The area (in sq unit) of the region bounded by the curves 2x = y2 - 1 and x = 0 is

Question

The area (in sq unit) of the region bounded by the curves 2x = y2 - 1 and x = 0 is (A) (1/3) (B) (2/3) (C) 1 (D) 2

Tardigrade Admin · Accepted Answer

Given curve can be rewritten as y 2=2 x +(1/2) Therefore, required area =∫ limits-11 xdy =2 ∫ limits01 ( y 2-1/2) dy =[( y 3/3)- y ]01 =(1/3)-1=(2/3) sq unit

Q. The area (in sq unit) of the region bounded by the curves $2 x = y^{2} - 1$ and $x = 0$ is

$\frac{1}{3}$

$\frac{2}{3}$

$1$

$2$

Given curve can be rewritten as
$y^{2} = 2 x + \frac{1}{2}$
Therefore, required area $= - 1 \int 1 x d y$
$= 2 0 \int 1 \frac{y ^{2} - 1}{2} d y$
$= [\frac{y ^{3}}{3} - y]_{0}^{1}$
$= \frac{1}{3} - 1 = \frac{2}{3}$ sq unit

Q. The area (in sq unit) of the region bounded by the curves 2x=y2−1 and x=0 is

31​

32​

1

2

Given curve can be rewritten as y2=2x+21​ Therefore, required area =−1∫1​xdy =20∫1​2y2−1​dy =[3y3​−y]01​ =31​−1=32​ sq unit

Q. The area (in sq unit) of the region bounded by the curves $2 x = y^{2} - 1$ and $x = 0$ is

$\frac{1}{3}$

$\frac{2}{3}$

$1$

$2$

Given curve can be rewritten as
$y^{2} = 2 x + \frac{1}{2}$
Therefore, required area $= - 1 \int 1 x d y$
$= 2 0 \int 1 \frac{y ^{2} - 1}{2} d y$
$= [\frac{y ^{3}}{3} - y]_{0}^{1}$
$= \frac{1}{3} - 1 = \frac{2}{3}$ sq unit