Question
Evaluate the following integrals:
$\int\limits^{8}_2\frac{\sqrt{10-\text{x}}}{\sqrt{\text{x}}+\sqrt{10-\text{x}}}\text{ dx}$
$\int\limits^{8}_2\frac{\sqrt{10-\text{x}}}{\sqrt{\text{x}}+\sqrt{10-\text{x}}}\text{ dx}$
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Answer
Let $\text{I}=\int\limits^{8}_2\frac{\sqrt{10-\text{x}}}{\sqrt{\text{x}}+\sqrt{10-\text{x}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{8}_2\frac{\sqrt{10-(2+8-\text{x}})}{\sqrt{2+8-\text{x}}+\sqrt{10-(2+8-\text{x}})}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{8}_2\frac{\sqrt{\text{x}}}{\sqrt{10-\text{x}}+\sqrt{\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{8}_2\frac{\sqrt{10-\text{x}}}{\sqrt{\text{x}}+\sqrt{10-\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{8}_2\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^8_2$
$\Rightarrow2\text{I}=8-2=6$
$\Rightarrow\text{I}=3$
Then,
$\text{I}=\int\limits^{8}_2\frac{\sqrt{10-(2+8-\text{x}})}{\sqrt{2+8-\text{x}}+\sqrt{10-(2+8-\text{x}})}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{8}_2\frac{\sqrt{\text{x}}}{\sqrt{10-\text{x}}+\sqrt{\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{8}_2\frac{\sqrt{10-\text{x}}}{\sqrt{\text{x}}+\sqrt{10-\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{8}_2\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^8_2$
$\Rightarrow2\text{I}=8-2=6$
$\Rightarrow\text{I}=3$
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