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Math Help - integral addition with given area

  1. #1
    Member mybrohshi5's Avatar
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    integral addition with given area

    Let

     \displaystyle \int_{-4}^{3.5} f(x) dx =7, \ \int_{-4}^{-1.5} f(x) dx=10, \ \int_{1}^{3.5} f(x)dx =6.

    Find \displaystyle \int_{-1.5}^{1} f(x)dx=

    and \displaystyle \int_{1}^{-1.5} (7 f(x)- 10)dx=

    i found the first part to be -9 because 10+6+x=7 (solved for x)

    i just cant get the second part. im just not exactly sure what it wants me to do.
    thanks for any help =)
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  2. #2
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    Hello, mybrohshi5!

    Let: . \int_{\text{-}4}^{3.5}f(x)\,dx \:=\:7 \qquad \int_{\text{-}4}^{\text{-}1.5} f(x)\,dx\;=\;10 \qquad \int_{1}^{3.5} f(x)\,dx \:=\:6.

    Find \int_{\text{-}1.5}^{1} f(x)\,dx

    and \int_{1}^{\text{-}1.5} (7f(x) - 10)\,dx

    i found the first part to be -9 because 10+6+x=7 (solved for x)

    i just cant get the second part. im just not exactly sure what it wants me to do.

    The second one is confusing because of the switched limits.

    I will switch them myself.
    My answer will be the negative of their answer.


    I have: . \int^1_{\text{-}1.5}\bigg[7f(x) - 10\bigg]\,dx

    . . . . =\; 7\underbrace{\int^1_{\text{-}1.5}f(x)\,dx}_{\text{This is -9}} \,-\,\int^1_{\text{-}1.5}10\,dx

    . . . . = \;-63 - \bigg[10t\bigg]^1_{\text{-}1.5}

    . . . . =\; -63 - \bigg[10 - (-15)\bigg]

    . . . . =\; -63 - 25 \;=\;-88


    Therefore, their answer is: . +88

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  3. #3
    Member mybrohshi5's Avatar
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    oops i didnt even notice the a and b values switched. either way thank you for the great explanation
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