Let int_{-2}^{2} (|sin x| + [x sin x]) dx = 2 | vedclass

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

Question

(B) Let $I = \int_{-2}^2 (|\sin x| + [x \sin x]) dx$. Since $|\sin x|$ is an even function,$\int_{-2}^2 |\sin x| dx = 2 \int_0^2 \sin x dx = 2 [-\cos

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let $I = \int_{-2}^2 (|\sin x| + [x \sin x]) dx$. Since $|\sin x|$ is an even function,$\int_{-2}^2 |\sin x| dx = 2 \int_0^2 \sin x dx = 2 [-\cos x]_0^2 = 2(1 - \cos 2)$. Now,consider $f(x) = x \sin x$. Since $f(x)$ is even,$\int_{-2}^2 [x \sin x] dx = 2 \int_0^2 [x \sin x] dx$. For $x \in [0, 2]$,$x \sin x$ increases from $0$ to $2 \sin 2 \approx 1.818$. Thus,$[x \sin x] = 0$ for $x \in [0, x_0)$ where $x_0 \sin x_0 = 1$,and $[x \sin x] = 1$ for $x \in [x_0, 2]$. So,$2 \int_0^2 [x \sin x] dx = 2 \int_{x_0}^2 1 dx = 2(2 - x_0) = 4 - 2x_0$. Thus,$I = 2(1 - \cos 2) + 4 - 2x_0 = 2 - 2\cos 2 + 4 - 2x_0 = 2(3 - \cos 2) - 2x_0$. Comparing this with $2(3 - \cos 2) + \beta$,we get $\beta = -2x_0$. Given $x_0 \sin x_0 = 1$,we find $x_0 \approx 1.11$. However,evaluating the expression $\beta \sin(\frac{\beta}{2})$ with $\beta = -2x_0$ where $x_0 \sin x_0 = 1$,we have $\beta = -2x_0$. Then $\beta \sin(\frac{\beta}{2}) = -2x_0 \sin(-x_0) = 2x_0 \sin x_0 = 2(1) = 2$.

Let $\int_{-2}^{2} (|\sin x| + [x \sin x]) dx = 2(3 - \cos 2) + \beta$,where $[\cdot]$ denotes the greatest integer function. Then $\beta \sin(\frac{\beta}{2})$ equals:

Similar Questions

The value of the integral $\frac{48}{\pi^{4}} \int_{0}^{\pi} \left(\frac{3 \pi x^{2}}{2} - x^{3}\right) \frac{\sin x}{1 + \cos^{2} x} dx$ is equal to

$\int_{-1}^1 \left(\sqrt{1+x+x^2}-\sqrt{1-x+x^2}\right) dx =$

If $[x]$ is the greatest integer $\leq x$,then the value of the integral $\int_{-0.9}^{0.9} \left( [x^2] + \log \left( \frac{2-x}{2+x} \right) \right) dx$ is

The value of the integral $I = \int_{0}^{1} x(1 - x)^n dx$ is

Let $T > 0$ be a fixed number. Suppose $f$ is a continuous function such that for all $x \in \mathbb{R}, f(x + T) = f(x)$. If $I = \int_{0}^{T} f(x) dx$,then the value of $\int_{3}^{3 + 3T} f(2x) dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module