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$\int_{1}^{3} \frac{\sqrt{4-\mathrm{x}}}{\sqrt{\mathrm{x}}+\sqrt{4-\mathrm{x}}} \mathrm{dx}$ is equal to
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Verified Answer
The correct answer is:
1
Let $I=\int_{1}^{3} \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} d x
\quad \text{...(i)}$
$$
\Rightarrow \quad I=\int_{1}^{3} \frac{\sqrt{4-(4-x)}}{\sqrt{4-x}+\sqrt{4-(4-x)}} d x
$$
$$
\begin{gathered}
{\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]} \\
\Rightarrow \quad I=\int_{1}^{3} \frac{\sqrt{x}}{\sqrt{4-x}+\sqrt{x}} d x
\quad \text{...(ii)}
\end{gathered}
$$
On adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
2 \mathrm{I} &=\int_{1}^{3} 1 \mathrm{dx}=[\mathrm{x}]_{1}^{3} \\
\Rightarrow \quad \mathrm{I} &=\frac{2}{2}=1
\end{aligned}
$$
\quad \text{...(i)}$
$$
\Rightarrow \quad I=\int_{1}^{3} \frac{\sqrt{4-(4-x)}}{\sqrt{4-x}+\sqrt{4-(4-x)}} d x
$$
$$
\begin{gathered}
{\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]} \\
\Rightarrow \quad I=\int_{1}^{3} \frac{\sqrt{x}}{\sqrt{4-x}+\sqrt{x}} d x
\quad \text{...(ii)}
\end{gathered}
$$
On adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
2 \mathrm{I} &=\int_{1}^{3} 1 \mathrm{dx}=[\mathrm{x}]_{1}^{3} \\
\Rightarrow \quad \mathrm{I} &=\frac{2}{2}=1
\end{aligned}
$$
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