▶️ Answer/Explanation
Solution
Area = \( \int_{0}^{2} e^{\frac{x}{2}} \, dx \).
Antiderivative: \( \int e^{\frac{x}{2}} \, dx = 2 e^{\frac{x}{2}} \).
Evaluate: \( \left[ 2 e^{\frac{x}{2}} \right]_{0}^{2} = 2 e^1 – 2 e^0 = 2e – 2 = 2(e – 1) \).
✅ Answer: C
▶️ Answer/Explanation
Solution
Intersection points: Solve \( -x^2 + x + 6 = 4 \), so \( x^2 – x – 2 = 0 \), \( x = -1, 2 \).
Curve is above the line from \( x = -1 \) to \( x = 2 \).
Area = \( \int_{-1}^{2} ((-x^2 + x + 6) – 4) \, dx = \int_{-1}^{2} (-x^2 + x + 2) \, dx \).
Antiderivative: \( -\frac{1}{3} x^3 + \frac{1}{2} x^2 + 2x \).
Evaluate: \( \left[ -\frac{1}{3} x^3 + \frac{1}{2} x^2 + 2x \right]_{-1}^{2} = \frac{10}{3} – \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} \).
✅ Answer: C
▶️ Answer/Explanation
Solution
Product \( xy = x (2x – 8) = 2x^2 – 8x \).
Derivative: \( p'(x) = 4x – 8 \).
Set \( 4x – 8 = 0 \), so \( x = 2 \).
Second derivative \( p”(x) = 4 > 0 \), confirming a minimum.
At \( x = 2 \), \( p(2) = 2(2)^2 – 8(2) = 8 – 16 = -8 \).
✅ Answer: B