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Thread: Cal I Problem...

  1. #1
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    Cal I Problem...

    Given that (image attached)

    What is the value of (image attached)

    Drawing the line and plugging in the point distances, I thought the answer would be 6, but that doesn't seem to be an answer choice. I'd appreciate help with this. I'd be grateful for a step-by-step guide.
    Attached Thumbnails Attached Thumbnails Cal I Problem...-image.jpg  
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  2. #2
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    Re: Cal I Problem...

    Quote Originally Posted by JenniferC View Post
    Given that (image attached)

    What is the value of (image attached)

    Drawing the line and plugging in the point distances, I thought the answer would be 6, but that doesn't seem to be an answer choice. I'd appreciate help with this. I'd be grateful for a step-by-step guide.
    The idea is that f(x) has an anti-derivative F(x) such that

    $\displaystyle \int_a^b f(x)dx=F(b)-F(a)$

    from the first equation $\displaystyle F(2)-F(0)=1$

    and from the second $\displaystyle F(6)-F(0)=6$

    finally

    $\displaystyle \int_6^2 f(x)dx=F(2)-F(6)=F(2)-F(6)-F(0)+F(0)=$

    $\displaystyle \left(F(2)-F(0)\right)-\left(F(6)-F(0)\right)=1-6=-5$
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  3. #3
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    Re: Cal I Problem...

    Quote Originally Posted by JenniferC View Post
    Given that (image attached)
    What is the value of (image attached)
    Here is another way:
    $\displaystyle \int_0^6 {f(x)dx} = \int_0^2 {f(x)dx + } \int_2^6 {f(x)dx} = \int_0^2 {f(x)dx }\color{blue} - \int_6^2 {f(x)dx} $
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