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Math Help - area under the curve

  1. #1
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    area under the curve

    Find the area of the region bounded by the given curves.





    i got the answer


    is this correct? if not can u show me how u did it?
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  2. #2
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    Hello, ahawk1!

    I wish you'd shown us your work . . .


    Find the area of the region bounded by the given curves: . \begin{array}{cccc}y &=& 35\ln x & {\color{blue}[1]}\\ y&=&x\ln x & {\color{blue}[2]}\end{array}
    Find the intersections of the curves . . .

    x\ln x \:=\:35\ln x \quad\Rightarrow\quad x\ln x - 35\ln x \:=\:0 \quad\Rightarrow\quad \ln x\bigg(x - 35\bigg) \:=\:0

    And we have: . \begin{array}{ccccccc}\ln x \:=\:0 & \Rightarrow & x \:=\:1 \\ x-35 \:=\:0 & \Rightarrow & x \:=\:35 \end{array}

    On the interval [1, 35], [1] is the upper curve.


    The integral is: . \int^{35}_1\bigg[35\ln x - x\ln x\bigg]\,dx \;=\;35\int^{35}_1\ln x\,dx - \int^{35}_1 x\ln x\,dx<br /> <br />


    Both integrals require integration-by-parts.

    We get: . 35x(\ln x - 1) - \tfrac{1}{4}x^2(2\ln x - 1)\,\bigg]^{35}_1 . . . . etc.

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