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
- (1) $\frac{2}{3}(2\sqrt{2}+1)$
- (2) $\frac{2}{3}(4\sqrt{2}+1)$
- (3) $\frac{2}{3}(\sqrt{2}+1)$
- (4) $\frac{2}{3}(3\sqrt{2}+1)$
Solution:
**Calculating $A_1$:**
Intersection of $y=x^2+2$ and $x+y=8$ ($y=8-x$):
$$x^2+2 = 8-x \Rightarrow x^2+x-6=0 \Rightarrow (x+3)(x-2)=0$$
In 1st quadrant, $x=2$.
Area $A_1$ is bounded by y-axis ($x=0$), $x=2$, upper curve $y=8-x$ and lower curve $y=x^2+2$.
Wait, looking at the region in standard graph, $y=x^2+2$ is a parabola opening up from $(0,2)$. $x+y=8$ is a line.
The area enclosed by these two and y-axis ($x=0$) is:
$$A_1 = \int_{0}^{2} (\text{Upper Curve} – \text{Lower Curve}) dx$$
The line $y=8-x$ is above the parabola $y=x^2+2$ in $(0,2)$.
$$A_1 = \int_{0}^{2} ((8-x) – (x^2+2)) dx = \int_{0}^{2} (6 – x – x^2) dx$$
$$A_1 = \left[ 6x – \frac{x^2}{2} – \frac{x^3}{3} \right]_0^2$$
$$A_1 = 12 – 2 – \frac{8}{3} = 10 – \frac{8}{3} = \frac{22}{3}$$
**Calculating $A_2$:**
Region bounded by $y=x^2+2$, $y^2=x$ ($y=\sqrt{x}$), $x=2$ and y-axis ($x=0$).
$$A_2 = \int_{0}^{2} ((x^2+2) – \sqrt{x}) dx$$
$$A_2 = \left[ \frac{x^3}{3} + 2x – \frac{2}{3}x^{3/2} \right]_0^2$$
$$A_2 = \left( \frac{8}{3} + 4 – \frac{2}{3}(2\sqrt{2}) \right) – 0$$
$$A_2 = \frac{20}{3} – \frac{4\sqrt{2}}{3}$$
**Calculating $A_1 – A_2$:**
$$A_1 – A_2 = \frac{22}{3} – \left( \frac{20}{3} – \frac{4\sqrt{2}}{3} \right)$$
$$= \frac{2}{3} + \frac{4\sqrt{2}}{3}$$
$$= \frac{2}{3}(1 + 2\sqrt{2})$$
Ans. (1)