If the area of the region bounded by the curv | vedclass

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

Question

(D) The given curves are $y^2-2y=-x$ and $x+y=0$.From the second equation,$x=-y$.Substituting this into the first equation: $y^2-2y=-(-y) \Rightarrow

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(D) The given curves are $y^2-2y=-x$ and $x+y=0$.From the second equation,$x=-y$.Substituting this into the first equation: $y^2-2y=-(-y) \Rightarrow y^2-2y=y \Rightarrow y^2-3y=0$.Thus,$y(y-3)=0$,which gives $y=0$ and $y=3$.When $y=0$,$x=0$. When $y=3$,$x=-3$.The area $A$ is given by the integral of the difference between the curves with respect to $y$:$A = \int_{0}^{3} (x_{	ext{right}} - x_{	ext{left}}) dy = \int_{0}^{3} (-y^2+2y - (-y)) dy = \int_{0}^{3} (-y^2+3y) dy$.Evaluating the integral:$A = \left[ -\frac{y^3}{3} + \frac{3y^2}{2} ight]_{0}^{3} = \left( -\frac{27}{3} + \frac{3(9)}{2} ight) - 0 = -9 + 13.5 = 4.5 = \frac{9}{2}$.Therefore,$8A = 8 	imes \frac{9}{2} = 36$.

If the area of the region bounded by the curves $y^2-2y=-x$ and $x+y=0$ is $A$,then $8A$ is equal to

Similar Questions

The area of the region described by $\{(x, y) \mid x^2+y^2 \leq 1\}$ and $\{y^2 \leq 1-x\}$ is

The area bounded by $y-1=-|x|$ and $y+1=|x|$ is

The area of the region included between the parabolas $y^2=8x$ and $x^2=8y$ is

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

The area (in sq. units) of the region described by $A = \{ (x,y) | y \ge x^2 - 5x + 4, x + y \ge 1, y \le 0 \}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module