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

(A) The region is bounded by $x=0$,$y=1+\sqrt{x}$,$x+y=3$,and $y=\frac{x^2}{4}$.From the graph,the intersection points are $(1,2)$ and $(2,1)$.The are

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

(A) The region is bounded by $x=0$,$y=1+\sqrt{x}$,$x+y=3$,and $y=\frac{x^2}{4}$.From the graph,the intersection points are $(1,2)$ and $(2,1)$.The area is given by the integral:$A = \int_{0}^{1} (1+\sqrt{x} - \frac{x^2}{4}) dx + \int_{1}^{2} (3-x - \frac{x^2}{4}) dx$$A = \left[ x + \frac{2}{3}x^{3/2} - \frac{x^3}{12} \right]_{0}^{1} + \left[ 3x - \frac{x^2}{2} - \frac{x^3}{12} \right]_{1}^{2}$$A = (1 + \frac{2}{3} - \frac{1}{12}) + [(6 - 2 - \frac{8}{12}) - (3 - \frac{1}{2} - \frac{1}{12})]$$A = (\frac{12+8-1}{12}) + [(4 - \frac{2}{3}) - (2.5 - \frac{1}{12})]$$A = \frac{19}{12} + (\frac{10}{3} - \frac{29}{12}) = \frac{19}{12} + \frac{40-29}{12} = \frac{19+11}{12} = \frac{30}{12} = \frac{5}{2}$ sq. units.

The area (in sq. units) of the region $\{ (x,y) : x \ge 0, x + y \le 3, x^2 \le 4y \text{ and } y \le 1 + \sqrt{x} \}$ is:

Similar Questions

The area (in sq. units) of the region $\{x \in R : x \ge 0, y \ge 0, y \ge x - 2, \text{ and } y \le \sqrt{x}\}$ is

The area of the region above the $X$-axis included between the parabola $y^{2}=x$ and the circle $x^{2}+y^{2}=2x$ in square units is

The area (in square units) enclosed between the curves $y=\sin x$ and $y=\cos x$ for $\frac{\pi}{4} \leq x \leq \frac{5 \pi}{4}$ is

The area of the region bounded by the line $y=x$ and the curve $y=x^3$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module