The area bounded by y =x2 +3 and y =2x+3 is

Question

The area bounded by y =x2 +3 and y =2x+3 is (A) (12/7) (B) (4/3) (C) (3/4) (D) (8/3)

Tardigrade Admin · Accepted Answer

Required area =∫02(2 x+3)-(x2+3) d x <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/2a65f1dc49ff7d03c240faeea6d5d44b-.png" /> =∫02(2 x-x2) d x =[(2 x2/2)-(x3/3)]02 =[4-(8/3)]=(4/3) sq units

Q. The area bounded by $y = x^{2} + 3$ and $y = 2 x + 3$ is

$\frac{12}{7}$

$\frac{4}{3}$

$\frac{3}{4}$

$\frac{8}{3}$

$\frac{3}{8}$

Required area
$= \int_{0}^{2} (2 x + 3) - (x^{2} + 3) d x$

$= \int_{0}^{2} (2 x - x^{2}) d x$
$= [\frac{2 x ^{2}}{2} - \frac{x ^{3}}{3}]_{0}^{2}$
$= [4 - \frac{8}{3}] = \frac{4}{3}$ sq units

Q. The area bounded by y=x2+3 and y=2x+3 is

712​

34​

43​

38​

83​

Required area =∫02​(2x+3)−(x2+3)dx =∫02​(2x−x2)dx =[22x2​−3x3​]02​ =[4−38​]=34​ sq units