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Math Help - Area 1

  1. #1
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    Area 1

    Calculate the area by integration of the region limited by the curve: And build the graph

    y=e^(-x), u=x+1 and x=-1

    And build the graph


    answer:
    e-\frac{3}{2}
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  2. #2
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    Hello, Apprentice123!

    Did you sketch the graph?


    Calculate the area by integration of the region limited by the curves:

    . . . y\:=e\:^{-x},\quad y \:=\:x+1,\quad x\:=\:-1

    Answer: . e-\frac{3}{2}
    Code:
            *       |
                    |
             *      |       *
              *     |     *
              |*    |   *
              |::*  | *
              |:::::*
              |:::* |   * 
              |:*   |           *
        - - - * - - + - - - - - - - - -
            *       |
          *         |

    A \;=\;\int^0_{\text{-}1}\bigg[e^{-x} - (x + 1)\bigg]\,dx \;=\;\int^0_{\text{-}1}\bigg[e^{-x} - x - 1\bigg]\,dx \;\;=\;\;-e^{-x} - \frac{1}{2}x^2 - x\,\bigg]^0_{\text{-}1}

    . . = \;\;\bigg[-e^0 - 0 - 0\bigg] - \bigg[-e^{\text{-}(\text{-}1)} - \frac{1}{2}(\text{-}1)^2 - (\text{-}1)\bigg]

    . . = \;\;-1 - \left(-e - \frac{1}{2} + 1\right) \;\;=\;\;-1 + e - \frac{1}{2} \;\;=\;\;e - \frac{3}{2}



    I doubt if anyone will help you with the other two problems
    . . until you show some effort.

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  3. #3
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    which the points of intersection?
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  4. #4
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    You didn't sketch the graph, did you?

    Did you look at my diagram?
    Can't you FIND the intersections?

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