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Math Help - Distance traveled using integrals

  1. #1
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    Distance traveled using integrals

    Ok, I know what to do (take antiderivative and plug in points from a certain interval) but the final answer i get is 10.5 (which is wrong to the system). Can someone please check my work and let me know the correct answer? it's driving me nuts


    My work (the boxes show the antiderivative and the interval from 1 to 0. I know that you have to plug in 1 to the antiderivative and 0 and subtract the two. The same goes for the interval 3 to 1. You then would add the two together):


    3 |t2 + 3t4| dt
    0
    =
    1 t23t + 4 dt
    0
    +
    3 t2 + 3t4 dt
    1
    Antiderivatve= -1/3 t^3 -3/2 t^2 +4t |1 and 0 intervals

    Antiderivative= 1/3t^3 +3/2 t^2 -4t | 3 and 1 intervals





    Interval 1 subtracted by interval of 0:
    [-1/3 (1)^3 -3/2 (1)^2 +4(1)] - [0]


    Added with


    The interval of 3 subtracted by interval of 1:
    [1/3(3)^3 +3/2(3)^2-4(3) ] - [1/3(1)^3 +3/2(1)^2-4(1)
    Last edited by Steelers72; February 22nd 2013 at 12:15 PM.
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  2. #2
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    Re: Distance traveled using integrals

    I'd advise you to fix your post, seeing as it is indecipherable
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    Re: Distance traveled using integrals

    Lol my fault. Fixed
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    Re: Distance traveled using integrals

    anyone?
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    Re: Distance traveled using integrals

    To be frank, I can't see anything amiss with your work.

    You have the definite integral split at its proper place, so you'll always get a positive output when you evaluate the definite integral:

    \int_0^1(-t^2-3t+4)dt + \int_1^3(t^2+3t-4)dt

    Likewise, it seems that you proceeded with taking the antiderivative properly:

    \frac{-t^3}{3}- \frac{3}{2}t^2+4t \biggr|_0^1+ \frac{t^3}{3}+ \frac{3}{2}t^2 - 4t \biggr|_1^3

    Perhaps you've just had some sort of arithmetic error.
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    Re: Distance traveled using integrals

    Quote Originally Posted by Steelers72 View Post

    3 |t2 + 3t4| dt
    0
    =
    1 t23t + 4 dt
    0
    +
    3 t2 + 3t4 dt
    1
    Antiderivatve= -1/3 t^3 -3/2 t^2 +4t |1 and 0 intervals

    Antiderivative= 1/3t^3 +3/2 t^2 -4t | 3 and 1 intervals
    You might find it helpful to take a look at our LaTeX forum. It may look hard at the start but it is reasonably easy to learn.

    -Dan
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