integration Q

**-sonal-** · March 8th 2008, 06:24 AM

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

**Krizalid** · March 8th 2008, 06:45 AM

$\int_0^1 {\frac{1}<br /> {{1 + \sqrt x }}\,dx} .$

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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