The value of the integral int_{-1}^{1} left{ | vedclass

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

Question

(D) Let $I = \int_{-1}^{1} \left\{ \frac{x^{2013}}{e^{|x|}(x^{2}+\cos x)} + \frac{1}{e^{|x|}} \right\} dx$.We can split the integral into two parts:$I

Vedclass · Accepted Answer

$2(1-e^{-1})$

Vedclass · Answer

(D) Let $I = \int_{-1}^{1} \left\{ \frac{x^{2013}}{e^{|x|}(x^{2}+\cos x)} + \frac{1}{e^{|x|}} \right\} dx$.We can split the integral into two parts:$I = \int_{-1}^{1} \frac{x^{2013}}{e^{|x|}(x^{2}+\cos x)} dx + \int_{-1}^{1} \frac{1}{e^{|x|}} dx$.Let $f(x) = \frac{x^{2013}}{e^{|x|}(x^{2}+\cos x)}$. Since $f(-x) = \frac{(-x)^{2013}}{e^{|-x|}((-x)^{2}+\cos(-x))} = \frac{-x^{2013}}{e^{|x|}(x^{2}+\cos x)} = -f(x)$,the function $f(x)$ is an odd function. Therefore,$\int_{-1}^{1} f(x) dx = 0$.Let $g(x) = \frac{1}{e^{|x|}} = e^{-|x|}$. Since $g(-x) = e^{-|-x|} = e^{-|x|} = g(x)$,the function $g(x)$ is an even function. Therefore,$\int_{-1}^{1} g(x) dx = 2 \int_{0}^{1} e^{-x} dx$.Calculating the integral: $2 \int_{0}^{1} e^{-x} dx = 2 [-e^{-x}]_{0}^{1} = 2 (-e^{-1} - (-e^{0})) = 2(1 - e^{-1})$.Thus,$I = 0 + 2(1 - e^{-1}) = 2(1 - e^{-1})$.

The value of the integral $\int_{-1}^{1} \left\{ \frac{x^{2013}}{e^{|x|}(x^{2}+\cos x)} + \frac{1}{e^{|x|}} \right\} dx$ is equal to

Similar Questions

Let $[t]$ denote the greatest integer $\leq t$. Then $\frac{2}{\pi} \int_{\pi/6}^{5\pi/6} (8[\operatorname{cosec} x] - 5[\cot x]) \, dx$ is equal to

If $f(x) = \int_1^x \frac{1}{2+t^4} dt$,then

For $x > 0,$ if $f(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt,$ then $f(e) + f\left(\frac{1}{e}\right)$ is equal to:

$\int_{-5}^{5} \log \left(\frac{7-x}{7+x}\right) dx =$

Let $u = \int\limits_0^1 {\frac{{\ln (x + 1)}}{{{x^2} + 1}}} \,dx$ and $v = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin 2x)} \,dx$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module