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

Question

If the area of the region bounded by the curves y 2-2 y=-x, x+y=0 is A, then 8 A is equal to  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

y2-2 y=-x ⇒ y2-2 y+1=-x+1 (y-1)2=-(x-1) y=-x Points of intersection x2+2 x=-x x2+3 x=0 x=0,-3 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/817197e72231579d45e3a20e44fac67d-.png" /> A=∫ limits03(-y2+2 y+y) d y =(3 y2/2)-.(y3/3)|0 3=(9/2) 8 A=36

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

$y^{2} - 2 y = - x$
$\Rightarrow y^{2} - 2 y + 1 = - x + 1$
$(y - 1)^{2} = - (x - 1)$
$y = - x$
Points of intersection
$x^{2} + 2 x = - x$
$x^{2} + 3 x = 0$
$x = 0, - 3$

$A = 0 \int 3 (- y^{2} + 2 y + y) d y$
$= \frac{3 y ^{2}}{2} - \frac{y ^{3}}{3} ∣ ∣_{0}^{3} = \frac{9}{2}$
$8 A = 36$

Q. If the area of the region bounded by the curves y2−2y=−x,x+y=0 is A, then 8A is equal to _________

y2−2y=−x ⇒y2−2y+1=−x+1 (y−1)2=−(x−1) y=−x Points of intersection x2+2x=−x x2+3x=0 x=0,−3 A=0∫3​(−y2+2y+y)dy =23y2​−3y3​∣∣​03​=29​ 8A=36

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

$y^{2} - 2 y = - x$
$\Rightarrow y^{2} - 2 y + 1 = - x + 1$
$(y - 1)^{2} = - (x - 1)$
$y = - x$
Points of intersection
$x^{2} + 2 x = - x$
$x^{2} + 3 x = 0$
$x = 0, - 3$

$A = 0 \int 3 (- y^{2} + 2 y + y) d y$
$= \frac{3 y ^{2}}{2} - \frac{y ^{3}}{3} ∣ ∣_{0}^{3} = \frac{9}{2}$
$8 A = 36$