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Math Help - Problem with Integration

  1. #1
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    Problem with Integration

    Hi

    I am having trouble trying to integrate this equation:

    \int \sqrt(1+2x+2x^3)

    how would i approach an expression like this?

    Do i integrate the inside and then the outside?

    P.S
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  2. #2
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    Are you sure it's \int{\sqrt{1+2x+3x^3}}\;{dx}?
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  3. #3
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    Well, the original question asked to find the length of arc of f(x) = \frac{2}{3}(1+x^2)^(\frac{3}{2})<br />
    y=2-\frac{1}{2}x between x=0, y=3

    I used the length of arc rule:

    \int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx

    This is where i was stuck:

    \int_0^3 \sqrt{1+2x+2x^3}dx
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  4. #4
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    Quote Originally Posted by Paymemoney View Post
    Well, the original question asked to find the length of arc of f(x) = \frac{2}{3}(1+x^2)^(\frac{3}{2})<br />
    y=2-\frac{1}{2}x between x=0, y=3

    I used the length of arc rule:

    \int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx

    This is where i was stuck:

    \int_0^3 \sqrt{1+2x+2x^3}dx
    Dear Paymemoney,

    I have some problems regarding your question. You want to find the arc length of f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}} is'nt?

    What is the second equation y=2-\frac{1}{2}x ??? How did you get this? Do you want to find the arc length of this curve too??
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  5. #5
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    Quote Originally Posted by Sudharaka View Post
    Dear Paymemoney,

    I have some problems regarding your question. You want to find the arc length of f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}} is'nt?

    What is the second equation y=2-\frac{1}{2}x ??? How did you get this? Do you want to find the arc length of this curve too??

    opps srry my mistake the equation is only f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}}
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  6. #6
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    Quote Originally Posted by Paymemoney View Post
    opps srry my mistake the equation is only f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}}
    Dear Paymemoney,

    You have simplified incorrectly,

    \int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx

    =\int_0^3\sqrt{1+4x^2(1+x^2)}dx

    =\int_0^3\sqrt{1+4x^2+4x^4}dx

    =\int_0^3\sqrt{(1+2x^2)^2}dx

    Can you continue from here??
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  7. #7
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    Quote Originally Posted by Sudharaka View Post
    Dear Paymemoney,

    You have simplified incorrectly,

    \int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx

    =\int_0^3\sqrt{1+4x^2(1+x^2)}dx

    =\int_0^3\sqrt{1+4x^2+4x^4}dx

    =\int_0^3\sqrt{(1+2x^2)^2}dx

    Can you continue from here??
    yes,
    thank you, i worked it out
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