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Posted by on Thursday, March 7, 2013 at 12:03am.

Given that \displaystyle \int_0^4 x^3\sqrt{9+x^2} dx = a, what is the value of \lfloor a \rfloor?

  • Calculus - , Thursday, March 7, 2013 at 11:55am

    Put x = 3 sinh(t), then:

    Integral from 0 to 4 of

    x^3 sqrt{9+x^2} dx =

    Integral from 0 to arcsinh(4) of

    3^5 sinh^3(t) cosh^2(t) dt =

    3^5 Integral from 0 to arcsinh(4) of

    sinh(t) [cosh^2(t) - 1] cosh^2(t) dt =

    3^5 Integral from 0 to arcsinh(4) of

    [cosh^2(t) - 1] cosh^2(t) dcosh(t)

    Putting cosh(t) = u gives:

    3^5 Integral from 1 to sqrt(17) of

    [u^2 - 1] u^2 du =

    3^5 {sqrt(17)[1/5 17^2 - 1/3 17] + 2/15 }

    You then need to round this off to below.

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