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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS3 Marks
Question
By Using properties of definite integral, evaluate the following integral in Exercise:
$\int^{\text{a}}_{0}\frac{\sqrt{\text{x}}}{\sqrt{\text{x}+\sqrt{\text{a}-\text{x}}}}\text{dx}$
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Answer

$\text{Let}\ \text{I}= \int^{\text{a}}\limits_{0}\frac{\sqrt{\text{x}}}{\sqrt{\text{x}+\sqrt{\text{a}-\text{x}}}}\text{dx}\ \ ...(\text{i})$

$\Rightarrow\ \ \ \text{I}=\int^{\text{a}}\limits_{0}\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{a}-\text{x}}+\sqrt{\text{a}-\text{(a}-\text{x)}}}\text{dx}=\int^{\text{a}}\limits_{0}\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{a}-\text{x}}+\sqrt{\text{x}}}\text{dx}\ \ ...(\text{ii})$

Adding eq.(i) and (ii),

$ 21=\int^{\text{a}}\limits_{0}\bigg(\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}+\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{a}-\text{x}}+\sqrt{\text{x}}}\bigg)\text{dx}=\int^{\text{a}}\limits_{0}\bigg(\frac{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}\bigg)\text{dx}=\int^{\text{a}}\limits_{0}1\text{dx}=\text{(x)}^{\text{a}}_{0}=\text{a}$

$\Rightarrow\ \ \text{I}=\frac{\text{a}}{2}$

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