Let be the bounded area enclosed by the curves and -axis that lies in the first quadrant. Let be the bounded area enclosed by the curves , , , and -axis that lies in the first quadrant. Then is equal to [2026]

Question

Let $A_{1}$ be the bounded area enclosed by the curves $y = x^{2} + 2,$ $x + y = 8$ and $y$ -axis that lies in the first quadrant. Let $A_{2}$ be the bounded area enclosed by the curves $y = x^{2} + 2$ , $y^{2} = x$ , $x = 2$ , and $y$ -axis that lies in the first quadrant. Then $A_{1} - A_{2}$ is equal to [2026]

Smart Achievers · Accepted Answer

(4)

$A_{1} = \int_{0}^{2} ((8 - x) - (x^{2} + 2)) d x$

$= \int_{0}^{2} (6 - x - x^{2}) d x$

$A_{1} {(6 x - \frac{x^{2}}{2} - \frac{x^{3}}{3})}_{0}^{2}$ $= 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{22}{3}$

$A_{2} = \int_{0}^{2} (x^{2} + 2) d x - \frac{2}{3} (2 \sqrt{2})$

$A_{2} = {(\frac{x^{3}}{3} + 2 x)}_{0}^{2} - \frac{4 \sqrt{2}}{3}$

$A_{2} = \frac{8}{3} + 4 - \frac{4 \sqrt{2}}{3} = \frac{20}{3} - \frac{4 \sqrt{2}}{3}$

$A_{1} - A_{2} = \frac{2}{3} + \frac{4 \sqrt{2}}{3}$

Solution

1 $\frac{2}{3} (\sqrt{2} + 1)$

2 $\frac{2}{3} (4 \sqrt{2} + 1)$

3 $\frac{2}{3} (3 \sqrt{2} + 1)$

4 $\frac{2}{3} (2 \sqrt{2} + 1)$

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let A1 be the bounded area enclosed by the curves y=x2+2, x+y=8 and y-axis that lies in the first quadrant. Let A2 be the bounded area enclosed by the curves y=x2+2, y2=x, x=2, and y-axis that lies in the first quadrant. Then A1-A2 is equal to [2026]

1 23(2+1)

2 23(42+1)

3 23(32+1)

4 23(22+1)

Ans. (4) A1=∫02((8-x)-(x2+2))dx =∫02(6-x-x2)dx A1(6x-x22-x33)02=12-2-83=10-83=223 A2=∫02(x2+2)dx-23(22) A2=(x33+2x)02-423 A2=83+4-423=203-423 A1-A2=23+423

Want Us to Email you About Special Offers & Updates?

1 $\frac{2}{3} (\sqrt{2} + 1)$

2 $\frac{2}{3} (4 \sqrt{2} + 1)$

3 $\frac{2}{3} (3 \sqrt{2} + 1)$

4 $\frac{2}{3} (2 \sqrt{2} + 1)$