▶️ Answer/Explanation
Solution
Average value = \(\frac{1}{2 – (-1)} \int_{-1}^2 (3t^3 – t^2) \, dt = \frac{1}{3} \int_{-1}^2 (3t^3 – t^2) \, dt\).
Antiderivative: \(\int (3t^3 – t^2) \, dt = \frac{3}{4} t^4 – \frac{1}{3} t^3\).
Evaluate: \(\left[ \frac{3}{4} t^4 – \frac{1}{3} t^3 \right]_{-1}^2\).
At \( t = 2 \): \(\frac{3}{4} (2)^4 – \frac{1}{3} (2)^3 = 12 – \frac{8}{3} = \frac{28}{3}\).
At \( t = -1 \): \(\frac{3}{4} (-1)^4 – \frac{1}{3} (-1)^3 = \frac{3}{4} + \frac{1}{3} = \frac{13}{12}\).
Integral: \(\frac{28}{3} – \frac{13}{12} = \frac{112}{12} – \frac{13}{12} = \frac{99}{12} = \frac{33}{4}\).
Average: \(\frac{1}{3} \cdot \frac{33}{4} = \frac{33}{12} = \frac{11}{4}\).
✅ Answer: A
Question
The average value of \(\sqrt{x}\) over the interval 0 ≤ 2 ≤ x is
(A)\(\frac{1}{3}\sqrt{2}\) (B)\(\frac{1}{2}\sqrt{2}\) (C)\(\frac{2}{3}\sqrt{2} \) (D) 1 (E)\(\frac{4}{3}\sqrt{2}\)
▶️Answer/Explanation
Ans:C
▶️ Answer/Explanation
Solution
Average velocity = \(\frac{x(3) – x(0)}{3 – 0}\).
Position \( x(t) = -5t^2 \).
At \( t = 0 \): \( x(0) = -5 (0)^2 = 0 \).
At \( t = 3 \): \( x(3) = -5 (3)^2 = -5 \cdot 9 = -45 \).
Average velocity = \(\frac{-45 – 0}{3} = \frac{-45}{3} = -15\).
✅ Answer: C
▶️ Answer/Explanation
Solution
Average value = \(\frac{1}{2 – 0} \int_0^2 x^2 \sqrt{x^3 + 1} \, dx = \frac{1}{2} \int_0^2 x^2 \sqrt{x^3 + 1} \, dx\).
Substitute \( u = x^3 + 1 \), \( du = 3x^2 \, dx \), \( dx = \frac{du}{3x^2} \).
Integrand: \( x^2 \sqrt{u} \cdot \frac{du}{3x^2} = \frac{1}{3} u^{1/2} \, du\).
Integrate: \(\frac{1}{3} \int u^{1/2} \, du = \frac{1}{3} \cdot \frac{2}{3} u^{3/2} = \frac{2}{9} u^{3/2}\).
Evaluate from \( u = 1 \) to \( u = 9 \): \(\frac{2}{9} (9^{3/2} – 1^{3/2}) = \frac{2}{9} (27 – 1) = \frac{2}{9} \cdot 26 = \frac{52}{9}\).
Average: \(\frac{1}{2} \cdot \frac{52}{9} = \frac{26}{9}\).
✅ Answer: A
▶️ Answer/Explanation
Solution
The curve is in the first quadrant where \( 0 \leq x \leq 3 \) (since \( y = 3x – x^2 = 0 \) at \( x = 0 \) and \( x = 3 \)).
Average value = \(\frac{1}{3 – 0} \int_0^3 (3x – x^2) \, dx = \frac{1}{3} \int_0^3 (3x – x^2) \, dx\).
Antiderivative: \(\int (3x – x^2) \, dx = \frac{3}{2} x^2 – \frac{1}{3} x^3\).
Evaluate: \(\left[ \frac{3}{2} x^2 – \frac{1}{3} x^3 \right]_0^3\).
At \( x = 3 \): \(\frac{3}{2} (3)^2 – \frac{1}{3} (3)^3 = \frac{27}{2} – 9 = \frac{9}{2}\).
At \( x = 0 \): 0.
Integral: \(\frac{9}{2}\).
Average: \(\frac{1}{3} \cdot \frac{9}{2} = \frac{3}{2}\).
✅ Answer: C