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Math Help - Area Bound between curves

  1. #1
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    Question Area Bound between curves

    x+ 5 y = 19 , x + 5 = y^2

    I have already determined that this can be calculated with:

    <br />
\displaystyle \int_{-5}^4 2\sqrt{x+5},dx + \int_{4}^{59} \frac{(19-x)}{5}+\sqrt{x+5},dx<br />

    When I solve this I get 22.183

    The second part of this question wants me to put this into the form of a SINGLE integral with just one upper and lower limit.

    How do I do this?

    I assume I just add the two integrals like so:

    <br />
\displaystyle \int_A^B 3\sqrt{x+5} + \frac{19-x}{5} dx<br />

    Is this correct? How do I get the upper and lower limits? I assume its not just -5 & 59 right?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Vamz View Post
    x+ 5 y = 19 , x + 5 = y^2

    I have already determined that this can be calculated with:

    <br />
\displaystyle \int_{-5}^4 2\sqrt{x+5},dx + \int_{4}^{59} \frac{(19-x)}{5}+\sqrt{x+5},dx<br />

    When I solve this I get 22.183

    The second part of this question wants me to put this into the form of a SINGLE integral with just one upper and lower limit.

    How do I do this?

    I assume I just add the two integrals like so:

    <br />
\displaystyle \int_A^B 3\sqrt{x+5} + \frac{19-x}{5} dx<br />

    Is this correct? How do I get the upper and lower limits? I assume its not just -5 & 59 right?
    No you write it as:

    \displaystyle A=\int_{-8}^3 (19-5y)-(y^2-5)\ dy

    (I would suggest you sketch the curves/area to see why)

    CB
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  3. #3
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    Can you please explain how you got that expression?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Vamz View Post
    Can you please explain how you got that expression?
    I drew a picture, then saw that you could integrate with respect to y more conveniently.

    CB
    Attached Thumbnails Attached Thumbnails Area Bound between curves-gash.png  
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