La función cuya gráfica aparece en el dibujo es f x sen x Considera el recinto

La función cuya gráfica aparece en el dibujo es f x

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La función cuya gráfica aparece en el dibujo es f ( x ) = sen x . Considera el recinto sombreado en la figura y contesta a las siguientes preguntas. a) Halla de forma aproximada su área, y razona cuál puede ser su medida aproximada. b) Calcula su valor mediante el cálculo integral. a) El recinto sombreado corresponde aproximadamente a 4 rectángulos. = + + + + = = x x x x x cos cos π π π π π π π 2 2 2 5 2 1 2 2 5 2 2 1 2 2 4 28 + + = π π π π π π , π π π π π π 2 2 5 2 2 1 1 1 ( ) ( ) + + = sen x dx dx sen x dx b) sen x x x = = = 1 2 5 2 π π A = = 4 3 1 4 3 4 18 π π · , 1 Y X 2 π π 071 c) 5 5 5 1 5 1 1 1 7 7 7 8 ( ( )) ( ) + = + = + f x dx dx f x dx x = + = 28 8 36 b) 5 5 1 1 2 2 2 8 16 ( · ( )) ( ) · f x dx f x dx = = = a) 10 5 10 1 1 5 8 2 10 f x dx f x dx f x dx ( ) ( ) ( ) = + = + = c) 5 1 7 ( ( )) + f x dx b) 5 1 2 ( ( )) f x dx a) 10 1 f x dx ( ) 5 10 1 5 8 2 f x dx f x dx ( ) ( ) = = 070 11 SOLUCIONARIO
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526 Representa gráficamente las siguientes funciones, y halla sus integrales sin utilizar la regla de Barrow. Dada la función f ( x ) = sen x , compara los valores de y el área bajo la curva en el intervalo La integral vale 0 mientras que el área es 4. Representa gráficamente el recinto plano limitado por la parábola y = x 2 4 y la recta que pasa por los puntos A ( 1, 3) y B (3, 5). Calcula su área. 074 Área = + = − + − 0 2 2 2 0 0 π π π sen x dx sen x dx x cos cos x = + = 0 2 1 1 2 π sen x x = = 0 0 π π π π 2 2 2 2 0 = − = sen x dx x cos π π 2 2 , . π π 2 2 sen x dx 073 Y X 1 1 c) 1 2 2 0 1 1 4 4 = = x dx π π · Y X 1 1 b) 3 0 1 1 1 2 2 2 2 5 2 x dx = + = · · Y X 1 1 a) 2 0 1 4 2 2 4 ( ) x dx + = + = c) 1 2 0 1 x dx b) 3 0 1 x dx a) 2 0 1 ( ) x dx + 072 Integrales Y X 1 1
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527 La ecuación de la recta es: y = 2 x 1 Determina el área que encierra una parábola que pase por los puntos: ( 2, 0) (0, 6) y la recta y = x en el intervalo [1, 3]. La ecuación de la parábola es de la forma: y = ax 2 + bx + c Entonces, resulta que: PARA FINALIZAR… Comprueba que los siguientes pares de funciones son primitivas de la misma función. a) b) c) g x sen x sen x x sen x x ' ( ) · · = = = 2 2 2 2 4 cos cos c) f x sen x x sen x x ' ( ) · = = 2 2 4 cos cos g x e e e e e e x x x x x x ' ( ) ( )( ) ( ) = + = 2 2 2 2 b) f x e e e e e e x x x x x x ' ( ) ( )( ) ( ) = + = 2 2 2 2 g x ax a x ' ( ) · = = 1 1 a) f x x ' ( ) = 1 f x sen x g x x ( ) ( ) = = − 2 2 2 y cos f x e e g x e e x x x x ( ) ( ) ( ) ( ) = + = 2 2 y f x x g x ax ( ) ( ) = = ln y ln 076 3 2 6 2 3 6 2 1 1 6 6 6 ( ) ( ) ( ) + + = + + + = x x x dx x dx x dx = − + + − + = x x x x 3 1 6 3 6 3 3 6 3 6 4 6 17 3 9 4 6 10 3 + = + + = + = = ± x x x x x 2 2 6 6 0 6 y x x = − + + 2 6 ( , ) , + = + + 2 0 4 2 0 1 2 25 4 1 4 1 2 a b c a b c = = = − + = 25 4 0 6 6 2 3 2 1 ( , ) c a b a b = − = a b 1 1 1 2 25 4 ,
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