int_{-5}^{5} left[ frac{e^{x} + e^{-x}}{e^{x} | vedclass

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

Question

(A) Let $f(x) = \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}}$.Then,$f(-x) = \frac{e^{-x} + e^{x}}{e^{-x} - e^{x}} = \frac{e^{-x} + e^{x}}{-(e^{x} - e^{-x})}

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let $f(x) = \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}}$.Then,$f(-x) = \frac{e^{-x} + e^{x}}{e^{-x} - e^{x}} = \frac{e^{-x} + e^{x}}{-(e^{x} - e^{-x})} = -\frac{e^{x} + e^{-x}}{e^{x} - e^{-x}} = -f(x)$.Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.For an odd function,the property of definite integrals states that $\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is continuous on $[-a, a]$.However,note that $f(x)$ has a discontinuity at $x = 0$ because the denominator $e^{x} - e^{-x} = 0$ at $x = 0$.Therefore,the integral $\int_{-5}^{5} \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}} dx$ is an improper integral.Evaluating the principal value: $\int_{-5}^{5} \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}} dx = \left[ \ln|e^{x} - e^{-x}| \right]_{-5}^{5} = \ln|e^{5} - e^{-5}| - \ln|e^{-5} - e^{5}| = 0$.

$\int_{-5}^{5} \left[ \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}} \right] dx = $

Similar Questions

$\int_0^{\infty} (x^{12} + x^{-12}) \frac{\log x}{x} dx =$

The value of $\int_0^\pi {{e^{{{\cos }^2}x}}{{\cos }^5}3x} \,dx$ is

$\int_{0}^{1} \frac{8 \log(1+x)}{1+x^{2}} dx = $

$\int_{0}^{\pi} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} = $

The value of $\int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+3^{x}} d x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module