Area Under the Curve
Area Under min of Two Functions
nta_pyq_2024_jan
Grade 12
Question:
If the area of the region $\{(x,y): 0\le y\le\min\{2x,\,6x-x^2\}\}$ is $A$, then $12A$ is equal to
Step-by-Step Solution
Key Concept: $\min\{2x,6x-x^2\}$: $2x\le6x-x^2\Leftrightarrow x^2\le4x\Leftrightarrow0\le x\le4$. So min is $2x$ for $0\le x\le4$ and $6x-x^2$ for $4\le x\le6$. Area = area of triangle $(0,0)$–$(4,8)$ plus integral of $6x-x^2$ from 4 to 6.
$A=\frac{1}{2}\cdot4\cdot8+\int_4^6(6x-x^2)dx=16+[3x^2-x^3/3]_4^6=16+\frac{28}{3}=\frac{76}{3}$. $12A=304$.
Correct Answer: 304