The area (in sq. units) of the region {(x, y) | vedclass

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

Question

(C) The region is defined by $0 \leq y \leq \min(x^{2}+1, x+1)$ for $\frac{1}{2} \leq x \leq 2$.First,find the intersection of $y = x^{2}+1$ and $y =

Vedclass · Accepted Answer

$\frac{79}{24}$

Vedclass · Answer

(C) The region is defined by $0 \leq y \leq \min(x^{2}+1, x+1)$ for $\frac{1}{2} \leq x \leq 2$.First,find the intersection of $y = x^{2}+1$ and $y = x+1$:$x^{2}+1 = x+1 \implies x^{2}-x = 0 \implies x(x-1) = 0$.So,the curves intersect at $x = 0$ and $x = 1$.For $\frac{1}{2} \leq x \leq 1$,$x+1 \geq x^{2}+1$,so the area is $\int_{1/2}^{1} (x^{2}+1) dx$.For $1 \leq x \leq 2$,$x^{2}+1 \geq x+1$,so the area is $\int_{1}^{2} (x+1) dx$.Area $= \int_{1/2}^{1} (x^{2}+1) dx + \int_{1}^{2} (x+1) dx$.$= [\frac{x^{3}}{3} + x]_{1/2}^{1} + [\frac{x^{2}}{2} + x]_{1}^{2}$.$= ((\frac{1}{3} + 1) - (\frac{1}{24} + \frac{1}{2})) + ((2 + 2) - (\frac{1}{2} + 1))$.$= (\frac{4}{3} - \frac{13}{24}) + (4 - \frac{3}{2}) = \frac{32-13}{24} + \frac{5}{2} = \frac{19}{24} + \frac{60}{24} = \frac{79}{24}$.

The area (in sq. units) of the region $\{(x, y): 0 \leq y \leq x^{2}+1, 0 \leq y \leq x+1, \frac{1}{2} \leq x \leq 2\}$ is

Similar Questions

The area of the shorter region bounded by $|y| = 4 - x^2$ and $|y| = 3x$ is given by $\left( 3K + \frac{1}{3} \right)$ sq-unit,where $K$ is equal to:

The area (in square units) bounded by the curves $y = \sqrt{x}$,the line $2y - x + 3 = 0$,and the $X$-axis,lying in the first quadrant,is:

The area of the region $\{(x, y): xy \leq 8, 1 \leq y \leq x^2\}$ is

The area of the region $\{(x, y): x^2+4x+2 \leq y \leq |x+2|\}$ is equal to

The area of the region $S = \{(x, y): 3x^{2} \leq 4y \leq 6x + 24\}$ is $......$

Vedclass Test Series

Exam Paper Generator

Online Exam Module