Emplea sumas superiores para encontrar el area limitada por la curva, el eje X las rectas dadas o el intervalo indicado f(x) = 4x + 5 x = 2 x = 5.
En resumen
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Es una integral de la forma V = π ∫ (y² dx, entre x = a, x = b) Para este caso es y² = x⁴ ; a = 0, b = 2 V = π x⁵ / 5, entre 0 y 2 = π 32 / 5 = 20, 1 unidades de volumen Saludos Herminio.
Despejamos x : x = y² / 4 El área pedida es : A = int[f(y) dy, entre 0 y 4] A = int[y² / 4 dy, entre 0 y 4] = [y³ / 12, entre 0 y 4] = 64 / 12 = 5, 33 Saludos Herminio.