it should be $-2/b^2$.
Any help would be appreciated.

riemann-integration

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edited Jul 31 at 10:46

Yves Daoust

110k665203

asked Jul 31 at 10:41

Danny Colautti

12

closed as off-topic by John Ma, Isaac Browne, Mostafa Ayaz, amWhy, Xander Henderson Aug 1 at 2:11

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Isaac Browne, Mostafa Ayaz, amWhy, Xander Henderson

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Perhaps you could show your work? Also, this question will be much more readable if you use MathJax.
  – Michael Burr
  Jul 31 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you mean this integral $$int_(a-b)^2^(a+b)^2(a^2+b^2)x^-3/2-x^-1/2dx$$?
  – Dr. Sonnhard Graubner
  Jul 31 at 10:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For a quick check, set $a=b$. Then the antiderivative is correct and the definite integral is $-2(2b)^2/2b$.
  – Yves Daoust
  Jul 31 at 10:51
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

1

I'm wanting to integrate the following function: $[(a^2-b^2)x^-1.5-x^-0.5] dx$ with the upper limit $(a+b)^2$ and the lower limit $(a-b)^2$.
The result is $ -2 (a^2-b^2+x)/x^0.5$.

My problem arises when I plug in the limits and get zero as the result when it should be $-2/b^2$.
Any help would be appreciated.

riemann-integration

share|cite|improve this question

edited Jul 31 at 10:46

Yves Daoust

110k665203

asked Jul 31 at 10:41

Danny Colautti

12

closed as off-topic by John Ma, Isaac Browne, Mostafa Ayaz, amWhy, Xander Henderson Aug 1 at 2:11

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Isaac Browne, Mostafa Ayaz, amWhy, Xander Henderson

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Perhaps you could show your work? Also, this question will be much more readable if you use MathJax.
  – Michael Burr
  Jul 31 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you mean this integral $$int_(a-b)^2^(a+b)^2(a^2+b^2)x^-3/2-x^-1/2dx$$?
  – Dr. Sonnhard Graubner
  Jul 31 at 10:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For a quick check, set $a=b$. Then the antiderivative is correct and the definite integral is $-2(2b)^2/2b$.
  – Yves Daoust
  Jul 31 at 10:51
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

1

up vote
-1
down vote

favorite

1

1

I'm wanting to integrate the following function: $[(a^2-b^2)x^-1.5-x^-0.5] dx$ with the upper limit $(a+b)^2$ and the lower limit $(a-b)^2$.
The result is $ -2 (a^2-b^2+x)/x^0.5$.

My problem arises when I plug in the limits and get zero as the result when it should be $-2/b^2$.
Any help would be appreciated.

riemann-integration

share|cite|improve this question

edited Jul 31 at 10:46

Yves Daoust

110k665203

asked Jul 31 at 10:41

Danny Colautti

12

I'm wanting to integrate the following function: $[(a^2-b^2)x^-1.5-x^-0.5] dx$ with the upper limit $(a+b)^2$ and the lower limit $(a-b)^2$.
The result is $ -2 (a^2-b^2+x)/x^0.5$.

My problem arises when I plug in the limits and get zero as the result when it should be $-2/b^2$.
Any help would be appreciated.

riemann-integration

share|cite|improve this question

edited Jul 31 at 10:46

Yves Daoust

110k665203

asked Jul 31 at 10:41

Danny Colautti

12

share|cite|improve this question

share|cite|improve this question

edited Jul 31 at 10:46

Yves Daoust

110k665203

edited Jul 31 at 10:46

Yves Daoust

110k665203

edited Jul 31 at 10:46

Yves Daoust

110k665203

110k665203

asked Jul 31 at 10:41

Danny Colautti

12

asked Jul 31 at 10:41

Danny Colautti

12

asked Jul 31 at 10:41

Danny Colautti

12

12

closed as off-topic by John Ma, Isaac Browne, Mostafa Ayaz, amWhy, Xander Henderson Aug 1 at 2:11

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Isaac Browne, Mostafa Ayaz, amWhy, Xander Henderson

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

closed as off-topic by John Ma, Isaac Browne, Mostafa Ayaz, amWhy, Xander Henderson Aug 1 at 2:11

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Isaac Browne, Mostafa Ayaz, amWhy, Xander Henderson

