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 (Worried)

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- Mar 8th 2008, 06:24 AM-sonal-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 (Worried)

- Mar 8th 2008, 06:45 AMKrizalid
$\displaystyle \int_0^1 {\frac{1}

{{1 + \sqrt x }}\,dx} .$

By setting $\displaystyle u = 1 + \sqrt x \implies du = \frac{1}

{{2\sqrt x }}\,dx\,\therefore \,du = \frac{1}

{{2(u - 1)}}\,dx,$ hence

$\displaystyle 2\int_1^2 {\frac{{u - 1}}

{u}\,du} ,$ now this is routine.