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I) Hallar una función \( f \) un numero \( a \) tal que \[ \text { b) } \int_{a}^{2 x} \frac{t^{2}}{(f(t))^{3}} d t=x^{2}-81 \]

Ask by Spencer Frazier. in Colombia
Jan 21,2025

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La función \( f(t) \) es \( \sqrt[3]{2t} \) y el número \( a \) es \( 9\sqrt{2} \).

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Bonus Knowledge

Para resolver la integral dada, primero debemos diferenciar ambos lados de la ecuación con respecto a \( x \) utilizando el teorema fundamental del cálculo. Esto nos dará una ecuación en términos de \( f(t) \): \[ \frac{d}{dx} \int_{a}^{2x} \frac{t^{2}}{(f(t))^{3}} dt = \frac{d}{dx}(x^{2} - 81) \] La derivada del lado derecho es \( 2x \) y, aplicando la regla de Leibniz en el lado izquierdo, obtenemos: \[ \frac{(2x)^{2}}{(f(2x))^{3}} \cdot 2 = 2x \] Esto simplifica a: \[ \frac{4x^{2}}{(f(2x))^{3}} = 2x \] Dividiendo ambos lados por \( 2x \) (asumiendo \( x \neq 0 \)): \[ \frac{2x}{(f(2x))^{3}} = 1 \] De aquí, hallamos que \( (f(2x))^{3} = 2x \), por lo que \( f(2x) = (2x)^{1/3} \). Si sustituimos \( 2x \) por \( t \), podemos deducir que: \[ f(t) = \left(\frac{t}{2}\right)^{1/3} \] Por lo tanto, la función que satisface la condición es: \[ f(t) = \frac{t^{1/3}}{2^{1/3}} \]

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