a b Halla la solución de las integrales a b b x x dx arc tg x k 1 9 1 6 3 4 2 a

A b halla la solución de las integrales a b b x x dx

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a) b) Halla la solución de las integrales. a) b) b) x x dx arc tg x k 1 9 1 6 3 4 2 + = + a) 1 1 2 3 1 2 2 3 2 = + ( ) ( ) x dx arc sen x k x x dx 1 9 4 + 1 1 2 3 2 ( ) x dx 012 b) 1 1 3 3 2 + = + ( ) ( ) x dx arc tg x k a) 1 1 25 1 5 5 2 = + x dx arc sen x k 1 1 3 2 + ( ) x dx 1 1 25 2 x dx 011 d) x cos x dx tg x k 2 2 2 3 1 2 3 ( ) ( ) = + c) ( ) ( ) ( ) x cos x x dx sen x x k + + = + + 1 2 1 2 2 2 2 b) + = + + 3 2 1 3 2 2 1 sen x dx cos x k ( ) ( ) a) 1 1 1 2 cos x dx tg x k ( ) ( ) + = + + x cos x dx 2 2 3 ( ) ( ) ( ) x cos x x dx + + 1 2 2 + 3 2 1 sen x dx ( ) 1 1 2 cos x dx ( ) + 010 11 SOLUCIONARIO
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496 Calcula las siguientes integrales definidas. a) b) Comprueba la siguiente igualdad. Halla el área del triángulo rectángulo que tiene como vértices (0, 0), (4, 0) y (4, 5), utilizando integrales definidas. El triángulo está limitado por el eje X y las rectas x = 4 e . Así, el área es: 4 0 2 0 4 5 4 5 4 2 10 x dx x = = y x = 5 4 Y X 1 1 x = 4 y x = 5 4 015 4 2 2 6 4 2 1 1 4 2 6 ( ) ( ) ( ) ( ) ( ) x x dx x x dx F F F + + + + + = + F ( ) 4 100 3 20 3 96 100 3 268 3 = = + = 6 2 2 1 6 2 96 20 3 268 3 ( ) ( ) ( ) x x dx F F + + = = = F x x x dx x x x ( ) ( ) = + + = + + 2 3 2 1 3 2 6 2 2 4 2 2 6 4 2 1 1 1 ( ) ( ) ( ) x x dx x x dx x x dx + + = + + + + + 014 2 4 2 4 8 2 4 1 2 1 2 = = e dx F F e e x ( ) ( ) b) F x e dx e x x ( ) = = 2 2 1 2 7 2 2 2 1 7 2 469 3 14 3 455 3 ( ) ( ) ( ) x x dx F F + = = = a) F x x x dx x x x ( ) ( ) = + = + 2 3 2 2 1 3 2 4 2 e dx x 7 2 2 2 1 ( ) x x dx + 013 Integrales
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497 Calcula las integrales definidas y . Explica los resultados obtenidos. La función es simétrica respecto del origen de coordenadas. Así, la primera integral da como resultado cero porque el área corresponde a dos regiones iguales, una por encima del eje y otra por debajo. Los valores son iguales, pero de signo contrario, y se anulan. Determina el área limitada por la gráfica de la función f ( x ) = x 2 1 y el eje X en el intervalo [ 2, 2]. Halla el área limitada por la gráfica de la función f ( x ) = sen x y el eje X en el intervalo [0, π ]. Determina el área comprendida entre las funciones f ( x ) = x 2 4 y g ( x ) = x + 2 en el intervalo [ 3, 4]. + + 2 3 2 3 2 2 4 3 2 6 6 6 ( ) ( ) ( ) x x dx x x dx x x d x F F F F F F = + + + = ( ) ( ) ( ) ( ) ( ) ( ) 2 3 3 2 4 3 22 3 9 2 + − + − + = 27 2 22 3 32 3 27 2 53 2 F x x x dx x x dx x x x ( ) ( ( )) ( ) = + = = 2 2 3 2 4 2 6 3 2 6 f x g x x x x x x x ( ) ( ) = = + = = − = 2 2 4 2 6 0 2 3 019 Área = = = + = π π 0 0 1 1 2 sen x dx F F ( ) ( ) F x sen x dx cos x ( ) = = − f x sen x x x ( ) = = = = 0 0 0 π 018 Área = + + ( ) ( ) ( ) 1 2 2 1 1 2 2 1 2 1 1 1 x dx x dx x dx = = + + = = + + F F F F F F ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 1 2 1 2 3 2 3 + + = 2 3 2 3 2 3 2 3 4 F x x dx x x ( ) ( ) = = 2 3 1 3 f x x x ( ) = = = ± 0 1 0 1 2 017 Y X 1 1 1 0 3 4 0 1 4 1 4 x dx x = = 1 1 3 4 1 1 4 1 4 1 4 0 = = = x dx x 1 0 3 x dx 1 1 3 x dx 016 11 SOLUCIONARIO
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498 Halla el área comprendida entre las parábolas f ( x ) = x 2 y g ( x ) = − x 2 + 2.
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