Area Under the Curve
Area Between Parabola and Shifted Parabola
nta_pyq_2024_jan
Grade 12
Question:
The area of the region enclosed by the parabola $y=4x-x^2$ and $3y=(x-4)^2$ is equal to
$\frac{32}{9}$
4
6
$\frac{14}{3}$
Step-by-Step Solution
Key Concept: Find intersections: $4x-x^2=\frac{(x-4)^2}{3}\Rightarrow12x-3x^2=(x-4)^2\Rightarrow4x^2-20x+16=0\Rightarrow(x-1)(x-4)=0$. Intersect at $x=1$ and $x=4$. Area $=\int_1^4\left[(4x-x^2)-\frac{(x-4)^2}{3}\right]dx$.
$A=\int_1^4\left[4x-x^2-\frac{(x-4)^2}{3}\right]dx=\int_1^4\left[\frac{-4x^2+20x-16}{3}\right]dx=\frac{1}{3}\left[-\frac{4x^3}{3}+10x^2-16x\right]_1^4=6$.
Correct Answer: 3