The area of the region bounded by the curve y | vedclass

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(C) The area $A$ is given by the integral $\int_{0}^{2} 2^{kx} \, dx = \frac{2}{\log_e 2}$.Using the formula $\int a^x \, dx = \frac{a^x}{\log_e a}$,

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$\frac{1}{2}$

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(C) The area $A$ is given by the integral $\int_{0}^{2} 2^{kx} \, dx = \frac{2}{\log_e 2}$.Using the formula $\int a^x \, dx = \frac{a^x}{\log_e a}$, we get $\int_{0}^{2} 2^{kx} \, dx = \left[ \frac{2^{kx}}{k \log_e 2} ight]_{0}^{2}$.Substituting the limits: $\frac{1}{k \log_e 2} (2^{2k} - 2^0) = \frac{2}{\log_e 2}$.Canceling $\log_e 2$ from both sides: $\frac{2^{2k} - 1}{k} = 2$.$2^{2k} - 1 = 2k$.By inspection, if $k = 1$, $2^2 - 1 = 3 
eq 2(1)$.If $k = 1/2$, $2^{2(1/2)} - 1 = 2^1 - 1 = 1$, and $2k = 2(1/2) = 1$.Since both sides are equal, $k = 1/2$ is the solution.

The area of the region bounded by the curve $y = 2^{kx}$ and the lines $x = 0$ and $x = 2$ in the first quadrant is $\frac{2}{\log_e 2}$. Find the value of $k$.

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