int_{0}^{2} [x^{2}] , dx | vedclass

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

Question

(C) Consider the given integral $I = \int_{0}^{2} [x^{2}] \, dx$. Since the greatest integer function $[x^2]$ changes its value at $x^2 = 1, 2, 3$,we

Vedclass · Accepted Answer

$ 5-\sqrt{2}-\sqrt{3} $

Vedclass · Answer

(C) Consider the given integral $I = \int_{0}^{2} [x^{2}] \, dx$. Since the greatest integer function $[x^2]$ changes its value at $x^2 = 1, 2, 3$,we split the interval $[0, 2]$ based on $x = 1, \sqrt{2}, \sqrt{3}$. For $x \in [0, 1)$,$x^2 \in [0, 1)$,so $[x^2] = 0$. For $x \in [1, \sqrt{2})$,$x^2 \in [1, 2)$,so $[x^2] = 1$. For $x \in [\sqrt{2}, \sqrt{3})$,$x^2 \in [2, 3)$,so $[x^2] = 2$. For $x \in [\sqrt{3}, 2]$,$x^2 \in [3, 4]$,so $[x^2] = 3$. Thus,$I = \int_{0}^{1} 0 \, dx + \int_{1}^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^{2} 3 \, dx$. $I = 0 + [x]_{1}^{\sqrt{2}} + [2x]_{\sqrt{2}}^{\sqrt{3}} + [3x]_{\sqrt{3}}^{2}$. $I = (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3})$. $I = \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3}$. $I = 5 - \sqrt{2} - \sqrt{3}$.

$ \int_{0}^{2} [x^{2}] \, dx $

Similar Questions

The value of $\int_0^{\pi /4} \frac{1 + \tan x}{1 - \tan x} \, dx$ is

$\int_0^{\pi /4} [\sqrt{\tan x} + \sqrt{\cot x}] \, dx$ equals

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot ^9 x \, dx =$

Evaluate the definite integral $\int_0^3 [x] \, dx$,where $[x]$ denotes the greatest integer function.

$\int_{2}^{3} \frac{dx}{x^2 - x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module