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1) The document contains two parts. The first part asks to solve several indefinite integrals by simplifying the results as much as possible. The second part asks to solve application problems involving integrals, such as determining the velocity of an object over time and calculating population growth over time. 2) One application problem involves calculating the average volume of a respiratory cycle over 5 seconds using an integral. Another asks to find the average value of a memory function for a child between their first and second birthdays. 3) The solutions involve setting up and evaluating the appropriate indefinite integrals using techniques like substitution and integration by parts, as well as determining constants based on initial conditions.
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The document contains solutions to several math problems from a CBSE X Mathematics exam in India from 2012.
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Εξισώσεις και προβλήματα, , Α.4.1, Η έννοια της εξίσωσης, Οι εξισώσεις, α + x = β, x – α = β,
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Άρθρο του μαθήματος: https://wp.me/PJcUD-8DN/
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Hermes
- 1. PARTE I Resolver las siguientes integrales indefinidas. Favor simplificar al máximo los resultados. (𝑥 + 1)(3𝑥 − 2) 𝑑𝑥 = (3𝑥 + 𝑥 − 3) 𝑑𝑥 = 3 𝑥 𝑑𝑥 + 𝑥 𝑑𝑥 − 3 𝑑𝑥 = 𝑥 + 𝑥 2 − 3𝑥 + 𝐶 𝑥 − 3𝑥 + 5 𝑥 𝑑𝑥 = 𝑥 𝑥 𝑑𝑥 − 3 𝑥 𝑥 𝑑𝑥 + 5 1 𝑥 𝑑𝑥 = 𝑑𝑥 − 3 1 𝑥 𝑑𝑥 + 5 1 𝑥 𝑑𝑥 = 𝑥 + 3 𝑥 − 5 3𝑥 + 𝐶 6𝑥 (4𝑥 − 9) 𝑑𝑥 𝑢 = 4𝑥 − 9 ⇒ 𝑑𝑢 = 12𝑥 𝑑𝑥 = 1 2 1 𝑢 𝑑𝑢 = − 1 4𝑢 = − 1 4(4𝑥 − 9) + 𝐶 2𝑥 (𝑥 − 1) 𝑑𝑥 𝑢 = 𝑥 − 1 ⇒ 𝑑𝑢 = 𝑑𝑥 𝑢 + 1 = 𝑥
- 2. = 2𝑥 𝑢 𝑑𝑢 = 2 𝑢 + 1 𝑢 𝑑𝑢 = 2 𝑢 𝑢 𝑑𝑢 + 2 1 𝑢 𝑑𝑢 = 2 1 𝑢 𝑑𝑢 + 2 1 𝑢 𝑑𝑢 = 2𝑙𝑛(𝑢) − 2 𝑢 = 2𝑙𝑛(𝑥 − 1) − 2 𝑥 − 1 + 𝐶 𝑙𝑛(𝑥) 𝑥 𝑑𝑢 𝑢 = 𝑙𝑛(𝑥) ⇒ 𝑑𝑢 = 1 𝑥 𝑑𝑥 𝑒 = 𝑥 ⇒ 𝑥 = 𝑒 = 𝑢 𝑥 𝑑𝑢 = 𝑢 𝑒 𝑑𝑢 = 𝑢𝑒 𝑑𝑢 Aplicando integración por partes se obtiene 𝑧 = 𝑢; 𝑑𝑣 = 𝑒 𝑑𝑢 𝑑𝑧 = 𝑑𝑢; 𝑣 = − 𝑒 3 = − 𝑢𝑒 3 − − 𝑒 3 𝑑𝑢 = − 𝑢𝑒 3 + 1 3 𝑒 𝑑𝑢 = − 𝑢𝑒 3 − 𝑒 9 = − 𝑢 3𝑒 − 1 9𝑒
