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Evaluate the following definite integral:\(\int_1^2 \frac {\sqrt x}{\sqrt{3-x}+ \sqrt x}\) dx |
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Answer» Let I = \(\int_1^2 \frac {\sqrt x}{\sqrt{3-x}+ \sqrt x}\) ....(1) ∴ I = \(\int_1^2 \frac {\sqrt{(1+2-x)}}{\sqrt{3-(1+2-x)} + \sqrt{1+2-x}}dx\) = \(\int_1^2 \frac {\sqrt{3-x}}{\sqrt x + \sqrt{3-x}}dx\) = \(\int_1^2 dx = (x)_1^2\) = 2-1 = 1 ∴ I = 1/2 = \(\int_1^2 \frac {\sqrt x}{\sqrt{3-x}+\sqrt x}dx = \frac 12\) |
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