int_0^1(sqrt{10})^{2x} dx= | vedclass

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(B) Let $I = \int_0^1 (\sqrt{10})^{2x} dx$. Since $(\sqrt{10})^{2x} = (10^{1/2})^{2x} = 10^x$,the integral becomes: $I = \int_0^1 10^x dx$. Using the

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$\frac{9}{\log 10}$

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(B) Let $I = \int_0^1 (\sqrt{10})^{2x} dx$. Since $(\sqrt{10})^{2x} = (10^{1/2})^{2x} = 10^x$,the integral becomes: $I = \int_0^1 10^x dx$. Using the standard integral formula $\int a^x dx = \frac{a^x}{\ln a} + C$,we get: $I = \left[ \frac{10^x}{\ln 10} \right]_0^1$. Evaluating at the limits: $I = \frac{10^1}{\ln 10} - \frac{10^0}{\ln 10} = \frac{10}{\ln 10} - \frac{1}{\ln 10} = \frac{9}{\ln 10}$.

$\int_0^1(\sqrt{10})^{2x} dx=$

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