- 3. = − 𝑙𝑛(𝑥) 3𝑥 − 1 9𝑥 PARTE II Resolver los siguientes problemas sobre aplicaciones de las integrales. Determine la función de velocidad de una piedra que es lanzada verticalmente hacia arriba y en la tierra (𝑎 = −9.8 𝑚/𝑠 ). Para tal caso considérese que la pelota alcanza una velocidad de 31 𝑚/𝑠 en 5 segundos. Determine la velocidad de la pelota cuando han pasado 8 segundos. Solución 𝑎 = 𝑑𝑣 𝑑𝑡 𝑎𝑑𝑡 = 𝑑𝑣 𝑎𝑑𝑡 = 𝑑𝑣 −9.8𝑑𝑡 = 𝑑𝑣 −9.8 𝑑𝑡 = 𝑑𝑣 −9.8𝑡 = 𝑣(𝑡) + 𝐶 𝑣(𝑡) = −9.8𝑡 + 𝐶 La condición del problema es, la pelota alcanza una velocidad de 31 𝑚/𝑠 en 5 segundos. Entonces, 𝑣(5) = 31 𝑚 𝑠 𝑣(5) = −9.8(5) + 𝐶 −9.8(5) + 𝐶 = 31 𝐶 = 31 + 9.8(5) = 80 Por lo tanto, la velocidad de la pelota en función del tiempo es 𝑣(𝑡) = −9.8𝑡 + 80 Determine la velocidad de la pelota cuando han pasado 8 segundos. 𝑣(8) = −9.8(8) + 80 𝑣(8) = 1.6 𝑚 𝑠
- 4. La razón de cambio del crecimiento de una población de peces es proporcional a la siguiente expresión: 𝑑𝑁 𝑑𝑡 = 100 √𝑡 Donde N es el tamaño de la población y t es el tiempo en días. Considérese que el tamaño inicial (t=0) de la población es igual a 500 peces. Integrar para encontrar aproximadamente la cantidad de peces después de 7 días. Solución 𝑑𝑁 𝑑𝑡 = 100√𝑡 𝑑𝑁 = 100√𝑡𝑑𝑡 𝑑𝑁 = 100 √𝑡𝑑𝑡 𝑁(𝑡) = 200𝑡 3 + 𝐶 𝑁(𝑡) = 200√𝑡 3 + 𝐶 el tamaño inicial (t=0) de la población es igual a 500 peces. Entonces, 𝑁(0) = 500 Por lo tanto, 𝑁(0) = 200√0 3 + 𝐶 200√0 3 + 𝐶 = 500 𝐶 = 500 Es decir, la función que representa el crecimiento de peces es 𝑁(𝑡) = 200√𝑡 3 + 500 la cantidad de peces después de 7 días. 𝑁(7) = 200√7 3 + 500 𝑁(7) ≈ 1734.68 Peces.
- 5. Solución Un ciclo respiratorio es de 5 segundos, entonces, 0 < 𝑡 < 5 𝑉 = 0.1729𝑡 + 0.1522𝑡 − 0.0374𝑡 Por lo tanto, el volumen medio es: 𝑉 = 1 𝑏 − 𝑎 𝑓(𝑡) 𝑑𝑡 𝑉 = 1 5 − 0 (0.1729𝑡 + 0.1522𝑡 − 0.0374𝑡 ) 𝑑𝑡 𝑉 = 1 5 0.1729𝑡 2 + 0.1522𝑡 3 − 0.0374𝑡 4 𝑉 = 1 5 0.1729(5) 2 + 0.1522(5) 3 − 0.0374(5) 4 𝑉 ≈ 0.5318 Solución Para encontrar el valor medio de la función de memoria de un niño entre el primer y segundo cumpleaños, entonces, 1 < 𝑡 < 2 𝑀(𝑡) = 1 + 1.6𝑙𝑛(𝑡) Por lo tanto, 𝑀 = 1 𝑏 − 𝑎 𝑓(𝑡) 𝑑𝑡
- 6. 𝑀 = 1 2 − 1 1 + 1.6𝑙𝑛(𝑡) 𝑑𝑡 𝑀 = (1.6(𝑡)𝑙𝑛(𝑡) − 1.6𝑡) 𝑀 = 1.6(2)𝑙𝑛(2) − 0.6(2) − 1.6(1)𝑙𝑛(1) − 0.6(1) 𝑀 = 1.6180