Find the area bounded by the curve x = 2 - y - y2 and y-axis.

Question

Find the area bounded by the curve x = 2 - y - y2 and y-axis. (A) -(9/2) (B) (9/2) (C) 9 (D) -9

Tardigrade Admin · Accepted Answer

Put 2 - y - y2 = 0 ⇒ y = 1, - 2 This means, the curve intersects the y-axis at y = 1 and y = - 2. Hence required area = ∫ limits1-2 xdy = ∫ limits1-2 (2-y-y2)dy <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/22bc93e113eeee378dc67fb55d4ebf65-.png" /> = [2y-(y2/2)-(y3/3)]1-2 = (9/2) sq. units

Q. Find the area bounded by the curve $x = 2 - y - y^{2}$ and y-axis.

$- \frac{9}{2}$

$\frac{9}{2}$

$9$

$- 9$

Put $2 - y - y^{2} = 0$
$\Rightarrow y = 1, - 2$
This means, the curve intersects the y-axis at $y = 1$ and $y = - 2$ .
Hence required area $= - 2 \int 1 x d y$
$= - 2 \int 1 (2 - y - y^{2}) d y$

$= [2 y - \frac{y ^{2}}{2} - \frac{y ^{3}}{3}]_{- 2}^{1} = \frac{9}{2}$ sq. units

Q. Find the area bounded by the curve x=2−y−y2 and y-axis.

−29​

29​

9

−9

Put 2−y−y2=0 ⇒y=1,−2 This means, the curve intersects the y-axis at y=1 and y=−2. Hence required area =−2∫1​xdy =−2∫1​(2−y−y2)dy =[2y−2y2​−3y3​]−21​=29​ sq. units

Q. Find the area bounded by the curve $x = 2 - y - y^{2}$ and y-axis.

$- \frac{9}{2}$

$\frac{9}{2}$

$9$

$- 9$

Put $2 - y - y^{2} = 0$
$\Rightarrow y = 1, - 2$
This means, the curve intersects the y-axis at $y = 1$ and $y = - 2$ .
Hence required area $= - 2 \int 1 x d y$
$= - 2 \int 1 (2 - y - y^{2}) d y$

$= [2 y - \frac{y ^{2}}{2} - \frac{y ^{3}}{3}]_{- 2}^{1} = \frac{9}{2}$ sq. units