tutorial 6sol

# tutorial 6sol - 1 Solution The integral is split into...

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1. Solution: The integral is split into subintervals + + + + = 5 . 2 1 5 5 . 2 7 5 11 7 1 0 11 0 ) ( ) ( ) ( ) ( ) ( ) ( dx x f dx x f dx x f dx x f dx x f dx x f So, 11 0 ) ( dx x f =area of the rectanglein the interval (0,1) + area of the rectangle in the interval (1,2.5) +area of the triangle in the interval (2.5,5) - area of the triangle in the interval (1,2.5) +area of the circle in the interval (5,7) + area of the ellipse in the interval (7,11) 86 . 0 ) 1 )( 2 ( ) 1 ( 2 5 . 2 1 2 3 5 . 1 3 1 2 = - + × - × + × π 2. Solution: Let + = x dt t t x f 2 1 3 ) ln( ) ( and x u 2 = Then )) ( ( ) ln( ) ln( ) ( 1 3 2 1 3 x u g dt t t dt t t x f u x = + = + = So, 2 ). ln( ). ( )) ( ( ) ln( 3 2 1 3 u u dx du u g du d x u g dx d dt t t dx d x + = = = + ( by FTC part 1) ) 2 8 ln( 2 ) 2 ) 2 ln(( 2 3 3 x x x x + = + = + - + = + - + = + x x x x x x dt t t dx d dt t t dx d dt t t dt t t dx d dt t t dx d 1 3 2 1 3 1

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tutorial 6sol - 1 Solution The integral is split into...

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