int_{e^{-1}}^{e^2} left| frac{log x}{x} right | vedclass

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

Question

(D) Let $I = \int_{e^{-1}}^{e^2} \left| \frac{\log x}{x} \right| dx$. Since $\log x < 0$ for $x \in [e^{-1}, 1)$ and $\log x \ge 0$ for $x \in [1, e^2

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

(D) Let $I = \int_{e^{-1}}^{e^2} \left| \frac{\log x}{x} \right| dx$. Since $\log x $I = \int_{e^{-1}}^{1} \left( -\frac{\log x}{x} \right) dx + \int_{1}^{e^2} \left( \frac{\log x}{x} \right) dx$. Substitute $t = \log x$,so $dt = \frac{dx}{x}$. When $x = e^{-1}, t = -1$. When $x = 1, t = 0$. When $x = e^2, t = 2$. $I = -\int_{-1}^{0} t dt + \int_{0}^{2} t dt$. $I = -\left[ \frac{t^2}{2} \right]_{-1}^{0} + \left[ \frac{t^2}{2} \right]_{0}^{2}$. $I = -\left( 0 - \frac{(-1)^2}{2} \right) + \left( \frac{2^2}{2} - 0 \right)$. $I = -\left( -\frac{1}{2} \right) + \frac{4}{2} = \frac{1}{2} + 2 = \frac{5}{2}$. Thus,the correct option is $D$.

$\int_{e^{-1}}^{e^2} \left| \frac{\log x}{x} \right| dx =$

Similar Questions

The value of the integral $\int_{-1/2}^{1/2} \left[ \left( \frac{x+1}{x-1} \right)^2 + \left( \frac{x-1}{x+1} \right)^2 - 2 \right]^{1/2} dx$ is

If $\theta_{1}$ and $\theta_{2}$ are respectively the smallest and the largest values of $\theta$ in $(0, 2\pi) - \{\pi\}$ which satisfy the equation $2 \cot^{2} \theta - \frac{5}{\sin \theta} + 4 = 0$,then $\int_{\theta_{1}}^{\theta_{2}} \cos^{2} 3\theta \, d\theta$ is equal to

$\int_0^1 \frac{e^{-x}}{1 + e^{-x}} \,dx = $

$\int_0^3 |x^2 - 3x + 2| dx = $

If $\int_0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2\right)^{\frac{3}{2}}} \,d x=\frac{k}{6}$, then the value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module