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

• A. $\dfrac{2}{3}(4\sqrt{2}+1)$
• B. $\dfrac{2}{3}(3\sqrt{2}+1)$
• C. $\dfrac{2}{3}(2\sqrt{2}+1)$
• D. $\dfrac{2}{3}(\sqrt{2}+1)$

Hint:

Always sketch the curves to correctly identify limits when dealing with area between curves.

Answer 1

The Correct Option is D

Step 1: Finding $A_1$.
The region $A_1$ is bounded by \[ y=x^2+2,\quad y=8-x,\quad x=0 \] Points of intersection are obtained by \[ x^2+2 = 8-x \Rightarrow x^2+x-6=0 \Rightarrow x=2 \] Thus, \[ A_1 = \int_{0}^{2} \left[(8-x)-(x^2+2)\right]dx \] \[ A_1 = \int_{0}^{2} (6-x-x^2)dx \] \[ A_1 = \left[6x-\frac{x^2}{2}-\frac{x^3}{3}\right]_{0}^{2} = \frac{22}{3} \] Step 2: Finding $A_2$.
The region $A_2$ is bounded by \[ y=x^2+2,\quad x=y^2,\quad x=2 \] Changing variables, area is \[ A_2 = \int_{0}^{\sqrt{2}} \left[(x^2+2)-y^2\right]dy \] \[ A_2 = \int_{0}^{\sqrt{2}} (2+y^2-y^4)dy \] \[ A_2 = \left[2y+\frac{y^3}{3}-\frac{y^5}{5}\right]_{0}^{\sqrt{2}} \] \[ A_2 = \frac{22}{3}-\frac{2\sqrt{2}}{3} \] Step 3: Calculating $A_1-A_2$.
\[ A_1-A_2=\frac{2}{3}(\sqrt{2}+1) \]