Let f(x) = int_{0}^{x} e^{x+t} dt. Then the a | vedclass

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(C) Given $f(x) = \int_{0}^{x} e^{x+t} dt$. We can rewrite the integral as $f(x) = e^x \int_{0}^{x} e^t dt$. Evaluating the integral,we get $f(x) = e^

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$-\ln\ 2$

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(C) Given $f(x) = \int_{0}^{x} e^{x+t} dt$. We can rewrite the integral as $f(x) = e^x \int_{0}^{x} e^t dt$. Evaluating the integral,we get $f(x) = e^x [e^t]_{0}^{x} = e^x (e^x - 1) = e^{2x} - e^x$. For the tangent to be parallel to the $x$-axis,the derivative $f'(x)$ must be equal to $0$. $f'(x) = \frac{d}{dx} (e^{2x} - e^x) = 2e^{2x} - e^x$. Setting $f'(x) = 0$,we have $2e^{2x} - e^x = 0$. $e^x (2e^x - 1) = 0$. Since $e^x$ is never $0$,we must have $2e^x - 1 = 0$,which implies $e^x = \frac{1}{2}$. Taking the natural logarithm on both sides,$x = \ln(\frac{1}{2}) = \ln(1) - \ln(2) = -\ln 2$.

Let $f(x) = \int_{0}^{x} e^{x+t} dt$. Then the abscissa of the point where the tangent to $f(x)$ is parallel to the $x$-axis is:

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