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integration of absolute value

Exposition On An Integral Of An Absolute Value FunctionFinding the average of the absolute value of a function?Integration by parts: $int xln x^2 ,dx$Finding the derivative of an absolute valueDouble Integral $intlimits_0^pi intlimits_0^pi|cos(x+y)|,dx,dy$Evaluating definite integral U substitutionAbsolute value and inequality in integralAbsolute max and minimum valuesContour integration with absolute valueantiderivatives of $frac1x-1$

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4

$begingroup$

$$int_0^6 |x^2 - 6x +8| dx =$$
I know the answer to this problem is $frac443$. But, how do you get that? How does the solution change based on the absolute value? Without the absolute value I got an answer of 12 - I found the anti-derivative:$fracx^33-3x^2+8x$. I then solved it which got me 12. However, this is not the answer to the above problem. What should I be doing?

calculus integration

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asked 8 hours ago

burtburt

1951010 bronze badges

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Try graphing the function and see where does it take positive/negative values. For instance, begin by checking the roots.
  $endgroup$
  – Rick
  8 hours ago
  
  

  
  
  
  

add a comment | 

4

$begingroup$

$$int_0^6 |x^2 - 6x +8| dx =$$
I know the answer to this problem is $frac443$. But, how do you get that? How does the solution change based on the absolute value? Without the absolute value I got an answer of 12 - I found the anti-derivative:$fracx^33-3x^2+8x$. I then solved it which got me 12. However, this is not the answer to the above problem. What should I be doing?

calculus integration

share|cite|improve this question

asked 8 hours ago

burtburt

1951010 bronze badges

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Try graphing the function and see where does it take positive/negative values. For instance, begin by checking the roots.
  $endgroup$
  – Rick
  8 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

$$int_0^6 |x^2 - 6x +8| dx =$$
I know the answer to this problem is $frac443$. But, how do you get that? How does the solution change based on the absolute value? Without the absolute value I got an answer of 12 - I found the anti-derivative:$fracx^33-3x^2+8x$. I then solved it which got me 12. However, this is not the answer to the above problem. What should I be doing?

calculus integration

share|cite|improve this question

asked 8 hours ago

burtburt

1951010 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

burtburt

1951010 bronze badges

share|cite|improve this question

asked 8 hours ago

burtburt

1951010 bronze badges

asked 8 hours ago

burtburt

1951010 bronze badges

asked 8 hours ago

burtburt

1951010 bronze badges

4 Answers
4

active

oldest

votes

6

$begingroup$

The absolute value is a function that's very difficult to work with directly. But there's an easy way to deal with it:

$$
|y|=begincases y &textif ygeq0\ -y &textif yleq0endcases
$$

So your first step is to find where the term inside the absolute value is positive and where negative. Then you break up the integral into pieces.

share|cite|improve this answer

answered 8 hours ago

Y. FormanY. Forman

11.8k55 silver badges2323 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What is the best way to do that? Finding the roots and checking the values?
  $endgroup$
  – burt
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @burt Yes, as in Rick's comment
  $endgroup$
  – Y. Forman
  8 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

Hint : $x^2-6x+8=(x-4)(x-2)$

Can you break domain $ [0,6]$ to get rid of absolute value notation?

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Praphulla KoushikPraphulla Koushik

26311 silver badge1919 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is modulus
  $endgroup$
  – burt
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @burt It refers to the magnitude $| |$
  $endgroup$
  – Ak19
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @burt "Modulus," "magnitude," and "absolute value" all mean the same thing.
  $endgroup$
  – Y. Forman
  8 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

If the integrand were given without the absolute value then it would be negative in the interval $(2,4)$ hence the given integral is equal to
$$int_0^2 f(x)mathrmdx-int_2^4f(x)mathrmdx+int_4^6f(x)mathrmdx$$
where $f(x)=x^2-6x+8$.

share|cite|improve this answer

answered 8 hours ago

Peter ForemanPeter Foreman

13k11 gold badge55 silver badges2929 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

If $f(x)>0$ on an interval, then $|f(x)|=f(x)$ on that interval. If $f(x)<0$ on an interval, then $|f(x)|=-f(x)$ on that interval. That's basically the actual definition of the absolute value function.

