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Math Help - Area between Curves

  1. #1
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    Area between Curves

    Hey I have a calc exam Thursday and have been practicing for it. I came across this area problem that has me stumped, usually area is no problem for me but I can't seem to figure this one out.

    Area between Curves-integral.jpg

    I figured that you would break the area into two from 0 to 1 and 1 to 3 since the function and y=127 intersect at x=1 but when I went to integrate everything the computer told me that my answer was wrong. Please help
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  2. #2
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    The question, as posted, doesn't make sense to me. The graph of the cubic, the x-axis, and x= 1 form a single region. I don't see what "y= 127" has to do with that.
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  3. #3
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    Hello, rawkstar!

    Since you didn't show your work, we have no idea where your error is.


    \text{Consider the region bounded by the graph of:}
    y\:=\:x^3 + 18x^2 + 108x,\;(0 \le x \le 1)\,\text{ on the left,}
    x\text{-axis below, }\,y = 127\text{ above, and }x = 3\text{ on the right.}

    \text{(1) Find the area of the region.}

    The graph looks like this:

    Code:
    
            |
         127+       * - - - - - - - *
            |      .|:::::::::::::::|
            |     .*|:::::::::::::::|
            |   .*B:|::::::A::::::::|
            | .*::::|:::::::::::::::|
        - - * - - - + - - - + - - - + -
            |       1       2       3
            |

    Area \,A is a 2-by-127 rectangle: . A = 254


    Area \,B requires an integral: . \displaystyle B \;=\;\int^1_0(x^3 + 18x^2 + 108x)\,dx

    We have: . B \;=\;\frac{1}{4}x^4 + 6x^3 + 54x^2\,\bigg]^1_0 \;=\;\frac{1}{4} + 6 + 54
    . . . . . . . . B \;=\;60\frac{1}{4}

    Therefore, the total area is: . 254 + 60\frac{1}{4} \;=\;314\frac{1}{4}\text{ units}^2.

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