If the area bounded by the parabola y=2-x2 and the line y=-x is (k/2) sq. units, then the value of 2k is equal to

Question

If the area bounded by the parabola y=2-x2 and the line y=-x is (k/2) sq. units, then the value of 2k is equal to (A) 9 (B) 27 (C) 18 (D) 32

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-kz2vaheqmnfj.png" /> Solving y=2-x2 y=-x; we get -x=2-x2 ⇒ x=2,-1 ⇒ A(- 1,1) B(2 , - 2) Thus, the required area A=∫ limits - 12 (2 - x2) - (- x) d x =[2 x - (x3/3) + (x2/2)]- 12 =(9/2) ⇒ k=9 ∴ 2k=18

Q. If the area bounded by the parabola $y = 2 - x^{2}$ and the line $y = - x$ is $\frac{k}{2}$ sq. units, then the value of $2 k$ is equal to

$9$

$27$

$18$

$32$

Solving $y = 2 - x^{2}$ & $y = - x;$ we get $- x = 2 - x^{2}$
$\Rightarrow x = 2, - 1$
$\Rightarrow A (- 1, 1) & B (2, - 2)$
Thus, the required area
$A = - 1 \int 2 (2 - x^{2}) - (- x) d x$
$= [2 x - \frac{x ^{3}}{3} + \frac{x ^{2}}{2}]_{- 1}^{2}$
$= \frac{9}{2}$
$\Rightarrow k = 9$
$∴ 2 k = 18$

Q. If the area bounded by the parabola y=2−x2 and the line y=−x is 2k​ sq. units, then the value of 2k is equal to

9

27

18

32

Solving y=2−x2 & y=−x; we get −x=2−x2 ⇒x=2,−1 ⇒A(−1,1)&B(2,−2) Thus, the required area A=−1∫2​(2−x2)−(−x)dx =[2x−3x3​+2x2​]−12​ =29​ ⇒k=9 ∴2k=18

Q. If the area bounded by the parabola $y = 2 - x^{2}$ and the line $y = - x$ is $\frac{k}{2}$ sq. units, then the value of $2 k$ is equal to

$9$

$27$

$18$

$32$

Solving $y = 2 - x^{2}$ & $y = - x;$ we get $- x = 2 - x^{2}$
$\Rightarrow x = 2, - 1$
$\Rightarrow A (- 1, 1) & B (2, - 2)$
Thus, the required area
$A = - 1 \int 2 (2 - x^{2}) - (- x) d x$
$= [2 x - \frac{x ^{3}}{3} + \frac{x ^{2}}{2}]_{- 1}^{2}$
$= \frac{9}{2}$
$\Rightarrow k = 9$
$∴ 2 k = 18$