▶️ Answer/Explanation
Solution
Area from \( x = -\frac{\pi}{2} \) to \( x = k \): \( \int_{-\frac{\pi}{2}}^{k} \cos x \, dx = \sin k -inf \sin\left(-\frac{\pi}{2}\right) = \sin k + 1 \).
Area from \( x = k \) to \( x = \frac{\pi}{2} \): \( \int_{k}^{\frac{\pi}{2}} \cos x \, dx = \sin\left(\frac{\pi}{2}\right) – \sin k = 1 – \sin k \).
Given: \( \sin k + 1 = 3 (1 – \sin k) \).
Solve: \( \sin k + 1 = 3 – 3 \sin k \), \( 4 \sin k = 2 \), \( \sin k = \frac{1}{2} \).
In \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), \( k = \frac{\pi}{6} \).
✅ Answer: C
▶️ Answer/Explanation
Solution
Intersection points: \( 2x = 4x – x^2 \), \( x^2 – 2x = 0 \), \( x = 0, 2 \).
\( y = 4x – x^2 \) is above \( y = 2x \) from \( x = 0 \) to \( x = 2 \).
Area = \( \int_{0}^{2} [(4x – x^2) – 2x] \, dx = \int_{0}^{2} (2x – x^2) \, dx \).
Antiderivative: \( x^2 – \frac{1}{3} x^3 \).
Evaluate: \( \left[ x^2 – \frac{1}{3} x^3 \right]_{0}^{2} = (4 – \frac{8}{3}) – 0 = \frac{4}{3} \).
✅ Answer: A
▶️ Answer/Explanation
Solution
Shaded region is under \( y = x^2 \) from \( x = 1 \) to \( x = 2 \), divided at \( x = \frac{4}{3} \), \( x = \frac{5}{3} \).
Points: \( x = 1, \frac{4}{3}, \frac{5}{3}, 2 \); \( h = \frac{1}{3} \).
Values: \( f(1) = 1 \), \( f\left(\frac{4}{3}\right) = \frac{16}{9} \), \( f\left(\frac{5}{3}\right) = \frac{25}{9} \), \( f(2) = 4 \).
Trapezoidal rule: \( \text{Area} \approx \frac{1}{6} \left( 1 + 2 \left(\frac{16}{9}\right) + 2 \left(\frac{25}{9}\right) + 4 \right) = \frac{1}{6} \left( \frac{127}{9} \right) = \frac{127}{54} \).
✅ Answer: D
▶️ Answer/Explanation
Solution
1. Intersections: \( x^3 – x = 0 \) at \( x = 0, 1 \).
2. \( y = x + 8 \) above \( y = x^3 + 8 \) in \( [0, 1] \).
3. Area: \( \int_{0}^{1} (x – x^3) \, dx = \left[ \frac{x^2}{2} – \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{4} \).
✅ A) \( \frac{1}{4} \).