int_{-2}^2 |[x]| , dx is equal to | vedclass

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

Question

(D) We have $\int_{-2}^2 |[x]| \, dx = \int_{-2}^{-1} |[x]| \, dx + \int_{-1}^0 |[x]| \, dx + \int_0^1 |[x]| \, dx + \int_1^2 |[x]| \, dx$. Since $[x]

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) We have $\int_{-2}^2 |[x]| \, dx = \int_{-2}^{-1} |[x]| \, dx + \int_{-1}^0 |[x]| \, dx + \int_0^1 |[x]| \, dx + \int_1^2 |[x]| \, dx$. Since $[x]$ is the greatest integer function: For $x \in [-2, -1)$,$[x] = -2$,so $|[x]| = |-2| = 2$. For $x \in [-1, 0)$,$[x] = -1$,so $|[x]| = |-1| = 1$. For $x \in [0, 1)$,$[x] = 0$,so $|[x]| = |0| = 0$. For $x \in [1, 2)$,$[x] = 1$,so $|[x]| = |1| = 1$. Substituting these values: $\int_{-2}^2 |[x]| \, dx = \int_{-2}^{-1} 2 \, dx + \int_{-1}^0 1 \, dx + \int_0^1 0 \, dx + \int_1^2 1 \, dx$. $= 2[x]_{-2}^{-1} + [x]_{-1}^0 + 0 + [x]_1^2$. $= 2(-1 - (-2)) + (0 - (-1)) + (2 - 1)$. $= 2(1) + 1 + 1 = 4$.

$\int_{-2}^2 |[x]| \, dx$ is equal to

Similar Questions

The $P-V$ diagram of $2$ g of helium gas for a certain process $A \to B$ is shown in the figure. What is the heat given to the gas during the process $A \to B$ (in ${P_0}{V_0}$)?

The horizontal acceleration that should be given to a smooth inclined plane of angle $\theta = \sin^{-1}\left(\frac{1}{l}\right)$ to keep an object stationary on the plane, relative to the inclined plane is:

$A$ fish looking up through the water sees the outside world,contained in a circular horizon. If the refractive index of water is $\frac{4}{3}$ and the fish is $12 \ cm$ below the water surface,the radius of this circle in $cm$ is

The decomposition of $H_2O_2$ can be prevented by the addition of which of the following?

The solution set of the equation $x^{\log_x(1-x)^2} = 9$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module