int_{e^{-1}}^{e^2} left| frac{log_e x}{x} rig | vedclass

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

Question

(B) माना $I = \int_{e^{-1}}^{e^2} \left| \frac{\log_e x}{x} \right| dx$. चूँकि $x \in [e^{-1}, 1)$ के लिए $\log_e x < 0$ और $x \in [1, e^2]$ के लिए $\

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

(B) माना $I = \int_{e^{-1}}^{e^2} \left| \frac{\log_e x}{x} \right| dx$. चूँकि $x \in [e^{-1}, 1)$ के लिए $\log_e x $I = \int_{e^{-1}}^{1} -\frac{\log_e x}{x} dx + \int_{1}^{e^2} \frac{\log_e x}{x} dx$. माना $z = \log_e x$,तब $dz = \frac{1}{x} dx$. जब $x = e^{-1}, z = -1$. जब $x = 1, z = 0$. जब $x = e^2, z = 2$. $I = \int_{-1}^{0} -z dz + \int_{0}^{2} z dz$. $I = \left[ -\frac{z^2}{2} \right]_{-1}^{0} + \left[ \frac{z^2}{2} \right]_{0}^{2}$. $I = (0 - (-1/2)) + (4/2 - 0) = \frac{1}{2} + 2 = \frac{5}{2}$.

$\int_{e^{-1}}^{e^2} \left| \frac{\log_e x}{x} \right| dx$ का मान है

Similar Questions

$\int_0^{\frac{\pi}{2}} \frac{x \tan x \sec^2 x}{\tan^4 x + 1} dx =$

$\int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\cos \theta }}\,d\theta = } $

$\int_{-2}^2(x \cos x+\sin x+1) d x$ का मान है

$\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$ का मान ज्ञात कीजिए।

Vedclass Test Series

Exam Paper Generator

Online Exam Module