Let A_1 be the bounded area enclosed by the curves y=x^2+2, x+y=8 and the y-axis in the fi…

Question

Let $A_1$ be the bounded area enclosed by the curves $y=x^2+2$, $x+y=8$ and the $y$-axis in the first quadrant. Let $A_2$ be the bounded area enclosed by $y=x^2+2$, $y^2=x$, $x=2$ and the $y$-axis in the first quadrant. Then $A_1-A_2$ is equal to:

Accepted Answer

Intersection of $y=x^2+2$ and $x+y=8$: $x^2+x-6=0 \Rightarrow x=2$ (first quadrant), point $(2,6)$.

$$A_1 = \int_0^2\big[(8-x)-(x^2+2)\big]dx = \left[-\frac{x^3}{3}-\frac{x^2}{2}+6x\right]_0^2 = \frac{22}{3}.$$

For $A_2$: $y=x^2+2$, $y=\sqrt x$ don't intersect in the region (discriminant negative), so bounded by $y$-axis, $x=2$, $y=x^2+2$, $y=\sqrt x$:
$$A_2 = \int_0^2\big(x^2+2-\sqrt x\big)\,dx = \left[\frac{x^3}{3}+2x-\frac{2x^{3/2}}{3}\right]_0^2 = \frac{20}{3}-\frac{4\sqrt2}{3}.$$

$$A_1-A_2 = \frac{22}{3}-\frac{20}{3}+\frac{4\sqrt2}{3} = \frac23(2\sqrt2+1).$$

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