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📅 May 6, 2026 👤 scraface

May 06, 2026

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📌 The area \( A \) between the curves from \( x = 0 \) to \( x = 2 \) is given by:

📌 A = \int_0^2 \left((4x - x^2) - x^2\right) \, dx = \int_0^2 (4x - 2x^2) \, dx.

📌 Evaluating the integral:

📌 A = \left[ 2x^2 - \frac{2}{3}x^3 \right]_0^2 = \left( 2(2)^2 - \frac{2}{3}(2)^3 \right) - \left( 2(0)^2 - \frac{2}{3}(0)^3 \right).

📌 Calculating each term:

📌 = \left( 8 - \frac{16}{3} \right) = \frac{24}{3} - \frac{16}{3} = \frac{8}{3}.

📌 \boxed{\frac{8}{3}}.

📌 Find the remainder when \( v^4 + 3v + 1 \) is divided by \( v^2 + 1 \).

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