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Question

(A) We need to evaluate the integral $I = \int a^x \, dx$. Recall the derivative of the exponential function $a^x$ with respect to $x$: $\frac{d}{dx}(

Vedclass · Accepted Answer

$\frac{a^x}{\log a} + c$

Vedclass · Answer

(A) We need to evaluate the integral $I = \int a^x \, dx$. Recall the derivative of the exponential function $a^x$ with respect to $x$: $\frac{d}{dx}(a^x) = a^x \log a$. Dividing both sides by $\log a$ (where $a > 0$ and $a 
eq 1$),we get: $\frac{d}{dx} \left( \frac{a^x}{\log a} ight) = a^x$. By the definition of the indefinite integral,if $\frac{d}{dx} F(x) = f(x)$,then $\int f(x) \, dx = F(x) + c$. Therefore,$\int a^x \, dx = \frac{a^x}{\log a} + c$.

$\int a^x \, dx = $

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