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Math Help - Integrals. Area.

  1. #1
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    Integrals. Area.

    sketch the region enclosed by y = 3x and y = 5x^2. Decide wether to integrate with respect to x or y. What is the area of the region?
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  2. #2
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    Quote Originally Posted by B-lap View Post
    Sketch the region enclosed by y = 3x and y = 5x^2. Decide whether to integrate with respect to x or y. The find the area of the region.
    Your limits will be where the two graphs intersect to find there solve 3x=5x^2

    (I get x=0 and x=0.6).

    You'll nearly always integrate with respect to x and this is no exception

    If you sketch a graph it will help give you an idea of the area. From what I can gather between 0 and 0.6: 3x will increase faster than 5x^2 so the integral becomes

    \int^{0.6}_0 (3x-5x^2) \, .dx
    Last edited by mr fantastic; December 12th 2009 at 04:17 AM. Reason: No edit. Just flagging the reply as having been moved from another thread.
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  3. #3
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    Talking

    You learned how to sketch back in algebra, and also how to solve systems of equations. So you should have a nice picture with labelled intersection points.

    Where are you stuck after that?

    Please be complete. Thank you!
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  4. #4
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    I was trying to solve this problem too and I'm just wondering if this would be right:

    5x^2 = 3x

    5x^2 - 3x = 0

    x(5x - 3) = 0; x = 0; x = \frac{3}{5}

    \int_0^\frac{3}{5} 5x^2 - 3x \,dx = \frac{5}{3}x^3 - \frac{3}{2}x^2

    \frac{5}{3}(\frac{3}{5})^3 - \frac{3}{2}(\frac{3}{5})^2 = \frac{9}{25} - \frac{27}{50} = -\frac{9}{50}

    so the area is:

    \frac{9}{50}

    Is that right?
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  5. #5
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    looks fine at a glance Korupt
    Last edited by mr fantastic; December 12th 2009 at 04:18 AM.
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  6. #6
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    Quote Originally Posted by Korupt View Post
    I was trying to solve this problem too and I'm just wondering if this would be right:

    5x^2 = 3x

    5x^2 - 3x = 0

    x(5x - 3) = 0; x = 0; x = \frac{3}{5}

    \int_0^\frac{3}{5} 5x^2 - 3x \,dx = \frac{5}{3}x^3 - \frac{3}{2}x^2

    \frac{5}{3}(\frac{3}{5})^3 - \frac{3}{2}(\frac{3}{5})^2 = \frac{9}{25} - \frac{27}{50} = -\frac{9}{50}

    so the area is:

    \frac{9}{50}

    Is that right?
    Everything is correct except your use of integration. You cannot subtract right minus left unless you're going to use horizontal strips. You must use vertical strips therefore you take the top function minus the bottom function.

    Therefore your integral should be

    \int_0^\frac{3}{5} 3x - 5x^2 dx

    This gives the answer \frac{9}{50} units^2
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  7. #7
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    Quote Originally Posted by millerst View Post
    Everything is correct except your use of integration. You cannot subtract right minus left unless you're going to use horizontal strips. You must use vertical strips therefore you take the top function minus the bottom function.

    Therefore your integral should be

    \int_0^\frac{3}{5} 3x - 5x^2 dx

    This gives the answer \frac{9}{50} units^2
    Well, he didn't "subtract right minus left", he subtracted "bottom minus top", probably because he did not graph this, as stapel originally suggested, and did not realize that 3x\ge 5x^2 on this interval.
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