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Evaluate the following repeated integral:

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\\[ I = \\int_0^1 \\; \\mathrm{d}x \\; \\int_0^{\\simplify[all]{x^{m-1}}}\\var{m} \\var{fun} \\; \\mathrm{d}y \\]

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Calculate a repeated integral of the form $\\displaystyle I=\\int_0^1\\;dx\\;\\int_0^{x^{m-1}}mf(x^m+a)dy$

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The $y$ integral is trivial, and the $x$ integral is of the form $g'(x)f'(g(x))$, so it straightforwardly integrates to $f(g(x))$.

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$I=\\;$[[0]]

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Input your answer to 3 decimal places.

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We want to find

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\\[ I=\\int_0^1 \\; dx \\; \\int_0^{\\simplify[all]{x^{m-1}}} \\var{m} \\var{fun} \\; \\mathrm{d}y \\]

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The innermost integral gives:

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\\[ \\int_0^{\\simplify[all]{x^{m-1}}}\\var{m} \\var{fun} \\; \\mathrm{d}y = \\left[\\var{m}y \\; \\var{fun} \\right]_0^{\\simplify[all]{x^{m-1}}}=\\simplify[all]{{m}x^{m-1}} \\var{fun} \\]

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So we have to find  $\\displaystyle I=\\int_0^1\\simplify[all]{{m}x^{m-1}} \\var{fun} \\; \\mathrm{d}x$.

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Note that if we use the substitution $u=\\simplify[all]{x^{m}+{a}}$ then it is easy to find this last definite integral and we find that:

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$I=\\var{ans}$ to 3 decimal places.

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