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Math Help - Properties of Integrals Question that I keep getting wrong

  1. #1
    s3a
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    Properties of Integrals Question that I keep getting wrong

    I know how to manipulate integrals algebraically if I had the numerical value of each integral however this question wants me to find the values of intervals of certain integrals which I don't know how to do!

    Could someone please explain it to me? (Question #14 on "temp.pdf" - my (useless) work is attached as "mywork.pdf" to just to show that I know how to manipulate the integrals)

    Any help would be greatly appreciated!
    Thanks in advance!

    P.S. I forgot to put dx on the first line of my work (not that it necessarily matters).
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  2. #2
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    do you understand why this definite integral has to be broken up into three parts?
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by s3a View Post
    I know how to manipulate integrals algebraically if I had the numerical value of each integral however this question wants me to find the values of intervals of certain integrals which I don't know how to do!

    Could someone please explain it to me? (Question #14 on "temp.pdf" - my (useless) work is attached as "mywork.pdf" to just to show that I know how to manipulate the integrals)

    Any help would be greatly appreciated!
    Thanks in advance!

    P.S. I forgot to put dx on the first line of my work (not that it necessarily matters).
    Note: 10x^2-x^3-21x=-x(x^2-10x+21)=-x(x-3)(x-7)



    But |-x(x-3)(x-7)|=\left\{\begin{array}{cc}-x(x-3)(x-7),&\mbox{ if }<br />
x\in(-\infty,3]\\\\-x(x-3)(x-7),&\mbox{ if }x\in(3,7)\\\\x(x-3)(x-7),&\mbox{ if }x\in[7,\infty)\end{array}\right.
    Last edited by VonNemo19; January 30th 2010 at 08:42 PM.
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  4. #4
    s3a
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    Before I go on, is this called "Integration by parts?" Because, if it is, my teacher didn't go over that yet and I should go read it in the book before attempting this problem again.

    Edit: Actually, having looked at the book; this doesn't seem to be integration by parts.
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  5. #5
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    No, it isn't related to integration by parts, although I can see how it might sound like it is.

    What it ultimately breaks down to is that you have to recognize that an absolute value function is actually a piecewise function in disguise. For example, suppose you were asked to solve:

    \int_{-1}^1 |x| dx

    Suppose you used traditional techniques and guessed that the antiderivative of |x| is \left|\frac{x^2}{2}\right|. If you did this, your solution would be

    \int_{-1}^1 |x| dx =^? \left[ \left|\frac{x^2}{2}\right| \right]_{-1}^1 = \left|\frac{1}{2}\right| - \left|\frac{1}{2}\right| = 0

    But this is clearly not true, as you can look at the graph and see there has to be positive area.

    In reality, we have to realize that the definition of the absolute value is like this:

    |x| = \begin{cases} -x, & \mbox{if }x<0 \\ x, & \mbox{if }x\ge0 \end{cases}

    Then you realize you must break the integral apart into two pieces:

    \int_{-1}^1 |x| dx = \int_{-1}^0 (-x) dx + \int_0^1 x dx

    You can then finish evaluating each of the two integrals in the usual way and get the correct answer of "1".

    Now, apply this to your problem, and we first identify that

    |10x^2-x^3-21x| = |(-x)(x-3)(x-7)| =<br />
\begin{cases}<br />
20x^2-x^3-21x, & \mbox{if } x<0 \\<br />
-20x^2+x^3+21x, & \mbox{if } 0 \le x<3 \\<br />
20x^2-x^3-21x, & \mbox{if } 3 \le x<7 \\<br />
-20x^2+x^3+21x, & \mbox{if } x \ge 7<br />
\end{cases}

    The problem asks you to integrate from -1 to 4. Does it make sense why this integral must be broken into 3 separate integrals?

    By the way, VonNemo19 made a small error when showing what the piecewise function looks like. Please disregard that and use this one.
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  6. #6
    s3a
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    I had written a long message but my post got lost! Anyways, thanks to everyone (I got the correct answers)!
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