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Find:I = \(\int\limits_0^2\frac{\sqrt x}{\sqrt x+\sqrt{2-x}}dx\) |
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Answer» I = \(\int\limits_0^2\frac{\sqrt x}{\sqrt x+\sqrt{2-x}}dx\)----(1) \(\therefore\) I = \(\int\limits_0^2\frac{\sqrt {2-x}}{\sqrt {2-x}+\sqrt{2-(2-x)}}dx\) ⇒ I = \(\int\limits_0^2\frac{\sqrt {2-x}}{\sqrt x+\sqrt{2-x}}dx\) By adding equations (1) and (2) we get, 2I = \(\int\limits_0^2\frac{\sqrt x+\sqrt{2-x}}{\sqrt x+\sqrt{2-x}}dx\) = \(\int\limits_0^2dx\) = \((x)_0^2\) = 2 - 0 = 2 \(\therefore\) I = 2/2 = 1 ⇒ \(\int\limits_0^2\frac{\sqrt x}{\sqrt x+\sqrt{2-x}}dx\) = 1 |
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