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Math Help - Integral

  1. #1
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    Integral

    Find the integral of sqrt(x)|2-x| dx from 1 to 4.

    What do I do with the absolute value? Is it true that I separate it from integral of sqrt(x)(2-x) from 1 to 2 + integral of sqrt(x)(x-2) from 2 to 4?

    If that is true, why does the 2-x change into x-2 for the 2nd part of the integral?

    Thanks
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  2. #2
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by math252 View Post
    Find the integral of sqrt(x)|2-x| dx from 1 to 4.

    What do I do with the absolute value? Is it true that I separate it from integral of sqrt(x)(2-x) from 1 to 2 + integral of sqrt(x)(x-2) from 2 to 4?

    If that is true, why does the 2-x change into x-2 for the 2nd part of the integral?

    Thanks
    Yes you are correct but I believe this integral diverges.
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  3. #3
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    Quote Originally Posted by math252 View Post
    Find the integral of sqrt(x)|2-x| dx from 1 to 4.

    What do I do with the absolute value? Is it true that I separate it from integral of sqrt(x)(2-x) from 1 to 2 + integral of sqrt(x)(x-2) from 2 to 4?

    If that is true, why does the 2-x change into x-2 for the 2nd part of the integral?

    Thanks

    we have  \int \sqrt{x} ( 2-x)~dx

     = \int 2x^{\frac{1}{2}} - x^{\frac{3}{2}} ~dx


     =  \frac{4}{3} x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}} + C

    we also have

     (2-x) \geq 0  for  x \leq 2

     (2-x) \leq 0 for  x \geq 2

    the integral

      \int_1^4 \sqrt{x} | 2-x|~dx

     = (\int_1^2 + \int_2^4 ) \sqrt{x} | 2-x|~dx

     = \int_1^2 \sqrt{x} ( 2-x) ~dx  - \int_2^4 \sqrt{x} ( 2-x)~dx

     = [\frac{4}{3} x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}} ]_1^2 - [\frac{4}{3} x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}} ]_2^4

     =  \frac{32\sqrt{2}}{15} + \frac{6}{5}
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  4. #4
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    @ 1st post:
    I believe this integral converges...

    @ 2nd post:
    Aren't you not supposed to ignore the absolute value? the end value I got was, {4/3[x^(3/2)]-2/5[x^(5/2)]} from 1 to 2 PLUS {2/5[x^(5/2)]-4/3[x^(3/2)]} from 2 to 4. I was told it was this way because of the absolute value...can somebody clarify/explain this?
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  5. #5
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    Quote Originally Posted by math252 View Post
    @ 1st post:
    I believe this integral converges...

    @ 2nd post:
    Aren't you not supposed to ignore the absolute value? the end value I got was, {4/3[x^(3/2)]-2/5[x^(5/2)]} from 1 to 2 PLUS {2/5[x^(5/2)]-4/3[x^(3/2)]} from 2 to 4. I was told it was this way because of the absolute value...can somebody clarify/explain this?

    OH , i think there is a step which is very important i have skipped it !


    <br />
= (\int_1^2 + \int_2^4 ) \sqrt{x} | 2-x|~dx<br />
     = \int_1^2 \sqrt{x} | 2-x|~dx + \int_2^4 \sqrt{x} | 2-x|~dx<br /> <br />

     = \int_1^2 \sqrt{x} ( 2-x ) ~dx + \int_2^4 \sqrt{x} [-( 2-x)]~dx<br /> <br />
    <br />
= \int_1^2 \sqrt{x} ( 2-x) ~dx - \int_2^4 \sqrt{x} ( 2-x)~dx<br />
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  6. #6
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    Quote Originally Posted by math252 View Post
    Find the integral of sqrt(x)|2-x| dx from 1 to 4.

    What do I do with the absolute value? Is it true that I separate it from integral of sqrt(x)(2-x) from 1 to 2 + integral of sqrt(x)(x-2) from 2 to 4?

    If that is true, why does the 2-x change into x-2 for the 2nd part of the integral?

    Thanks
    Because |x|= x if x\ge 0 and |x|= -x if x< 0.
    If x\ge 2, then x- 2\ge 0 and so |x- 2|= x- 2. If x< 2, then x- 2< 0 and so |x- 2|= -(x-2)= 2- x.
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  7. #7
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    Quote Originally Posted by 11rdc11 View Post
    Yes you are correct but I believe this integral diverges.
    ??? \sqrt{x}|x-2| is continuous and bounded over the finite interval 1\le x\le 4. It can't possibly diverge!
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