On the interval $[2,4]$, $x^2-6x+8<0$. Everywhere else it's positive. Thus, your original integral is equivalent to the sum of the following three integrals:

$$
int_0^2(x^2-6x+8),dx+int_2^4left[-(x^2-6x+8)right],dx+int_4^6(x^2-6x+8),dx.
$$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Michael RybkinMichael Rybkin

5,95922 gold badges66 silver badges2525 bronze badges

$endgroup$

add a comment | 

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6

$begingroup$

The absolute value is a function that's very difficult to work with directly. But there's an easy way to deal with it:

$$
|y|=begincases y &textif ygeq0\ -y &textif yleq0endcases
$$

So your first step is to find where the term inside the absolute value is positive and where negative. Then you break up the integral into pieces.

share|cite|improve this answer

answered 8 hours ago

Y. FormanY. Forman

11.8k55 silver badges2323 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What is the best way to do that? Finding the roots and checking the values?
  $endgroup$
  – burt
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @burt Yes, as in Rick's comment
  $endgroup$
  – Y. Forman
  8 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

The absolute value is a function that's very difficult to work with directly. But there's an easy way to deal with it:

$$
|y|=begincases y &textif ygeq0\ -y &textif yleq0endcases
$$

So your first step is to find where the term inside the absolute value is positive and where negative. Then you break up the integral into pieces.

share|cite|improve this answer

answered 8 hours ago

Y. FormanY. Forman

11.8k55 silver badges2323 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What is the best way to do that? Finding the roots and checking the values?
  $endgroup$
  – burt
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @burt Yes, as in Rick's comment
  $endgroup$
  – Y. Forman
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

The absolute value is a function that's very difficult to work with directly. But there's an easy way to deal with it:

$$
|y|=begincases y &textif ygeq0\ -y &textif yleq0endcases
$$

So your first step is to find where the term inside the absolute value is positive and where negative. Then you break up the integral into pieces.

share|cite|improve this answer

answered 8 hours ago

Y. FormanY. Forman

11.8k55 silver badges2323 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Y. FormanY. Forman

11.8k55 silver badges2323 bronze badges

answered 8 hours ago

Y. FormanY. Forman

11.8k55 silver badges2323 bronze badges

answered 8 hours ago

Y. FormanY. Forman

11.8k55 silver badges2323 bronze badges

6

$begingroup$

Hint : $x^2-6x+8=(x-4)(x-2)$

Can you break domain $ [0,6]$ to get rid of absolute value notation?

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Praphulla KoushikPraphulla Koushik

26311 silver badge1919 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is modulus
  $endgroup$
  – burt
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @burt It refers to the magnitude $| |$
  $endgroup$
  – Ak19
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @burt "Modulus," "magnitude," and "absolute value" all mean the same thing.
  $endgroup$
  – Y. Forman
  8 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

Hint : $x^2-6x+8=(x-4)(x-2)$

Can you break domain $ [0,6]$ to get rid of absolute value notation?

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Praphulla KoushikPraphulla Koushik

26311 silver badge1919 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is modulus
  $endgroup$
  – burt
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @burt It refers to the magnitude $| |$
  $endgroup$
  – Ak19
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @burt "Modulus," "magnitude," and "absolute value" all mean the same thing.
  $endgroup$
  – Y. Forman
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Hint : $x^2-6x+8=(x-4)(x-2)$

Can you break domain $ [0,6]$ to get rid of absolute value notation?

