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$\int_2^8 \frac{5^{\sqrt{10-x}}}{5^{\sqrt{x}}+5^{\sqrt{10-x}}} d x$ is equals to
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The correct answer is:
$3$
Let $I=\int_2^8 \frac{5^{\sqrt{10-x}}}{5^{\sqrt{x}}+5^{\sqrt{10-x}}} d x$
Using property $\int_0^a f(x) d x=\int_0^a f(0+a-x) d x$
$\begin{aligned} & I=\int_2^8 5^{\sqrt{10-(2+8-x)}} \\ & I=\int_2^8 \frac{5^{\sqrt{10-10+x}}}{5^{\sqrt{10-x}}+5^{\sqrt{10-10+x}}} \\ & I=\int_2^8 \frac{5^{\sqrt{x}}}{5^{\sqrt{10-x}}+5^{\sqrt{x}}}\end{aligned}$
Adding Eqs. (i) and (ii), we get
$\begin{aligned} & 2 I=\int_2^8 \frac{5^{\sqrt{10-x}}+5^{\sqrt{x}}}{5^{\sqrt{x}}+5^{\sqrt{10-x}}} d x \\ & I I=\int_2^8 d x \\ & I=\frac{1}{2}[x]_2^8=\frac{1}{2} \times 6=3\end{aligned}$
Using property $\int_0^a f(x) d x=\int_0^a f(0+a-x) d x$
$\begin{aligned} & I=\int_2^8 5^{\sqrt{10-(2+8-x)}} \\ & I=\int_2^8 \frac{5^{\sqrt{10-10+x}}}{5^{\sqrt{10-x}}+5^{\sqrt{10-10+x}}} \\ & I=\int_2^8 \frac{5^{\sqrt{x}}}{5^{\sqrt{10-x}}+5^{\sqrt{x}}}\end{aligned}$
Adding Eqs. (i) and (ii), we get
$\begin{aligned} & 2 I=\int_2^8 \frac{5^{\sqrt{10-x}}+5^{\sqrt{x}}}{5^{\sqrt{x}}+5^{\sqrt{10-x}}} d x \\ & I I=\int_2^8 d x \\ & I=\frac{1}{2}[x]_2^8=\frac{1}{2} \times 6=3\end{aligned}$
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