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Math Help - Definite Integrals

  1. #1
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    Definite Integrals

    Hey guys,
    just confused on a couple problems, not sure how some of the numbers showed up in the answer

    1.) (t^2+2)/(t^3+6t+3)dt
    The answer is 1/3 LN| t^3+6t+3| +c
    Where did the natural log come from, is there any type of formula I am missing for this? That's the only part I'm confused about from the problem.


    2.) (-x^2-3x+5)dx defined between 3 and -2
    The answer is 35/6. For my math, I did the following:

    ((-x^3/3)-3x^2+5x) and then plugged in the defined points, (3) - (-2), however I'm not getting anything close to 35/6.

    Any tips and advice is appreciated, also if you have any special formulas that you use to solve definite integrals would also be nice,

    Thanks alot!
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  2. #2
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    Quote Originally Posted by cokeclassic View Post
    Hey guys,
    just confused on a couple problems, not sure how some of the numbers showed up in the answer

    1)

    (t^2+2)/(t^3+6t+3)dt

    The answer is 1/3 LN| t^3+6t+3| +c

    Where did the natural log come from, is there any type of formula I am missing for this? That's the only part I'm confused about from the problem.

    Any tips and advice is appreciated, also if you have any special formulas that you use to solve definite integrals would also be nice,

    Thanks alot!
    Notice that the derivative of the denominator is 3 times the numerator.


    \displaystyle\int{f(t)}dt=\int{\frac{t^2+2}{t^3+6t  +3}}dt

    u=t^3+6t+3

    \displaystyle\frac{d}{dt}\left(t^3+6t+3\right)=3t^  2+6=3\left(t^2+2\right)

    \displaystyle\Rightarrow\frac{du}{3}=\left(t^2+2\r  ight)dt

    Then you end up with

    \displaystyle\frac{1}{3}\int{\frac{du}{u}
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  3. #3
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    Where are the bounds/limits, if it's a definite integral?

    \displaystyle 1. ~ \int\frac{t^2+2}{t^3+6t+3}\;{dt} = \frac{1}{3}\int\frac{(t^3+6t+3)'}{t^3+6t+3}\;{dt}.
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  4. #4
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    Quote Originally Posted by cokeclassic View Post
    Hey guys,
    just confused on a couple problems, not sure how some of the numbers showed up in the answer

    2.) (-x^2-3x+5)dx defined between 3 and -2
    The answer is 35/6. For my math, I did the following:

    ((-x^3/3)-3x^2+5x) and then plugged in the defined points, (3) - (-2), however I'm not getting anything close to 35/6.
    Thanks alot!
    \displaystyle \int_{-3}^2 -x^2-3x+5\;dx = \bigg[\frac{-x^3}{3}-\frac{3x^2}{2}+5x\bigg]_{-3}^{2}
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  5. #5
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    The first one is just a integral that's supposed to be evaluated I guess, not sure if it's called something else :P. The answer is said to be 1/3 LN| t^3+6t+3| +c , however I'm not sure how to get there because I don't know where the natural log comes in.


    Also, why did you flip the limits for #2?
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  6. #6
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    Understood #2, just a typo, thanks a lot, still having trouble with #1
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  7. #7
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    2. should be \displaystyle \int_{-2}^3{-x^2 - 3x + 5\,dx} = \left[-\frac{x^3}{3} - \frac{3x^2}{2} + 5x\right]_{-2}^3...
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  8. #8
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    \displaystyle \int{\frac{t^2 + 2}{t^3 + 6t + 3}\,dt} = \frac{1}{3}\int{\frac{1}{t^3 + 6t + 3}(3t^2 + 6)\,dt}.

    Make the substitution \displaystyle u = t^3 + 6t + 3 so that \displaystyle \frac{du}{dt} = 3t^2 + 6 and the integral becomes

    \displaystyle \frac{1}{3}\int{\frac{1}{u}\,\frac{du}{dt}\,dt}

    \displaystyle = \frac{1}{3}\int{\frac{1}{u}\,du}...

    I'm sure you can go from here...
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  9. #9
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    Quote Originally Posted by cokeclassic View Post
    The first one is just a integral that's supposed to be evaluated I guess, not sure if it's called something else :P. The answer is said to be 1/3 LN| t^3+6t+3| +c , however I'm not sure how to get there because I don't know where the natural log comes in.


    Also, why did you flip the limits for #2?
    \displaystyle\int{\frac{1}{x}}dx=ln|x|+C\Rightarro  w\int{\frac{1}{u}}du=ln|u|+C

    Back-substitute for u.
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