Results 1 to 3 of 3

Math Help - Cal I Problem...

  1. #1
    Newbie
    Joined
    Jan 2014
    From
    Texas
    Posts
    3

    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  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,776
    Thanks
    1141

    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

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

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

    and from the second F(6)-F(0)=6

    finally

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

    \left(F(2)-F(0)\right)-\left(F(6)-F(0)\right)=1-6=-5
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,923
    Thanks
    1762
    Awards
    1

    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:
    \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}
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum