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Math Help - Evaluate the intergral

  1. #1
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    Evaluate the intergral

    1.

    2.

    3.

    4.

    Please and Thank You
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  2. #2
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    Quote Originally Posted by dizizviet View Post

    2.


    Please and Thank You
    \int_1^2 9x^2\times 2x^3 dx

    \int_1^2 9\times2x^{2+3} dx

    \int_1^2 18x^{5} dx

    now use the following formula

    \int ax^{n} dx = \frac{ax^{n+1}}{n+1}+c

    to find

    \int_1^2 18x^{5} dx = \frac{18x^{5+1}}{5+1}

     = \frac{18x^{6}}{6}

     = 3x^{6}

    We we are intergrating between 1 and 2 sub these values in and subtract the values

     = 3(2)^{6} - 3(1)^{6}

     = 3\times 32 - 3\times 1

     = 96- 3 = 93

    this will work for Q4.
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  3. #3
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    can anyone else help me figure out the last three equations?
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  4. #4
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    Quote Originally Posted by dizizviet View Post
    can anyone else help me figure out the last three equations?
    Post #2 showed you how to do Q2 and told you that Q4 was done in the same way.

    Q1 Substitute w = 7t and recognise a standard form involving a hyperbolic function.

    Q3 Substitute w = log(9x).
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  5. #5
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    Quote Originally Posted by pickslides View Post
    \int_1^2 9x^2\times 2x^3 dx

    \int_1^2 9\times2x^{2+3} dx

    \int_1^2 18x^{5} dx

    now use the following formula

    \int ax^{n} dx = \frac{ax^{n+1}}{n+1}+c

    to find

    \int_1^2 18x^{5} dx = \frac{18x^{5+1}}{5+1}

     = \frac{18x^{6}}{6}

     = 3x^{6}

    We we are intergrating between 1 and 2 sub these values in and subtract the values

     = 3(2)^{6} - 3(1)^{6}

     = 3\times 32 - 3\times 1

     = 96- 3 = 93

    this will work for Q4.
    I'm sorry i missed typed my equations for the the second problem and third problem
    but i firgured it out though thanks alot for your helps =)
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