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Math Help - definite integral

  1. #1
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    definite integral

    just want to ask what will i do first if i will be given a problem like this:

    \int_7^7 \sqrt[3](|t|+1)dt

    btw the lower limit must be -7 and the |t|+1 must be inside the cube root. I dont know how to put in in latex sorry

    thanks
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  2. #2
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    Quote Originally Posted by FailCalculus View Post
    just want to ask what will i do first if i will be given a problem like this:

    \int_7^7 \sqrt[3](|t|+1)dt

    btw the lower limit must be -7 and the |t|+1 must be inside the cube root. I dont know how to put in in latex sorry

    thanks
    You will need to treat |t| as a hybrid function.

    Note that

    |t| = \begin{cases}\phantom{-}t\textrm{ if }t \geq 0\\ -t\textrm{ if }t < 0\end{cases}.


    Therefore

    \int_{-7}^7{\sqrt[3]{|t| + 1}\,dt} = \int_{-7}^0{\sqrt[3]{-t + 1}\,dt} + \int_0^7{\sqrt[3]{t + 1}\,dt}.


    Can you go from here? You will need to use u substitutions...
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  3. #3
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    ok thanks just to clarify things. so i need to use u=|t|+1 and use the general integration formula?
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  4. #4
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    \displaystyle \int_{-7}^7 \sqrt[3]{|t|+1}dt

    note that for f(t) = \sqrt[3]{|t|+1} , f(t) = f(-t) , an even function.

    using symmetry about the y-axis ...

    \displaystyle \int_{-7}^7 \sqrt[3]{|t|+1}\,dt = 2 \int_0^7 \sqrt[3]{t+1}\,dt = \frac{3}{2}\left[(t+1)^{\frac{4}{3}\right]_0^7 = \frac{45}{2}
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  5. #5
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    Quote Originally Posted by FailCalculus View Post
    ok thanks just to clarify things. so i need to use u=|t|+1 and use the general integration formula?
    No, like I said, you have to break it into two integrals.

    So for the first, use u = t + 1 and in the second use v = -t + 1.
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