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Math Help - Area enclosed by a curve?.

  1. #1
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    Area enclosed by a curve?.

    Hey can anybody help me by giving a step to step guide on how to work out an area enclosed by a curve please?.

    Find the area enclosed by the curve: Y=4x^2 (x^2-2x+1). The axis and the ordinates x=1 and x=2.

    thanks
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    Re: Area enclosed by a curve?.

    Quote Originally Posted by MstrKurt View Post
    Hey can anybody help me by giving a step to step guide on how to work out an area enclosed by a curve please?.

    Find the area enclosed by the curve: Y=4x^2 (x^2-2x+1). The axis and the ordinates x=1 and x=2.
    I assume the area between the curve and the x-axis ... what do you know about definite integrals ?

    A = \int_1^2 4x^2(x^2-2x+1) \, dx
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    Re: Area enclosed by a curve?.

    Hi, thanks for the reply.
    I know that definite integrals have limits, and always end up with a numerical answer with no need for +c at the end.
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    Re: Area enclosed by a curve?.

    I have worked out that the integrated equation would look like
    4x^5-8x^4+4x^3
    However, apparently the -8x^3 turns to -2x^4
    Can anybody explain why?. Thanks.
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  5. #5
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    Re: Area enclosed by a curve?.

    A = \int_1^2 4x^2(x^2-2x+1) \, dx

    A = \int_1^2 4x^4 - 8x^3 + 4x^2 \, dx

    the antiderivative of k x^n is \frac{k x^{n+1}}{n+1}

    \left[\frac{4x^5}{5} - 2x^4 + \frac{4x^3}{3}\right]_1^2

    \left(\frac{128}{5} - 32 + \frac{32}{3}\right) - \left(\frac{4}{5} - 2 + \frac{4}{3}\right)

    I leave you to finish the arithmetic
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