Area Under the Curve
Area Under Curves
nta_pyq_2025_apr
Grade 12
Question:
The area (in sq. units) of the region $\{(x,y): 0\leq y\leq 2|x|+1,\; 0\leq y\leq x^2+1,\; |x|\leq 3\}$ is
$\dfrac{80}{3}$
$\dfrac{64}{3}$
$\dfrac{32}{3}$
$\dfrac{17}{3}$
Step-by-Step Solution
Key Concept: Determine where $2|x|+1 \leq x^2+1$ (i.e., $|x|\geq 2$) and where $x^2+1 \leq 2|x|+1$ (i.e., $|x|\leq 2$); the effective upper boundary switches between the two curves at $|x|=2$.
By symmetry in $x$, area $= 2\times$(area for $x\geq 0$).
For $0\leq x\leq 2$: upper boundary is $y=x^2+1$.
For $2\leq x\leq 3$: upper boundary is $y=2x+1$.
$$\text{Area} = 2\left[\int_0^2(x^2+1)dx + \frac{1}{2}(5+7)\times 1\right] = 2\left[\frac{8}{3}+2+6\right] = 2\left[\frac{8}{3}+8\right] = 2\cdot\frac{32}{3} = \frac{64}{3}.$$
Correct Answer: 2