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Math Help - The Difference Between Two Integration Problems

  1. #1
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    The Difference Between Two Integration Problems

    1.

    \int (x^{2} + 4)^{5} 2xdx

    Do the anti-derivative on the integrand.

    \frac{(x^{2} + 4)^{6}}{6} + C Final Answer


    and this problem

    2.

    \int (x^{3} - 3x)^{1/2}(x^{2} - 1)dx

    Next taking the derivative of the highest power in the integrand x^{3} (BUT WAIT - 3 is not the derivative)

    (x^{3} - 3x)^{1/2} 3(x^{2} - 1)dx

    Next flipping the 3 so it becomes \frac{1}{3}

    \frac{1}{3} \int (x^{3} - 3x)^{1/2} 3(x^{2} - 1)dx

    The stuff to the right of the left linear factor (integrand) disappears. Now we do the anti-derivative on the integrand.

    \frac{1}{3} [\frac{(x^{3} - 3x)^{3/2}}{\frac{3}{2}} + C]

    \frac{2}{9}(x^{3} - 3x)^{3/2} + C Final Answer


    As you can see that in the 2nd problem the \frac{1}{3} is a flipped version of 3 which came from the x^{3} in the beginning (since that was the largest power).

    But in the 1st problem the highest power x^{3} is not made into 3 and flipped.

    But I see a problem in the reasoning: x^{3} differentiated is [tex]3x^{2}[tex] not 3. So why does the 3 come about? Where did it come from?
    Last edited by Jason76; November 19th 2012 at 07:26 PM.
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    Re: The Difference Between Two Integration Problems

    Sorry for double post, but I think I figured it out.

    The 1st problem is complete, so the anti-derivative can be taken. However, the 2nd problem is incomplete. Threfore, the 3 has to put to the left of x^{2} (in the linear factor to the right of the integrand), to make 3x^{2} which is the derivative of x^{3}.

    So now the 2nd equation is complete and the anti-derivative can be taken. Of course, when you complete an integration problem, then you also have to flip the 3 and put the result \frac{1}{3} in front of the integrand (for some balancing reason).
    Last edited by Jason76; November 19th 2012 at 09:04 PM.
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    Re: The Difference Between Two Integration Problems

    Quote Originally Posted by Jason76 View Post
    ... then you also have to flip the 3 and put the result \frac{1}{3} in front of the integrand (for some balancing reason).
    ... and that "balancing reason" is because you're effectively multiplying the integrand by the constant 1 .

    note 1 = \frac{1}{3} \cdot 3
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    Re: The Difference Between Two Integration Problems

    In the first example the composite function is (x^2+4)^5 and the derivative of the function x^2+4, which is 2x, appears exactly as is as a factor in the intergrand. Hence the substitution rule applies.

    In the second example the composite function is (x^3-3x)^{1/2} and the derivative of the function x^3-3x is 3(x^2-1). This derivative does not appear exactly as is in the integrand but something close does, which is x^2-1. To get the derivative exactly as a factor in the integrand we meed to multiply it by 3, which means we must also divide by 3 so nothing is changed.
    Last edited by tsquires; November 20th 2012 at 10:40 AM.
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    Re: The Difference Between Two Integration Problems

    The derivative of the starting left linear factor is larger than the dx of the starting right linear factor. So multiplication is needed. However, if the derivative of the starting left linear factor was smaller than division would be needed. Is that right? If the derivative of the starting left linear factor matches the dx of the starting right linear factor, then you can go ahead and take the anti-derivative without any manipulation.
    Last edited by Jason76; November 21st 2012 at 12:26 PM.
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