Question ID: #451
The area (in sq. units) of the region $\{(x, y) : 0 \le y \le 2|x|+1, 0 \le y \le x^2+1, |x| \le 3\}$ is
- (1) $\frac{80}{3}$
- (2) $\frac{64}{3}$
- (3) $\frac{17}{3}$
- (4) $\frac{75}{4}$
Solution:
The region is symmetric about the y-axis, so $A = 2 \times \text{Area}(x \ge 0)$.
The region is bounded by $y = \min(2x+1, x^2+1)$ for $x \in [0, 3]$.
Find intersection of $y = 2x+1$ and $y = x^2+1$:
$$ x^2 + 1 = 2x + 1 \Rightarrow x(x-2) = 0 \Rightarrow x = 0, x = 2 $$
Check which curve is lower (min):
For $0 \le x \le 2$: Take $x=1 \Rightarrow y_{parabola}=2, y_{line}=3$. So $y = x^2+1$ is the boundary.
For $2 \le x \le 3$: Take $x=2.5 \Rightarrow y_{parabola}=7.25, y_{line}=6$. So $y = 2x+1$ is the boundary.
Total Area $A$:
$$ A = 2 \left[ \int_{0}^{2} (x^2+1) dx + \int_{2}^{3} (2x+1) dx \right] $$
Calculate Integrals:
$$ \int_{0}^{2} (x^2+1) dx = \left[ \frac{x^3}{3} + x \right]_0^2 = \frac{8}{3} + 2 = \frac{14}{3} $$
$$ \int_{2}^{3} (2x+1) dx = \left[ x^2 + x \right]_2^3 = (9+3) – (4+2) = 12 – 6 = 6 $$
$$ A = 2 \left( \frac{14}{3} + 6 \right) = 2 \left( \frac{14+18}{3} \right) = 2 \left( \frac{32}{3} \right) $$
$$ A = \frac{64}{3} $$
Ans. (2)
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