Find the area bounded by \( 0 \le y \le 6 - x \), \( y^2 + 3 \le 4x \) and \( x>0 \):
Correct Answer: 9
Solution and Explanation
Step 1: Understanding the Concept:
The area is bounded by a straight line \( x + y = 6 \), a parabola \( y^2 = 4x - 3 \) (which is \( y^2 = 4(x - 3/4) \)), and the axes. We need to find the intersection points to set the limits for integration.
Step 2: Key Formula or Approach:
1. Intersection of \( y = 6 - x \) and \( x = \frac{y^2 + 3}{4} \). 2. Area \( = \int_{y_1}^{y_2} (x_{line} - x_{parabola}) \, dy \).
Step 3: Detailed Explanation:
1. Substitute \( x = 6 - y \) into the parabola equation: \[ y^2 + 3 = 4(6 - y) \implies y^2 + 4y - 21 = 0 \] \[ (y + 7)(y - 3) = 0 \] 2. Since \( y \ge 0 \), the intersection is at \( y = 3 \). At \( y = 3 \), \( x = 3 \). 3. The parabola starts at \( x = 3/4 \) (when \( y = 0 \)). 4. Area \( = \int_0^3 \left( (6 - y) - \frac{y^2 + 3}{4} \right) dy \) \[ = \left[ 6y - \frac{y^2}{2} - \frac{y^3}{12} - \frac{3y}{4} \right]_0^3 \] \[ = \left( 18 - 4.5 - 2.25 - 2.25 \right) = 9 \] (Note: Depending on the specific region boundary \( x>0 \), the area between \( x=0 \) and the parabola vertex \( x=3/4 \) may be added, resulting in \( 9 + (3/4 \times 3) = 11.25 \) or similar.)
Step 4: Final Answer:
The area is 9 sq. units.
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