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    https://gyazo.com/4bf8069ae8fa3d9876...2eec891f50020e

    Part two of the question, I dont know what im doing wrong, so i find the points they intersect which are x = -1 and -3

    Then i guess i do the integration of the equation of the curve between points -3 and -1 take away the integration of the equation of the line between the same points as before.

    Could someone provide me with the working out? Or something I'm doing wrong.

    Thanks!
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    (Original post by Jian17)
    https://gyazo.com/4bf8069ae8fa3d9876...2eec891f50020e

    Part two of the question, I dont know what im doing wrong, so i find the points they intersect which are x = -1 and -3

    Then i guess i do the integration of the equation of the curve between points -3 and -1 take away the integration of the equation of the line between the same points as before.

    Could someone provide me with the working out? Or something I'm doing wrong.

    Thanks!
    Seems like the correct approach.

    You should have \displaystyle \int_{-3}^{-1} 8-2x-x^2 .dx - \int_{-3}^{-1} 2x+11 .dx = \int_{-3}^{-1} -3-4x-x^2.dx
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    (Original post by RDKGames)
    Seems like the correct approach.

    You should have \displaystyle \int_{-3}^{-1} 8-2x-x^2 .dx - \int_{-3}^{-1} 2x+11 .dx = \int_{-3}^{-1} -3-4x-x^2.dx
    Okay thanks, one thing how do you know if you do it from -1 to -3 and not from -3 to -1?
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    As you say, the points of intersection are at (-3,5) and (-1,9).

    Your method is right, so I'm not sure why you're not getting the required answer? Be careful with the signs!

    By my rough calculation I get the answer as 4/3.
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    (Original post by Jian17)
    Okay thanks, one thing how do you know if you do it from -1 to -3 and not from -3 to -1?
    Can be either way, the norm is to go from lower to upper bound such that there is an increase rather than decrease.

    Though if you are going from a higher number to lower one, so here that would be from -1 to -3, then x is decreasing and hence dx changes sign from +ve to -ve, and hence the whole integral changes sign, ie

    \displaystyle \int_{-3}^{-1} -3-4x-x^2 .dx = - \int_{-1}^{-3} -3-4x-x^2.dx
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    (Original post by Jian17)
    Okay thanks, one thing how do you know if you do it from -1 to -3 and not from -3 to -1?
    Because -1 is more positive than -3! Therefore -1 is your upper limit. Just had to put it out there!
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    (Original post by Jian17)
    Okay thanks, one thing how do you know if you do it from -1 to -3 and not from -3 to -1?
    Because it is usual to integrate from the smaller number to the larger number (think in terms of a number line containing all the integers from -infinity to +infinity). You are free to set the limits the other way around so that -1 is now on the bottom but that would impart an unnecessary minus sign in your result.
 
 
 
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