intlimits_{0}^{5} cos left(pileft(x-left[frac | vedclass

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

Question

(D) Let $I = \int\limits_{0}^{5} \cos \left(\pi x - \pi \left[\frac{x}{2}\right]\right) d x$.We split the integral based on the intervals where $\left

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let $I = \int\limits_{0}^{5} \cos \left(\pi x - \pi \left[\frac{x}{2}\right]\right) d x$.We split the integral based on the intervals where $\left[\frac{x}{2}\right]$ is constant:For $x \in [0, 2)$,$\left[\frac{x}{2}\right] = 0$.For $x \in [2, 4)$,$\left[\frac{x}{2}\right] = 1$.For $x \in [4, 5]$,$\left[\frac{x}{2}\right] = 2$.Thus,$I = \int\limits_{0}^{2} \cos(\pi x) d x + \int\limits_{2}^{4} \cos(\pi x - \pi) d x + \int\limits_{4}^{5} \cos(\pi x - 2\pi) d x$.Evaluating each integral:$\int\limits_{0}^{2} \cos(\pi x) d x = \left[\frac{\sin(\pi x)}{\pi}\right]_{0}^{2} = \frac{\sin(2\pi) - \sin(0)}{\pi} = 0$.$\int\limits_{2}^{4} \cos(\pi x - \pi) d x = \left[\frac{\sin(\pi x - \pi)}{\pi}\right]_{2}^{4} = \frac{\sin(3\pi) - \sin(\pi)}{\pi} = 0$.$\int\limits_{4}^{5} \cos(\pi x - 2\pi) d x = \left[\frac{\sin(\pi x - 2\pi)}{\pi}\right]_{4}^{5} = \frac{\sin(3\pi) - \sin(2\pi)}{\pi} = 0$.Therefore,$I = 0 + 0 + 0 = 0$.

$\int\limits_{0}^{5} \cos \left(\pi\left(x-\left[\frac{x}{2}\right]\right)\right) d x$,where $[t]$ denotes the greatest integer less than or equal to $t$,is equal to:

Similar Questions

If $\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$, then $b$ is equal to

$\int_0^1 {(1 - x)^9} dx = $

If $\frac{d}{dx}\{f(x)\} = g(x)$, then $\int_a^b f(x) g(x) dx$ is equal to

Let $a, b$ and $c$ be positive constants. The value of $a$ in terms of $c$ if the value of the integral $\int_{0}^{1} (acx^{b+1} + a^3bx^{3b+5}) \, dx$ is independent of $b$ is:

If $I_n = \int_0^{\pi/4} \tan^n \theta \, d\theta$ for $n = 1, 2, 3, \ldots$,then $I_{n-1} + I_{n+1}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module