Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Integration Maths Help?

Not getting the correct answer with this question. Can anyone help? greatly appreciated

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  • 1 month ago
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    You can get the correct answer from Wolfram Alpha, if that's all you needed.  (Yes, it will do "area between" computation, but you may need to play with it a bit to find out how it expects you to ask the question.

    To solve (a), I'd integrate the terms separately.  The integral of -1 dx from x=2 to 4 is clearly -2.  It's a rectangle.

    For the 2/(x-1) term substitute u=x-1, du = dx.  The integral becomes du/u integrated from u=1 to 3.  That's ln(3) - ln(1) = ln(3)

    Put both terms together to get ln(3) - 2 as the answer for part a.

    For part (b) there's more work because f(x) = 2/(x-1) - 1 goes from positive to negative at x=3.  You need to add the absolute values of the integrals from 2 to 3 and from 3 to 4, and add the absolute values.  The procedure for each of those integrals is the same as above, with u=x-1 each time.

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  • Philip
    Lv 6
    1 month ago

    (a) J = Int'l over(2,4) of [2/(x-1) -1]dx, x =/= 1. Put u=(x-1). Then dx=du &

    (b) J = Int'l over(1,3) of [2/u -1]du = f(3)-f(1), where f(u) = 2lnu - u.

    f(3) = 2ln3 -3, f(1) = 2ln1 -1 = -1 and f(3)-f(1) = 2ln3 -2 = 2(ln3 -1) =

    o.1972245773.

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  • Remember, int((f(x) + g(x)) * dx) = int(f(x) * dx) + int(g(x) * dx)

    int(2 * dx / (x - 1))  -  int(1 * dx)

    u = x - 1

    du = dx

    int(2 * du / u)  -  int(dx)

    Integrate

    2 * ln|u| - x + C =>

    2 * ln|x - 1| - x + C

    Now, before we go crazy and just plug in our domain endpoints, we need to determine that 2 / (x - 1) was both continuous and differentiable along the domain.  There's a discontinuity when x = 1, and that's it, so we're good.  Just plug in your domain endpoints now

    2 * ln|4 - 1| - 4 - 2 * ln|2 - 1| + 2 =>

    2 * (ln|3| - ln|1|) - 2 =>

    2 * (ln(3) - 0) - 2 =>

    2 * ln(3) - 2 =>

    2 * (ln(3) - 1)

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  • Bryce
    Lv 7
    1 month ago

    The function goes below the x-axis at x= 3, so you have to make two calculations: one on the interval 2≤ x ≤3 and another on 3≤ x ≤4.  Take the negative of the second calculation because y is below the x-axis.

    2ln|x - 1| - x= ln4 - 3 - (-2) - (ln9 - 4)

    = ln(4/9) + 3≈ 2.189069784

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