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Math Help - Area bounded by curve and tangent line

  1. #1
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    Area bounded by curve and tangent line

    Find the area of the plane region bounded by the curve y=x^3 and the tangent line to the curve at the point (1,1). So far, I have found the tangent line to be y = 3x - 2, and since the point x=1 was an intersection, it's one of the limits in the integral. However, I am having trouble finding the second intersection in order to know both the upper and lower limit. How would I calculate it?
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  2. #2
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    Hello, zer0e!

    It should be simple Algebra . . .(okay, maybe not so simple).


    Find the area of the plane region bounded by the curve y\:=\:x^3
    and the tangent line to the curve at the point (1,1).
    Your tangent line is correct: . y \:=\:3x - 2


    To find the intersections, set the functions equal to each other:
    . . x^3 \:=\:3x - 2 \quad\Rightarrow\quad x^3 - 3x + 2 \:=\:0

    We see that x = 1 is a solution, so (x-1) is a factor.

    With long division: . x^3 - 2x + 2 \;=\;(x-1)(x^2+x-2) \;=\;(x-1)(x-1)(x-2)

    The intersections occurs at: . (-2,-8),\;(1,1)
    Code:
                  |
                  |      ...♥(1,1) 
                  | ...o::*
      - - - - - ..o:::::* - - - -
             o::::|:::*
           .::::::|:*
          o:::::::*
         .::::::* |
         o::::*   |
         :::*     |
         :*       |
        ♥(-2,-8)  |
                  |

    Therefore: . A \;=\;\int^1_{-2}\bigg[x^3 - (3x-2)\bigg]\,dx

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  3. #3
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    Ahhh I see. I had forgotten about the long division. Thanks!
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