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Perhaps you could show your work? Also, this question will be much more readable if you use MathJax.
  – Michael Burr
  Jul 31 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you mean this integral $$int_(a-b)^2^(a+b)^2(a^2+b^2)x^-3/2-x^-1/2dx$$?
  – Dr. Sonnhard Graubner
  Jul 31 at 10:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For a quick check, set $a=b$. Then the antiderivative is correct and the definite integral is $-2(2b)^2/2b$.
  – Yves Daoust
  Jul 31 at 10:51
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Perhaps you could show your work? Also, this question will be much more readable if you use MathJax.
  – Michael Burr
  Jul 31 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you mean this integral $$int_(a-b)^2^(a+b)^2(a^2+b^2)x^-3/2-x^-1/2dx$$?
  – Dr. Sonnhard Graubner
  Jul 31 at 10:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For a quick check, set $a=b$. Then the antiderivative is correct and the definite integral is $-2(2b)^2/2b$.
  – Yves Daoust
  Jul 31 at 10:51
  

  
  
  
  

2

2

Perhaps you could show your work? Also, this question will be much more readable if you use MathJax.
– Michael Burr
Jul 31 at 10:45

Perhaps you could show your work? Also, this question will be much more readable if you use MathJax.
– Michael Burr
Jul 31 at 10:45

Do you mean this integral $$int_(a-b)^2^(a+b)^2(a^2+b^2)x^-3/2-x^-1/2dx$$?
– Dr. Sonnhard Graubner
Jul 31 at 10:49

Do you mean this integral $$int_(a-b)^2^(a+b)^2(a^2+b^2)x^-3/2-x^-1/2dx$$?
– Dr. Sonnhard Graubner
Jul 31 at 10:49

For a quick check, set $a=b$. Then the antiderivative is correct and the definite integral is $-2(2b)^2/2b$.
– Yves Daoust
Jul 31 at 10:51

For a quick check, set $a=b$. Then the antiderivative is correct and the definite integral is $-2(2b)^2/2b$.
– Yves Daoust
Jul 31 at 10:51

add a comment | 

2 Answers
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$$-2 (a^2-b^2+x)/x^0.5=-2 (a^2-b^2)x^-0.5-2x^0.5$$

$$ -2 (a^2-b^2)[(a+b)^2(-.5)-(a-b)^2(-.5)]-2[(a+b)-(a-b)]$$

$$= -2 (a^2-b^2)[(a+b)^-1-(a-b)^-1]-4b$$

$$ -2 (a^2-b^2)[frac -2ba^2-b^2]-4b =0$$

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answered Jul 31 at 11:11

Mohammad Riazi-Kermani

27.3k41851

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up vote
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It is:
$$frac-2(a^2-b^2+x)x^0.5big_(a-b)^2^(a+b)^2=frac-2(a^2-b^2+(a+b)^2)a+b+frac2(a^2-b^2+(a-b)^2)a-b=\
frac-2(2a^2+2ab)a+b+frac2(2a^2-2ab)a-b=\
-4a+4a=0.$$
Alternatively, one can change $x=t^2$ in integral:
$$I=int_a-b^a+b (2(a^2-b^2)t^-2-2) dt=\
[-2(a^2-b^2)t^-1-2t]big_a-b^a+b=\
[-2(a-b)-2(a+b)]-[-2(a+b)-2(a-b)]=0.$$