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Praphulla KoushikPraphulla Koushik

26311 silver badge1919 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Praphulla KoushikPraphulla Koushik

26311 silver badge1919 bronze badges

answered 8 hours ago

Praphulla KoushikPraphulla Koushik

26311 silver badge1919 bronze badges

answered 8 hours ago

Praphulla KoushikPraphulla Koushik

26311 silver badge1919 bronze badges

3

$begingroup$

If the integrand were given without the absolute value then it would be negative in the interval $(2,4)$ hence the given integral is equal to
$$int_0^2 f(x)mathrmdx-int_2^4f(x)mathrmdx+int_4^6f(x)mathrmdx$$
where $f(x)=x^2-6x+8$.

share|cite|improve this answer

answered 8 hours ago

Peter ForemanPeter Foreman

13k11 gold badge55 silver badges2929 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

If the integrand were given without the absolute value then it would be negative in the interval $(2,4)$ hence the given integral is equal to
$$int_0^2 f(x)mathrmdx-int_2^4f(x)mathrmdx+int_4^6f(x)mathrmdx$$
where $f(x)=x^2-6x+8$.

share|cite|improve this answer

answered 8 hours ago

Peter ForemanPeter Foreman

13k11 gold badge55 silver badges2929 bronze badges

$endgroup$

add a comment | 

$begingroup$

If the integrand were given without the absolute value then it would be negative in the interval $(2,4)$ hence the given integral is equal to
$$int_0^2 f(x)mathrmdx-int_2^4f(x)mathrmdx+int_4^6f(x)mathrmdx$$
where $f(x)=x^2-6x+8$.

share|cite|improve this answer

answered 8 hours ago

Peter ForemanPeter Foreman

13k11 gold badge55 silver badges2929 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Peter ForemanPeter Foreman

13k11 gold badge55 silver badges2929 bronze badges

answered 8 hours ago

Peter ForemanPeter Foreman

13k11 gold badge55 silver badges2929 bronze badges

answered 8 hours ago

Peter ForemanPeter Foreman

13k11 gold badge55 silver badges2929 bronze badges

1

$begingroup$

If $f(x)>0$ on an interval, then $|f(x)|=f(x)$ on that interval. If $f(x)<0$ on an interval, then $|f(x)|=-f(x)$ on that interval. That's basically the actual definition of the absolute value function.

On the interval $[2,4]$, $x^2-6x+8<0$. Everywhere else it's positive. Thus, your original integral is equivalent to the sum of the following three integrals:

$$
int_0^2(x^2-6x+8),dx+int_2^4left[-(x^2-6x+8)right],dx+int_4^6(x^2-6x+8),dx.
$$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Michael RybkinMichael Rybkin

5,95922 gold badges66 silver badges2525 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

If $f(x)>0$ on an interval, then $|f(x)|=f(x)$ on that interval. If $f(x)<0$ on an interval, then $|f(x)|=-f(x)$ on that interval. That's basically the actual definition of the absolute value function.

On the interval $[2,4]$, $x^2-6x+8<0$. Everywhere else it's positive. Thus, your original integral is equivalent to the sum of the following three integrals:

$$
int_0^2(x^2-6x+8),dx+int_2^4left[-(x^2-6x+8)right],dx+int_4^6(x^2-6x+8),dx.
$$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Michael RybkinMichael Rybkin

5,95922 gold badges66 silver badges2525 bronze badges

$endgroup$

add a comment | 

$begingroup$

If $f(x)>0$ on an interval, then $|f(x)|=f(x)$ on that interval. If $f(x)<0$ on an interval, then $|f(x)|=-f(x)$ on that interval. That's basically the actual definition of the absolute value function.

On the interval $[2,4]$, $x^2-6x+8<0$. Everywhere else it's positive. Thus, your original integral is equivalent to the sum of the following three integrals:

$$
int_0^2(x^2-6x+8),dx+int_2^4left[-(x^2-6x+8)right],dx+int_4^6(x^2-6x+8),dx.
$$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Michael RybkinMichael Rybkin

5,95922 gold badges66 silver badges2525 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Michael RybkinMichael Rybkin

5,95922 gold badges66 silver badges2525 bronze badges

answered 8 hours ago

Michael RybkinMichael Rybkin

5,95922 gold badges66 silver badges2525 bronze badges

answered 8 hours ago

Michael RybkinMichael Rybkin

5,95922 gold badges66 silver badges2525 bronze badges