The value of intlimits_{frac{1}{2}}^2 frac{1} | vedclass

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

Question

(A) Let $I = \int\limits_{\frac{1}{2}}^2 \frac{1}{x} \sin \left( x - \frac{1}{x} \right) dx$.Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x)

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let $I = \int\limits_{\frac{1}{2}}^2 \frac{1}{x} \sin \left( x - \frac{1}{x} \right) dx$.Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,where $a = \frac{1}{2}$ and $b = 2$,we have $a+b = \frac{5}{2}$.Substituting $x = \frac{1}{t}$,we get $dx = -\frac{1}{t^2} dt$.When $x = \frac{1}{2}$,$t = 2$. When $x = 2$,$t = \frac{1}{2}$.$I = \int\limits_2^{\frac{1}{2}} t \sin \left( \frac{1}{t} - t \right) \left( -\frac{1}{t^2} \right) dt$$I = \int\limits_2^{\frac{1}{2}} \frac{1}{t} \sin \left( t - \frac{1}{t} \right) dt$$I = -\int\limits_{\frac{1}{2}}^2 \frac{1}{t} \sin \left( t - \frac{1}{t} \right) dt$$I = -I$Therefore,$2I = 0$,which implies $I = 0$.

The value of $\int\limits_{\frac{1}{2}}^2 \frac{1}{x} \sin \left( x - \frac{1}{x} \right) dx$ is equal to

Similar Questions

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x$.

$\int_0^\pi \frac{x \sin x}{\sin ^2 x+2 \cos ^2 x} d x=$

$\int_0^\pi x f(\sin x) dx = $

$\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos x} = \dots$

$\int_{-1}^{1} \log \left( \frac{1+x}{1-x} \right) \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module