Question

$A_1$ is the area bounded by $y=x^2+2$, $x+y=8$, and the $y$-axis in the first quadrant, and $A_2$ is the area bounded by $y=x^2+2$, $y^2=x$, $x=0$ and $x=2$ in the first quadrant. Find $(A_1-A_2)$.

• A. $\dfrac{2}{3}+\dfrac{4\sqrt{2}}{3}$
• B. $\dfrac{3}{2}+\dfrac{4\sqrt{2}}{3}$
• C. $\dfrac{3}{5}+\dfrac{4\sqrt{2}}{3}$
• D. None of these

Hint:

Always sketch the curves to identify upper and lower functions correctly before setting up area integrals.

Answer 1

The Correct Option is A

Step 1: Find $A_1$.
The curves intersect where: \[ x^2+2=8-x \Rightarrow x^2+x-6=0 \Rightarrow x=2 \quad (\text{first quadrant}) \] Thus, \[ A_1=\int_{0}^{2}\left[(8-x)-(x^2+2)\right]dx \] \[ =\int_{0}^{2}(6-x-x^2)\,dx \] \[ =\left[6x-\frac{x^2}{2}-\frac{x^3}{3}\right]_{0}^{2} =\frac{22}{3} \] Step 2: Find $A_2$.
For $A_2$, the region lies between $y=\sqrt{x}$ and $y=x^2+2$ from $x=0$ to $x=2$: \[ A_2=\int_{0}^{2}\left[(x^2+2)-\sqrt{x}\right]dx \] \[ =\left[\frac{x^3}{3}+2x-\frac{2}{3}x^{3/2}\right]_{0}^{2} = \frac{20}{3}-\frac{4\sqrt{2}}{3} \] Step 3: Compute $(A_1-A_2)$.
\[ A_1-A_2=\frac{22}{3}-\left(\frac{20}{3}-\frac{4\sqrt{2}}{3}\right) =\frac{2}{3}+\frac{4\sqrt{2}}{3} \]