The area of the region for which 0

Question

The area of the region for which 0 < y < 3-2x-x2 and x>0 is (A) (5/3) (B) 3 (C) (13/3) (D) 9

Tardigrade Admin · Accepted Answer

Area =∫ limits 01(3 - 2 x - x2)dx=(3 x - x2 - (x3/3))01 <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-7mvfjzblmwwldkzz.jpg" /> =3-1-(1/3)=(5/3)

Q. The area of the region for which $0 < y < 3 - 2 x - x^{2}$ and $x > 0$ is

$\frac{5}{3}$

$3$

$\frac{13}{3}$

$9$

Area $= 0 \int 1 (3 - 2 x - x^{2}) d x = (3 x - x^{2} - \frac{x ^{3}}{3})_{0}^{1}$

$= 3 - 1 - \frac{1}{3} = \frac{5}{3}$

Q. The area of the region for which 0<y<3−2x−x2 and x>0 is

35​

3

313​

9

Area =0∫1​(3−2x−x2)dx=(3x−x2−3x3​)01​ =3−1−31​=35​

Q. The area of the region for which $0 < y < 3 - 2 x - x^{2}$ and $x > 0$ is

$\frac{5}{3}$

$3$

$\frac{13}{3}$

$9$

Area $= 0 \int 1 (3 - 2 x - x^{2}) d x = (3 x - x^{2} - \frac{x ^{3}}{3})_{0}^{1}$

$= 3 - 1 - \frac{1}{3} = \frac{5}{3}$