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Math Help - definite integral

  1. #1
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    definite integral

    need help plz
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  2. #2
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    Quote Originally Posted by gracy View Post
    need help plz
    In shorthand:
    \int_0^3 = \int_0^1 + \int_1^3

    so
    \int_1^3 = \int_0^3 - \int_0^1 = 4 - 7 = -3

    -Dan
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  3. #3
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    Hello, gracy!

    Make some sketches and use baby-talk . . .


    \int^1_0f(x)\,dx\:=\:7 and \int^3_0f(x)\,dx\:=\:4
    Find: \int^3_1f(x)\,dx

    The first integral says: the area under f(x) from 0 to 1 is 7 square units.
    The graph might look like this.
    Code:
            |
            *
            |   *
            |     *
            |  7   *
            |
          --+-------*----
            |       1 
            |

    The second integral says: the area under f(x) from 0 to 3 is only 4.

    How did we lose some area?
    Some of the area from 1 to 3 must be negative.
    . . The graph must dip below the x-axis.

    The graph might look like this.
    Code:
            |
            *
            |   *
            |     *
            |  7   *
            |               3
          --+-------*-------+--
            |       1 *  -3 :
            |           *   :
            |               *

    And that is why: . \int^3_1f(x)\,dx\:=\:-3

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