If [x] denotes the greatest integer less than | vedclass

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

Question

(D) Let $I = \int_{-\pi/2}^{\pi/2} [[x] - \sin x] \, dx$. Using the property $[a - b] = [a] + [-b]$ is not generally true,but we can use the property

Vedclass · Accepted Answer

$-\pi$

Vedclass · Answer

(D) Let $I = \int_{-\pi/2}^{\pi/2} [[x] - \sin x] \, dx$. Using the property $[a - b] = [a] + [-b]$ is not generally true,but we can use the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$. Here $a+b = -\pi/2 + \pi/2 = 0$,so $I = \int_{-\pi/2}^{\pi/2} [[-x] - \sin(-x)] \, dx = \int_{-\pi/2}^{\pi/2} [[-x] + \sin x] \, dx$. Adding the two expressions for $I$: $2I = \int_{-\pi/2}^{\pi/2} ([[x] - \sin x] + [[-x] + \sin x]) \, dx$. Using the property $[y] + [-y] = -1$ if $y 
otin \mathbb{Z}$ and $0$ if $y \in \mathbb{Z}$. For $x \in [-\pi/2, \pi/2]$,$[x] + [-x] = -1$ except at $x=0$ where it is $0$. Similarly,$[\sin x] + [-\sin x] = -1$ except at $x=0$ where it is $0$. Thus,the integrand is $-2$ almost everywhere. $2I = \int_{-\pi/2}^{\pi/2} (-2) \, dx = -2 [x]_{-\pi/2}^{\pi/2} = -2(\pi/2 - (-\pi/2)) = -2(\pi) = -2\pi$. Therefore,$I = -\pi$.

If $[x]$ denotes the greatest integer less than or equal to $x$,then the value of the integral $\int_{-\pi / 2}^{\pi / 2} [[x] - \sin x] \, dx$ is equal to:

Similar Questions

Let $f: [2, 5] \to [2, 5]$ be a bijective function such that $\frac{d}{dx}(f^{-1}(x)) > 0$ for all $x \in [2, 5]$. Then $\int_{2}^{5} (f(x) + f^{-1}(x)) dx$ is

The value of the integral $\int_{-\pi/4}^{\pi/4} \left( \frac{32 \cos^4 x}{1 + e^{\sin x}} \right) dx$ is:

The value of $\int_{0}^{2\pi} [\sin 2x(1 + \cos 3x)] \,dx$,where $[t]$ denotes the greatest integer function,is

Evaluate the definite integral: $\int_0^\pi \frac{x \cos^2 x}{1+\sin x} dx$

$ \int_{-2}^{2} |x \cos \pi x| \, dx $ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module