ParabolaMedium
Question
The area of the region $\{(x,y):0 \leq y \leq 6 - x$, $\left. \ y^{2} \geq 4x - 3,x \geq 0 \right\}$ is :
Options
A.8
B.9
C.12
D.15
Solution
Sol. [Key Concept: Integrate with respect to $y$: $A = \int_{0}^{3}(6 - y - \frac{y^{2} + 3}{4})dy$, the area between the line $y = 6 - x$ and the parabola $y^{2} = 4x - 3$.]
$$A = \int_{0}^{3}\mspace{2mu}\left( 6 - y - \frac{y^{2} + 3}{4} \right)dy = 9 $$
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