Let [x] denote the greatest integer leq x. Co | vedclass

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

Question

(A) We need to evaluate $I = \int_0^2 \max \{x^2, 1 + [x]\} dx$.For $x \in [0, 1)$,$[x] = 0$,so $f(x) = \max \{x^2, 1\} = 1$.For $x \in [1, \sqrt{2})$

Vedclass · Accepted Answer

$\frac{5+4 \sqrt{2}}{3}$

Vedclass · Answer

(A) We need to evaluate $I = \int_0^2 \max \{x^2, 1 + [x]\} dx$.For $x \in [0, 1)$,$[x] = 0$,so $f(x) = \max \{x^2, 1\} = 1$.For $x \in [1, \sqrt{2})$,$[x] = 1$,so $f(x) = \max \{x^2, 2\} = 2$ (since $x^2 For $x \in [\sqrt{2}, 2)$,$[x] = 1$,so $f(x) = \max \{x^2, 2\} = x^2$ (since $x^2 \geq 2$ for $x \geq \sqrt{2}$).At $x=2$,$f(2) = \max \{4, 1+2\} = 4$.Thus,the integral is:$I = \int_0^1 1 dx + \int_1^{\sqrt{2}} 2 dx + \int_{\sqrt{2}}^2 x^2 dx$$I = [x]_0^1 + [2x]_1^{\sqrt{2}} + [\frac{x^3}{3}]_{\sqrt{2}}^2$$I = (1 - 0) + (2\sqrt{2} - 2) + (\frac{8}{3} - \frac{2\sqrt{2}}{3})$$I = 1 + 2\sqrt{2} - 2 + \frac{8}{3} - \frac{2\sqrt{2}}{3}$$I = (1 - 2 + \frac{8}{3}) + (2\sqrt{2} - \frac{2\sqrt{2}}{3})$$I = \frac{5}{3} + \frac{4\sqrt{2}}{3} = \frac{5+4\sqrt{2}}{3}$.

Let $[x]$ denote the greatest integer $\leq x$. Consider the function $f(x) = \max \{x^2, 1 + [x]\}$. Then the value of the integral $\int_0^2 f(x) dx$ is:

Similar Questions

The integral $\int_{\pi /6}^{\pi /4} {\frac{{dx}}{{\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} $ equals

$\int_0^{\pi / 2} \sin ^5\left(\frac{x}{2}\right) \cdot \sin x \, dx =$

If $\int_0^1 f(x) dx = 1$,$\int_0^1 x f(x) dx = a$,and $\int_0^1 x^2 f(x) dx = a^2$,then $\int_0^1 (x-a)^2 f(x) dx$ is equal to

$\int_{\pi /4}^{3\pi /4} \frac{dx}{1 + \cos x}$ is equal to

If $\int_{0}^{\frac{\pi}{3}} \frac{\tan \theta}{\sqrt{2 k \sec \theta}} d \theta = 1 - \frac{1}{\sqrt{2}}$,$(k > 0)$,then the value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module