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Math Help - integration Q

  1. #1
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    integration Q

    ind the area under the curve Y = 1 / 1 + (X)^0.5 between x = 0 and x = 1 using the substitution u = 1 + (x)^0.5
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  2. #2
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    \int_0^1 {\frac{1}<br />
{{1 + \sqrt x }}\,dx} .

    By setting u = 1 + \sqrt x  \implies du = \frac{1}<br />
{{2\sqrt x }}\,dx\,\therefore \,du = \frac{1}<br />
{{2(u - 1)}}\,dx, hence

    2\int_1^2 {\frac{{u - 1}}<br />
{u}\,du} , now this is routine.
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