Area Under the Curve
Area Between Absolute Value Curve and Line
nta_pyq_2023_apr
Grade 12
Question:
If the area of the region $\{(x,y):|x^2-2|\leq y\leq x\}$ is $A$, then $6A+16\sqrt{2}$ is equal to ______________.
Step-by-Step Solution
Key Concept: Find intersections of $y=x$ with $y=|x^2-2|$: at $x=1$ (where $y=x$ meets $y=2-x^2$) and at $x=2$ (where $y=x$ meets $y=x^2-2$). Split at $x=\sqrt{2}$.
$A=\int_1^{\sqrt{2}}(x-2+x^2)dx+\int_{\sqrt{2}}^2(x-x^2+2)dx=\frac{9}{2}-\frac{8\sqrt{2}}{3}$. $6A+16\sqrt{2}=27$.
Correct Answer: 27