The area bounded by the curves y=2x^2, y=max | vedclass

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

Question

(A) Given curves are $y=2x^2$ and $y=\max \{x-[x], x+|x|\}$.Since $x-[x] = \{x\}$, we have $y=\max \{\{x\}, x+|x|\}$.For $x \in [0, 2]$, $x+|x| = 2x$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given curves are $y=2x^2$ and $y=\max \{x-[x], x+|x|\}$.Since $x-[x] = \{x\}$, we have $y=\max \{\{x\}, x+|x|\}$.For $x \in [0, 2]$, $x+|x| = 2x$ and $\{x\} \in [0, 1)$.Since $2x \geq \{x\}$ for all $x \in [0, 2]$, the function simplifies to $y=2x$.We need the area bounded by $y=2x^2$ and $y=2x$ between $x=0$ and $x=2$.The curves intersect where $2x^2 = 2x$, which gives $x^2-x=0$, so $x=0$ and $x=1$.For $x \in [0, 1]$, $2x \geq 2x^2$. For $x \in [1, 2]$, $2x^2 \geq 2x$.The required area is $\int_0^1 (2x - 2x^2) dx + \int_1^2 (2x^2 - 2x) dx$.$= [x^2 - \frac{2x^3}{3}]_0^1 + [\frac{2x^3}{3} - x^2]_1^2$.$= (1 - \frac{2}{3}) + [(\frac{16}{3} - 4) - (\frac{2}{3} - 1)]$.$= \frac{1}{3} + \frac{4}{3} - (-\frac{1}{3}) = \frac{1}{3} + \frac{4}{3} + \frac{1}{3} = \frac{6}{3} = 2$.

The area bounded by the curves $y=2x^2$, $y=\max \{x-[x], x+|x|\}$ and the lines $x=0, x=2$ (in sq units) is

Similar Questions

Area of the region bounded by the curves $y = \sin x$ and $y = x$ between the lines $x = 0$ and $x = 2\pi$ is:

Let for $x \in R$; $f(x) = \frac{x+|x|}{2}$ and $g(x) = \begin{cases} x, & x < 0 \\ x^2, & x \geq 0 \end{cases}$. Then the area bounded by the curve $y = (f \circ g)(x)$ and the lines $y = 0$,$2y - x = 15$ is equal to $...........$.

The area of the region bounded by the curve $y = \sin(\pi x)$ and the $X$-axis for $x \in [-1, 2]$ is . . . . . . sq. units.

The area of the region common to the parabolas $4y^2 = 9x$ and $3x^2 = 16y$ is...

The area of the shaded region is ... sq. units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module