▶️ Answer/Explanation
Solution
First integral: \( \int_{0}^{1} x^n \, dx = \left[ \frac{x^{n+1}}{n+1} \right]_{0}^{1} = \frac{1}{n+1} \).
Second integral: Use \( u = 1-x \), so \( dx = -du \), limits \( x: 0 \to 1 \) become \( u: 1 \to 0 \).
\( \int_{0}^{1} (1-x)^n \, dx = \int_{1}^{0} u^n (-du) = \int_{0}^{1} u^n \, du = \frac{1}{n+1} \).
Both integrals equal \( \frac{1}{n+1} \) for all non-negative integers \( n \).
✅ Answer: E
▶️ Answer/Explanation
Solution
Rate of height increase: \( \frac{dh}{dt} = 4t^3 e^{-1.5t} \).
Height: \( h(t) = \int 4t^3 e^{-1.5t} \, dt \). After integration by parts (repeatedly):
\( h(t) = -\frac{8}{3} t^3 e^{-1.5t} – \frac{16}{3} t^2 e^{-1.5t} – \frac{64}{9} t e^{-1.5t} – \frac{128}{27} e^{-1.5t} + C \).
At \( t = 0 \), \( h(0) = 0 \), so \( C = \frac{128}{27} \).
Numerically, \( \int_{0}^{3} 4t^3 e^{-1.5t} \, dt \approx 2.111 \).
Height at \( t = 3 \): \( h(3) \approx 2.111 \) ft.
✅ Answer: D