If [x] is the greatest integer function not g | vedclass

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

Question

(A) Let $I = \int_0^8 [x] dx$. Since $[x]$ is the greatest integer function,it takes constant integer values on intervals $[n, n+1)$. We can split the

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(A) Let $I = \int_0^8 [x] dx$. Since $[x]$ is the greatest integer function,it takes constant integer values on intervals $[n, n+1)$. We can split the integral as: $I = \int_0^1 0 dx + \int_1^2 1 dx + \int_2^3 2 dx + \int_3^4 3 dx + \int_4^5 4 dx + \int_5^6 5 dx + \int_6^7 6 dx + \int_7^8 7 dx$. Evaluating each integral: $I = 0 + 1(2-1) + 2(3-2) + 3(4-3) + 4(5-4) + 5(6-5) + 6(7-6) + 7(8-7)$. $I = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7$. $I = \frac{7(7+1)}{2} = \frac{7 	imes 8}{2} = 28$.

If $[x]$ is the greatest integer function not greater than $x$,then $\int_0^8 [x] dx$ is equal to

Similar Questions

The value of $\int_{\pi}^{2\pi} [2\sin x] \, dx$,where $[\cdot]$ represents the greatest integer function,is

If $I = \int_0^{100\pi} \sqrt{1 - \cos 2x} \, dx$,then the value of $I$ is

If $f(x) = \begin{cases} e^{\cos x} \sin x, & \text{for } |x| \leq 2 \\ 2, & \text{otherwise} \end{cases}$,then $\int_{-2}^{3} f(x) dx$ is equal to

The value of the integral $\int_0^2 \frac{\sqrt{x}(x^2 + x + 1)}{(\sqrt{x}+1)(\sqrt{x^4+x^2+1})} dx$ is equal to:

The value of the definite integral $\int_0^1 \frac{dx}{x^2 + 2x\cos \alpha + 1}$ for $0 < \alpha < \pi$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module