Let y = {x}^{[x]} where {x} denotes the fract | vedclass

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

Question

(D) We need to evaluate the integral $I = \int_{0}^{3} \{x\}^{[x]} \, dx$. Since the function involves $[x]$,we split the integral at integer points:

Vedclass · Accepted Answer

$11/6$

Vedclass · Answer

(D) We need to evaluate the integral $I = \int_{0}^{3} \{x\}^{[x]} \, dx$. Since the function involves $[x]$,we split the integral at integer points: $I = \int_{0}^{1} \{x\}^{[x]} \, dx + \int_{1}^{2} \{x\}^{[x]} \, dx + \int_{2}^{3} \{x\}^{[x]} \, dx$. For $x \in [0, 1)$,$[x] = 0$,so $\{x\}^{[x]} = \{x\}^{0} = 1$ (for $x 
eq 0$). Thus,$\int_{0}^{1} 1 \, dx = 1$. For $x \in [1, 2)$,$[x] = 1$,so $\{x\}^{[x]} = \{x\}^{1} = x - 1$. Thus,$\int_{1}^{2} (x - 1) \, dx = [\frac{(x-1)^2}{2}]_{1}^{2} = \frac{1}{2} - 0 = \frac{1}{2}$. For $x \in [2, 3)$,$[x] = 2$,so $\{x\}^{[x]} = \{x\}^{2} = (x - 2)^2$. Thus,$\int_{2}^{3} (x - 2)^2 \, dx = [\frac{(x-2)^3}{3}]_{2}^{3} = \frac{1}{3} - 0 = \frac{1}{3}$. Adding these values: $I = 1 + \frac{1}{2} + \frac{1}{3} = \frac{6+3+2}{6} = \frac{11}{6}$.

Let $y = \{x\}^{[x]}$ where $\{x\}$ denotes the fractional part of $x$ and $[x]$ denotes the greatest integer $\le x$. Then $\int_{0}^{3} y \, dx = $

Similar Questions

$\int_0^4 ||x-2|-x| dx = $

Evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \cos ^{2} x \,d x$.

The value of $\int_0^{2\pi} |\sin^3 \theta| \, d\theta$ is

If $\int_3^5 \sqrt{8 x-x^2-15} d x=p$,then $\sin p+\operatorname{cosec} p=$

If $\int_{0}^{k} \frac{dx}{2 + 18x^2} = \frac{\pi}{24}$,then the value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module