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MAT Question with respect to Integration Watch

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    Given a function f(x), you are told that:

    \int^1_0 3f(x)\ dx + \int^2_1 2f(x)\ dx = 7
    and
    \int^2_0 f(x)\ dx + \int^2_1 f(x)\ dx = 1

    The mark scheme rearranges this to:

    3A + 2B = 7 (which is fine)
    (A + B) + B = 1 (the part I don't understand)

    The part I don't understand is why the \int^2_0 f(x)\ dx is rearranged as (A + B). Is there something that's missing in my understanding of integration?

    Could you please explain?
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    (Original post by UCASLord)
    Given a function f(x), you are told that:

    \int^1_0 3f(x)\ dx + \int^2_1 2f(x)\ dx = 7
    and
    \int^2_0 f(x)\ dx + \int^2_1 f(x)\ dx = 1

    The mark scheme rearranges this to:

    3A + 2B = 7 (which is fine)
    (A + B) + B = 1 (the part I don't understand)

    The part I don't understand is why the \int^2_0 f(x)\ dx is rearranged as (A + B). Is there something that's missing in my understanding of integration?

    Could you please explain?
    0 to 2 is 0 to 1 and then 1 to 2
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    (Original post by UCASLord)
    Given a function f(x), you are told that:

    \int^1_0 3f(x)\ dx + \int^2_1 2f(x)\ dx = 7
    and
    \int^2_0 f(x)\ dx + \int^2_1 f(x)\ dx = 1

    The mark scheme rearranges this to:

    3A + 2B = 7 (which is fine)
    (A + B) + B = 1 (the part I don't understand)

    The part I don't understand is why the \int^2_0 f(x)\ dx is rearranged as (A + B). Is there something that's missing in my understanding of integration?

    Could you please explain?
    A is the area under f between 0 and 1 and B is the area under f between 1 and 2. Adding these gives the area between 0 and 2.
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    (Original post by Integer123)
    A is the area under f between 0 and 1 and B is the area under f between 1 and 2. Adding these gives the area between 0 and 2.
    *Strong Facepalm*

    A massive thank you! I can't believe I didn't see that...
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    (Original post by UCASLord)
    *Strong Facepalm*
    A useful smillie:

    :facepalm2:
 
 
 
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