Area Under the Curve
Area Under min of Two Functions
nta_pyq_2024_jan
Grade 12
Question:
Let the area of the region $\{(x,y): 0\le x\le3,\, 0\le y\le\min\{x^2+2,\,2x+2\}\}$ be $A$. Then $12A$ is equal to
Step-by-Step Solution
Key Concept: $\min\{x^2+2,2x+2\}$: $x^2+2\le2x+2\Leftrightarrow x^2-2x\le0\Leftrightarrow0\le x\le2$. So for $0\le x\le2$: use $x^2+2$; for $2\le x\le3$: use $2x+2$.
$A=\int_0^2(x^2+2)dx+\int_2^3(2x+2)dx=\left[\frac{x^3}{3}+2x\right]_0^2+[x^2+2x]_2^3=\frac{20}{3}+7=\frac{41}{3}$. $12A=164$.
Correct Answer: 164