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answered Jul 31 at 14:46

farruhota

13.5k2632

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

$$-2 (a^2-b^2+x)/x^0.5=-2 (a^2-b^2)x^-0.5-2x^0.5$$

$$ -2 (a^2-b^2)[(a+b)^2(-.5)-(a-b)^2(-.5)]-2[(a+b)-(a-b)]$$

$$= -2 (a^2-b^2)[(a+b)^-1-(a-b)^-1]-4b$$

$$ -2 (a^2-b^2)[frac -2ba^2-b^2]-4b =0$$

share|cite|improve this answer

answered Jul 31 at 11:11

Mohammad Riazi-Kermani

27.3k41851

add a comment | 

up vote
0
down vote

$$-2 (a^2-b^2+x)/x^0.5=-2 (a^2-b^2)x^-0.5-2x^0.5$$

$$ -2 (a^2-b^2)[(a+b)^2(-.5)-(a-b)^2(-.5)]-2[(a+b)-(a-b)]$$

$$= -2 (a^2-b^2)[(a+b)^-1-(a-b)^-1]-4b$$

$$ -2 (a^2-b^2)[frac -2ba^2-b^2]-4b =0$$

share|cite|improve this answer

answered Jul 31 at 11:11

Mohammad Riazi-Kermani

27.3k41851

add a comment | 

up vote
0
down vote

up vote
0
down vote

$$-2 (a^2-b^2+x)/x^0.5=-2 (a^2-b^2)x^-0.5-2x^0.5$$

$$ -2 (a^2-b^2)[(a+b)^2(-.5)-(a-b)^2(-.5)]-2[(a+b)-(a-b)]$$

$$= -2 (a^2-b^2)[(a+b)^-1-(a-b)^-1]-4b$$

$$ -2 (a^2-b^2)[frac -2ba^2-b^2]-4b =0$$

share|cite|improve this answer

answered Jul 31 at 11:11

Mohammad Riazi-Kermani

27.3k41851

$$-2 (a^2-b^2+x)/x^0.5=-2 (a^2-b^2)x^-0.5-2x^0.5$$

$$ -2 (a^2-b^2)[(a+b)^2(-.5)-(a-b)^2(-.5)]-2[(a+b)-(a-b)]$$

$$= -2 (a^2-b^2)[(a+b)^-1-(a-b)^-1]-4b$$

$$ -2 (a^2-b^2)[frac -2ba^2-b^2]-4b =0$$

share|cite|improve this answer

answered Jul 31 at 11:11

Mohammad Riazi-Kermani

27.3k41851

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 11:11

Mohammad Riazi-Kermani

27.3k41851

answered Jul 31 at 11:11

Mohammad Riazi-Kermani

27.3k41851

answered Jul 31 at 11:11

Mohammad Riazi-Kermani

27.3k41851

27.3k41851

add a comment | 

add a comment | 

up vote
0
down vote

It is:
$$frac-2(a^2-b^2+x)x^0.5big_(a-b)^2^(a+b)^2=frac-2(a^2-b^2+(a+b)^2)a+b+frac2(a^2-b^2+(a-b)^2)a-b=\
frac-2(2a^2+2ab)a+b+frac2(2a^2-2ab)a-b=\
-4a+4a=0.$$
Alternatively, one can change $x=t^2$ in integral:
$$I=int_a-b^a+b (2(a^2-b^2)t^-2-2) dt=\
[-2(a^2-b^2)t^-1-2t]big_a-b^a+b=\
[-2(a-b)-2(a+b)]-[-2(a+b)-2(a-b)]=0.$$

share|cite|improve this answer

answered Jul 31 at 14:46

farruhota

13.5k2632

add a comment | 

up vote
0
down vote

It is:
$$frac-2(a^2-b^2+x)x^0.5big_(a-b)^2^(a+b)^2=frac-2(a^2-b^2+(a+b)^2)a+b+frac2(a^2-b^2+(a-b)^2)a-b=\
frac-2(2a^2+2ab)a+b+frac2(2a^2-2ab)a-b=\
-4a+4a=0.$$
Alternatively, one can change $x=t^2$ in integral:
$$I=int_a-b^a+b (2(a^2-b^2)t^-2-2) dt=\
[-2(a^2-b^2)t^-1-2t]big_a-b^a+b=\
[-2(a-b)-2(a+b)]-[-2(a+b)-2(a-b)]=0.$$

share|cite|improve this answer

answered Jul 31 at 14:46

farruhota

13.5k2632

add a comment | 

up vote
0
down vote

up vote
0
down vote

It is:
$$frac-2(a^2-b^2+x)x^0.5big_(a-b)^2^(a+b)^2=frac-2(a^2-b^2+(a+b)^2)a+b+frac2(a^2-b^2+(a-b)^2)a-b=\
frac-2(2a^2+2ab)a+b+frac2(2a^2-2ab)a-b=\
-4a+4a=0.$$
Alternatively, one can change $x=t^2$ in integral:
$$I=int_a-b^a+b (2(a^2-b^2)t^-2-2) dt=\
[-2(a^2-b^2)t^-1-2t]big_a-b^a+b=\
[-2(a-b)-2(a+b)]-[-2(a+b)-2(a-b)]=0.$$

share|cite|improve this answer

answered Jul 31 at 14:46

farruhota

13.5k2632

It is:
$$frac-2(a^2-b^2+x)x^0.5big_(a-b)^2^(a+b)^2=frac-2(a^2-b^2+(a+b)^2)a+b+frac2(a^2-b^2+(a-b)^2)a-b=\
frac-2(2a^2+2ab)a+b+frac2(2a^2-2ab)a-b=\
-4a+4a=0.$$
Alternatively, one can change $x=t^2$ in integral:
$$I=int_a-b^a+b (2(a^2-b^2)t^-2-2) dt=\
[-2(a^2-b^2)t^-1-2t]big_a-b^a+b=\
[-2(a-b)-2(a+b)]-[-2(a+b)-2(a-b)]=0.$$

share|cite|improve this answer

answered Jul 31 at 14:46

farruhota

13.5k2632

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 14:46

farruhota

13.5k2632

answered Jul 31 at 14:46

farruhota

13.5k2632

answered Jul 31 at 14:46

farruhota

13.5k2632

13.5k2632

add a comment | 

add a comment |