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Math Help - integration using the fundamental theorem of calc part1

  1. #1
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    Smile integration using the fundamental theorem of calc part1

    integrate t^(1/2)sintdt. let lower bound=(x^1/2) upper bound=(x^3). use the chain rule and fundamental theorem calculas part1.
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    Quote Originally Posted by myoplex11 View Post
    integrate t^(1/2)sintdt. let lower bound=(x^1/2) upper bound=(x^3). use the chain rule and fundamental theorem calculas part1.
    It seems to me that x should be at least x>0
    It seems you wish to integrate the function,
    f(x)=\int_{x^{1/2}}^{x^3}t^{1/2}\sin tdt

    Let use find whether x^{1/2}=x^3
    The only possible solution is x=0,1 but x>0 thus, x=1.
    In that case,
    f(1)=0

    If, 0<x<1 then, x^{1/2}>x^3
    Thus,
    \int_{x^{1/2}}^{x^3}t^{1/2}\sin t dt=-\int_{x^3}^{x^{1/2}}t^{1/2}\sin tdt
    Therefore, there exists a point in the interval [x^3,x^{1/2}]
    Using the integral subdivision rule,
    -\int_{x^3}^c t^{1/2}\sin tdt-\int_c^{x^{1/2}}t^{1/2}\sin tdt
    Thus,
    \int_c^{x^3} t^{1/2}\sin tdt-\int_c^{x^{1/2}}t^{1/2}\sin tdt
    Let,
    \mu (x)=\int_c^x t^{1/2}\sin tdt
    Therefore, those two functions can be expressed as composition functions,
    \mu (x^3)-\mu (x^{1/2})
    But by the Fundamental theorem,
    \mu ' (x)=x^{1/2}\sin x
    And since x^3,x^{1/2} are both differenciable on (0,1) we can use the chain rule.
    3x^2 \mu' (x^3)+\frac{1}{2}x^{-1/2}\mu'(x^{1/2})=3 x^{7/2}\sin(x^3)+\frac{1}{2}x^{-1/2}\sin (x^{-1/2})
    Something interesting to note.
    \lim_{x\to 1} of this expression if not equal to f(1) which means the function you mentioned is not everywhere differenciable.

    The other part of the problem is to consider the case x>1. Since it is late now I cannot complete the solution but I think it will be the same as for (0,1)
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