Evaluate the following integrals : _2^5 x x+7-x d x - Vidyadip

Question

Evaluate the following integrals : $\int_2^5 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}} d x$

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Answer

Let $I=\int_2^5 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}} d x$
We use the property, $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$.
Hence in $I$, we change $x$ by $2+5-x$.
$
\begin{aligned}
\therefore I & =\int_2^5 \frac{\sqrt{2+5-x}}{\sqrt{2+5-x}+\sqrt{7-2-5+x}} d x \\
& =\int_2^5 \frac{\sqrt{7-x}}{\sqrt{7-x}+\sqrt{x}} d x
\end{aligned}
$
Adding (1) and (2), we get
$
\begin{array}{rl}
2 I & =\int_2^5 \frac{\sqrt{7-x}}{\sqrt{x}+\sqrt{7-x}} d x+\int_2^5 \frac{\sqrt{x}}{\sqrt{7-x}+\sqrt{x}} d x \\
& =\int_2^5 \frac{\sqrt{x}+\sqrt{7-x}}{\sqrt{x}+\sqrt{7-x}} d x \\
& =\int_2^5 1 d x=[x]_2^5=5-2=3 \\
\therefore & I=\frac{3}{2}
\end{array}
$
Hence, $\int_2^5 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}} d x=\frac{3}{2}$.

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