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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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^

Vedclass · Accepted Answer

$-\ln\ 2$

Vedclass · Answer

(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:

Similar Questions

The number of distinct real roots of the equation $3x^{4} + 4x^{3} - 12x^{2} + 4 = 0$ is ..... .

Let $f(x) = \int_0^{x^2} \frac{t^2-8t+15}{e^t} dt$,$x \in R$. Then the numbers of local maximum and local minimum points of $f$,respectively,are:

The minimum value of the function $f(x) = 2x^2 - \ln|x|$ for $x \geq 1$ is:

If $x+y=6, x \geqslant 0, y \geqslant 0$,then the maximum value of $x^2 y$ is

$A$ right circular cone is inscribed in a sphere of radius $3$ units. If the volume of the cone is maximum, then the semi-vertical angle of the cone is

Vedclass Test Series

Exam Paper Generator

Online Exam Module