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

$A$ closed vessel tapers to a point both at its top $E$ and its bottom $F$ and is fixed with $EF$ vertical. When the depth of the liquid in it is $x \, \text{cm}$,the volume of the liquid in it is $V(x) = x^2 (15 - x) \, \text{cu. cm}$. The length $EF$ is ........ $\text{cm}$.

The volume of the largest possible right circular cylinder that can be inscribed in a sphere of radius $R = \sqrt{3}$ is

If $f(x) = 2x^6 - 3$,then:

Let $f(x)=3 \sqrt{x-2}+\sqrt{4-x}$ be a real-valued function. If $\alpha$ and $\beta$ are respectively the minimum and the maximum values of $f$,then $\alpha^2+2 \beta^2$ is equal to

If $f(x) = 7e^{\sin^2 x} - 7e^{\cos^2 x} + 2$,then $\sqrt{7f_{\min} + f_{\max}}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module