The area (in sq. units) of the region (x, y): 0 ≤ y ≤ x2+1,0 ≤ y ≤ x+1. .(1/2) ≤ x ≤ 2 is

Question

The area (in sq. units) of the region (x, y): 0 ≤ y ≤ x2+1,0 ≤ y ≤ x+1. .(1/2) ≤ x ≤ 2 is (A) (79/16) (B) (23/6) (C) (79/24) (D) (23/16)

Tardigrade Admin · Accepted Answer

0 ≤ y ≤ x2+1,0 ≤ y ≤ x+1, (1/2) ≤ x ≤ 2 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/4a002784ec66bd913feb8d5662ff2409-.png" /> Required area =∫1 / 21(x2+1) d x+(1/2)(2+3) × 1 =(19/24)+(5/2)=(79/24)

Q. The area (in sq. units) of the region
${(x, y) : 0 \leq y \leq x^{2} + 1, 0 \leq y \leq x + 1$ $\frac{1}{2} \leq x \leq 2}$ is

$\frac{79}{16}$

$\frac{23}{6}$

$\frac{79}{24}$

$\frac{23}{16}$

$0 \leq y \leq x^{2} + 1, 0 \leq y \leq x + 1, \frac{1}{2} \leq x \leq 2$

Required area $= \int_{1/2}^{1} (x^{2} + 1) d x + \frac{1}{2} (2 + 3) \times 1$ $< b r / >= \frac{19}{24} + \frac{5}{2} = \frac{79}{24}$

Q. The area (in sq. units) of the region {(x,y):0≤y≤x2+1,0≤y≤x+1 21​≤x≤2} is

1679​

623​

2479​

1623​