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Integration: ares under two curves Watch

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    Did an exam recently with a question like the title says.

    Used the ( integral of y1 - y2 ) dx rule, however, I integrated fully before working out y1 - y2. However, someone else smarter than me worked out y1-y2 before integrating.

    Does that affect the overall answer or are the steps interchangeable?
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    (Original post by Regret786)
    Did an exam recently with a question like the title says.

    Used the ( integral of y1 - y2 ) dx rule, however, I integrated fully before working out y1 - y2. However, someone else smarter than me worked out y1-y2 before integrating.

    Does that affect the overall answer or are the steps interchangeable?
    Can you post an example of what you mean?
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    Integrating before subtracting is just taking the area under y2 away from the area under y1.

    Subtracting before integrating gives you a different function, but the area under that curve is equivalent to the difference above.

    In other words, yes, they're interchangeable.
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    I am curious too
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    (Original post by notnek)
    Can you post an example of what you mean?
    (Original post by TeeEm)
    I am curious too
    http://qualifications.pearson.com/co...-June-2014.pdf

    Question 7C. This isn't the exact question since the exam was on Fridays. The link is volume instead of curve.
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    (Original post by Regret786)
    http://qualifications.pearson.com/co...-June-2014.pdf

    Question 7C. This isn't the exact question since the exam was on Fridays. The link is volume instead of curve.
    \displaystyle \int_a^b (y_1 - y_2)\mathrm{d}x  = \int_a^b y_1 \mathrm{d}x - \int_a^b y_2 \, \mathrm{d}x in general for continuous y_1, y_2.
 
 
 
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