If I_1 = intlimits_0^1 {{e^{ - x}}} {cos ^2}x | vedclass

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

Question

(D) For $x \in (0, 1)$,we have $x^3  -x^2 > -x$. Since the exponential function $f(t) = e^t$ is strictly

Vedclass · Accepted Answer

$I_3 > I_2 > I_1$

Vedclass · Answer

(D) For $x \in (0, 1)$,we have $x^3 Multiplying by $-1$,we get $-x^3 > -x^2 > -x$. Since the exponential function $f(t) = e^t$ is strictly increasing,we have $e^{-x^3} > e^{-x^2} > e^{-x}$. Now,consider the integrands: For $I_3$,the integrand is $f_3(x) = e^{-x^3}$. For $I_2$,the integrand is $f_2(x) = e^{-x^2} \cos^2 x$. Since $0 \le \cos^2 x \le 1$,we have $e^{-x^2} \cos^2 x \le e^{-x^2}$. For $I_1$,the integrand is $f_1(x) = e^{-x} \cos^2 x$. Since $0 \le \cos^2 x \le 1$,we have $e^{-x} \cos^2 x \le e^{-x}$. Comparing the functions on $(0, 1)$: $e^{-x^3} > e^{-x^2} \ge e^{-x^2} \cos^2 x$ implies $I_3 > I_2$. Also,$e^{-x^2} \cos^2 x > e^{-x} \cos^2 x$ because $e^{-x^2} > e^{-x}$ for $x \in (0, 1)$. Thus,$I_3 > I_2 > I_1$.

If $I_1 = \int\limits_0^1 {{e^{ - x}}} {\cos ^2}x\,dx$,$I_2 = \int\limits_0^1 {{e^{ - {x^2}}}} {\cos ^2}x\,dx$ and $I_3 = \int\limits_0^1 {{e^{ - {x^3}}}} dx$; then

Similar Questions

$\int_{0}^{\frac{\pi}{2}} \log \left[\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}\right] dx =$

$\int_0^\pi x \sin^3 x \, dx = $

Let the domain of the function $f(x) = \log_{4}(\log_{5}(\log_{3}(18x - x^{2} - 77)))$ be $(a, b)$. Then the value of the integral $\int_{a}^{b} \frac{\sin^{3} x}{\sin^{3} x + \sin^{3}(a + b - x)} dx$ is equal to $.....$

The value of $\int_{0}^{2 \pi} \frac{x \sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$ is equal to

The value of $\int_{-\pi}^{\pi} \frac{\cos^2 x}{1+\alpha^x} \, dx$ for $\alpha > 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module