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Math Help - Prove that ... Help!!!

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    Prove that ... Help!!!

    Prove that
    9 < integral (0,3) (4x + 1)^(1/4)dx + integral (1,3) (4x − 1)^(1/4)dx < 9.0001
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  2. #2
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    Quote Originally Posted by mathdanny View Post
    Prove that
    9 < integral (0,3) (4x + 1)^(1/4)dx + integral (1,3) (4x − 1)^(1/4)dx < 9.0001
    \int_0^3{(4x + 1)^{\frac{1}{4}}\,dx} + \int_1^3{(4x - 1)^{\frac{1}{4}}\,dx} = \frac{1}{4}\int_0^3{4(4x + 1)^{\frac{1}{4}}\,dx} + \frac{1}{4}\int_1^3{(4x - 1)^{\frac{1}{4}}\,dx}

     = \frac{1}{4}\left[\frac{4}{5}(4x + 1)^{\frac{5}{4}}\right]_0^3 +\frac{1}{4}\left[\frac{4}{5}(4x - 1)^{\frac{5}{4}}\right]_1^3

     = \frac{1}{5}\left[(4x + 1)^{\frac{5}{4}}\right]_0^3 + \frac{1}{5}\left[(4x - 1)^{\frac{5}{4}}\right]_1^3

     = \frac{1}{5}\left[13^{\frac{5}{4}} - 1^{\frac{5}{4}}\right] + \frac{1}{5}\left[11^{\frac{5}{4}} - 3^{\frac{5}{4}}\right]

    What does the calculator say this is?
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  3. #3
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    Quote Originally Posted by Prove It View Post
    \int_0^3{(4x + 1)^{\frac{1}{4}}\,dx} + \int_1^3{(4x - 1)^{\frac{1}{4}}\,dx} = \frac{1}{4}\int_0^3{4(4x + 1)^{\frac{1}{4}}\,dx} + \frac{1}{4}\int_1^3{(4x - 1)^{\frac{1}{4}}\,dx}

     = \frac{1}{4}\left[\frac{4}{5}(4x + 1)^{\frac{5}{4}}\right]_0^3 +\frac{1}{4}\left[\frac{4}{5}(4x - 1)^{\frac{5}{4}}\right]_1^3

     = \frac{1}{5}\left[(4x + 1)^{\frac{5}{4}}\right]_0^3 + \frac{1}{5}\left[(4x - 1)^{\frac{5}{4}}\right]_1^3

     = \frac{1}{5}\left[13^{\frac{5}{4}} - 1^{\frac{5}{4}}\right] + \frac{1}{5}\left[11^{\frac{5}{4}} - 3^{\frac{5}{4}}\right]

    What does the calculator say this is?
    I calculated this value and I got 7.9538... This value is too small! So, there is should the other the way to prove it! I tried many ways, but all gave me so small value!
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