To find the area \( A \), we need to evaluate the integral over the region bounded by \( y = x^2 + 2 \) and \( y = 2x + 2 \) from \( x = 0 \) to \( x = 3 \).
Step 1. Set up the area as the sum of two integrals based on the intersection point of \( y = x^2 + 2 \) and \( y = 2x + 2 \), which is \( x = 2 \):**
\(A = \int_{0}^{2} (x^2 + 2) \, dx + \int_{2}^{3} (2x + 2) \, dx\)
Step 2. Evaluate each integral:
- For \( \int_{0}^{2} (x^2 + 2) \, dx \):
\(= \left[ \frac{x^3}{3} + 2x \right]_0^2 = \frac{8}{3} + 4 = \frac{20}{3}\)
- For \( \int_{2}^{3} (2x + 2) \, dx \):
\(= \left[ x^2 + 2x \right]_2^3 = (9 + 6) - (4 + 4) = 7\)
Step 3. Combine the results:
\(A = \frac{20}{3} + 7 = \frac{41}{3}\)
Step 4. Calculate \( 12A \):
12A = \(12 \times \frac{41}{3}\) = 164
The Correct Answer is: \( 12A = 164 \)
The portion of the line \( 4x + 5y = 20 \) in the first quadrant is trisected by the lines \( L_1 \) and \( L_2 \) passing through the origin. The tangent of an angle between the lines \( L_1 \) and \( L_2 \) is: