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$\int_0^1(\sqrt{10})^{2 x} d x=$
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Verified Answer
The correct answer is:
$\frac{9}{\log 10}$
Let $I=\int_0^1(\sqrt{10})^{2 x} d x$
$\begin{aligned} & =\int_0^1 10^{\frac{2 x}{2}} d x=\int_0^1 10^x d x=\left(\frac{10^x}{\log 10}\right)_0^1 \\ & =\frac{10}{\log 10}-\frac{1}{\log 10}=\frac{9}{\log 10} .\end{aligned}$
$\begin{aligned} & =\int_0^1 10^{\frac{2 x}{2}} d x=\int_0^1 10^x d x=\left(\frac{10^x}{\log 10}\right)_0^1 \\ & =\frac{10}{\log 10}-\frac{1}{\log 10}=\frac{9}{\log 10} .\end{aligned}$
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