The value of int_0^2 frac{3^{sqrt{x}}}{sqrt{x | vedclass

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(A) Let $I = \int_0^2 \frac{3^{\sqrt{x}}}{\sqrt{x}} \, dx$. Substitute $\sqrt{x} = t$. Then,$\frac{1}{2\sqrt{x}} \, dx = dt$,which implies $\frac{1}{\

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$\frac{2}{\log 3}(3^{\sqrt{2}} - 1)$

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(A) Let $I = \int_0^2 \frac{3^{\sqrt{x}}}{\sqrt{x}} \, dx$. Substitute $\sqrt{x} = t$. Then,$\frac{1}{2\sqrt{x}} \, dx = dt$,which implies $\frac{1}{\sqrt{x}} \, dx = 2 \, dt$. When $x = 0$,$t = 0$. When $x = 2$,$t = \sqrt{2}$. Substituting these into the integral,we get: $I = \int_0^{\sqrt{2}} 3^t \cdot 2 \, dt$ $I = 2 \int_0^{\sqrt{2}} 3^t \, dt$ Using the formula $\int a^x \, dx = \frac{a^x}{\log a} + C$,we have: $I = 2 \left[ \frac{3^t}{\log 3} \right]_0^{\sqrt{2}}$ $I = \frac{2}{\log 3} (3^{\sqrt{2}} - 3^0)$ $I = \frac{2}{\log 3} (3^{\sqrt{2}} - 1)$.

The value of $\int_0^2 \frac{3^{\sqrt{x}}}{\sqrt{x}} \, dx$ is

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