Calculus

There is a diagram with the curve y=(2x-5)^4.The point P has coordinates (4,81) and the tangent to the curve at P meets the x-axis at Q.Find the area of the region enclosed between curve ,PQ and the x-axis.

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  1. y = (2x-5)^4
    dy/dx = 4(2x-5)^3 (2) = 8(2x-5)^3
    P(4,81) lies on the curve, and the slope of the tangent at that point is
    8(3^3) = 216
    and the equation of the tangent is
    y-81 = 216(x-4)
    y = 216x - 783
    At Q, the intercept with the x-axis, x = 783/216 = 29/8

    the curve makes contact with the x-axis at 5/2

    Let a vertical from P cut the x-axis at R

    So find the area of the shape between the x-axis the vertical PR and the curve, then subtract the area of triangle PRQ.

    you will need the integral of (2x-5)^4 which would be (1/10)(2x-5)^5

    take over

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    Reiny
  2. How come At Q, the intercept with the x-axis, x = 783/216 = 29/8

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  3. How does the vertical cut x-axis?

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  4. From the tangent equation, y = 216x - 783
    at the x-axis, y = 0
    so 0 = 216x - 783
    216x = 783 ----> x = 783/216 = 29/8

    your 2nd post: the x-axis is horizontal, any vertical line would cut it, wouldn't it?

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    Reiny
  5. I found area of triangle but couldn't get area of the shape between the x-axis the vertical PR and the curve

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  6. I got it.Thanks

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