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alguien ke me ayude con esta inegral?
∫ (3/500)(25-y^2)
y que metodo sera el apropiado?
1 Answer
- 1 decade agoFavorite Answer
(te falto poner el diferencial de dy)
f(x) = ∫ (3/500)(25-y^2)dy (supongo que es asi)
Debes saber que 3/500 es una constante, entonces la puedes sacar de la integral
f(x) = 3/500∫(25-y^2)dy
Ademas:
∫ a dx = a
∫ x^n dx = 1/(n+1) x^(n+1)
Entonces, Aplicando esos conceptos
f(y) = 3/500 [25y - 3y^3]
(No olvidar que son integrales indefinidas y se debe poner una constante C, la cual no se considera aqui porque si se asume un y=0, entonces C = 0)
