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Math Help - Please help, integration

  1. #1
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    Please help, integration

    The area under the graph y= 4 + 2x from x=0 to x=4 is divided into two equal parts by the line x=k. Find the value of k.


    Please ive been stuck on this for years
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  2. #2
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    You're asking for a value of k such that \int_0^k 4+2x \, dx = \frac12 \int_0^4 4+2x \, dx. Doing the integration,  \left[4x+x^2\right]_0^k = \frac12 \left[4x+x^2\right]_0^4, that is, k^2 + 4k = \frac12 (4.4+4^2) = 16. This is a quadratic equation in k which has a root near 2.47.
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  3. #3
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    Quote Originally Posted by ify00
    The area under the graph y= 4 + 2x from x=0 to x=4 is divided into two equal parts by the line x=k. Find the value of k.

    Please ive been stuck on this for years
    Hello,

    I assume that 0 < k < 4.

    From the title of your post I assume that you want to calculate the area by integration (which isn't necessary with your problem). You calculate 2 areas which should be equal:
    A_1=\int^{k}_{0}(2x+4)dx=\left[x^2+4x \right]^{k}_{0}=k^2+4k-0

    A_2=\int^{4}_{k}(2x+4)dx=\left[x^2+4x \right]^{4}_{k}=32-(k^2+4k)

    It is now

    A_1=A_2 \Longrightarrow k^2+4k=32-(k^2+4k)

    You'll get a quadratic equation:

    2k^2+8k-32=0 \Longleftrightarrow k=-2+2\cdot \sqrt{5}\ \vee \ k= -2-2\cdot \sqrt{5}

    The negative solution doesn't fit into your problem. So k = 2.47...

    Greetings

    EB
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