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Math Help - Definite Intergral

  1. #1
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    Definite Intergral

    Hi, could someone tell me if this is the correct answer?

    <br /> <br />
\int_0^2 {3x^5e^{x^3}dx}<br />

    I got:

    <br />
7e^8 +1
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  2. #2
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    Quote Originally Posted by millerst View Post
    Hi, could someone tell me if this is the correct answer?

    <br /> <br />
\int_0^2 {3x^5e^{x^3}dx}<br />

    I got:

    <br />
7e^8 +1
    yes that is correct
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  3. #3
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    Thanks, what about this one? It's rather strange:

    <br /> <br />
\int_{e}^{1} \ \frac{(ln(x))^{1/n}}{x} \ dx <br />
    where n is a natural number

    My answer:

    (n^(1/n +1))/(1+n)
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  4. #4
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    Quote Originally Posted by millerst View Post
    Thanks, what about this one? It's rather strange:

    <br /> <br />
\int_{e}^{1} \ \frac{(ln(x))^{1/n}}{x} \ dx <br />
    where n is a natural number

    My answer:

    (n^(1/n +1))/(1+n)

    Note quite the final answer should be

    -\frac{n}{n+1}
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  5. #5
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    Quote Originally Posted by millerst View Post
    Thanks, what about this one? It's rather strange:

    <br /> <br />
\int_{e}^{1} \ \frac{(ln(x))^{1/n}}{x} \ dx <br />
    where n is a natural number

    My answer:

    (n^(1/n +1))/(1+n)
    Here is what I did, can you point out where i am going wrong:
    <br /> <br />
\int_{e}^{1} \ \frac{(ln(x))^{1/n}}{x} \ dx <br /> <br />

    let ln(x) = u
    du = 1/x dx

    change limits of integration
    x=e then u=1
    x=1, u=0

    =\int_{1}^{0} {u}^{1/n}du<br />

    = \frac {n{u}^{1/n +1}}{1+n}\ <br />
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  6. #6
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    Quote Originally Posted by millerst View Post
    Here is what I did, can you point out where i am going wrong:
    <br /> <br />
\int_{e}^{1} \ \frac{(ln(x))^{1/n}}{x} \ dx <br /> <br />

    let ln(x) = u
    du = 1/x dx

    change limits of integration
    x=e then u=1
    x=1, u=0

    =\int_{1}^{0} {u}^{1/n}du<br />

    = \frac {n{u}^{1/n +1}}{1+n}\ <br />
    \int_{1}^{0}u^{\frac{1}{n}}du=\frac{n}{n+1}u^{\fra  c{n+1}{n}} \bigg|_{1}^{0}=\frac{n}{n+1}(0^{\frac{n+1}{n}}-1^{\frac{n+1}{n}})=-\frac{n}{n+1}
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