Calculus 1 - Assignment 3 Solutions

Calculus 1 - Assignment 3 Solutions - Pawesuppressl Lorek,...

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Unformatted text preview: Pawesuppressl Lorek, University of Ottawa, MAT 1330D, Winter 2009 Assignment 3, due April 1, 17:30 in class Student Name Student Number Problem 1: [6 points] Find the integral F ( t ) of the function f ( t ) = 6 t 4- 3 t- 6 + 1 t 2 . Find the antiderivative of f ( t ) that satisfies F (1) = 26 5 . Solution F ( t ) = integraldisplay f ( t ) dt = integraldisplay parenleftbigg 6 t 4- 3 t- 6 + 1 t 2 parenrightbigg dt = integraldisplay parenleftbigg 6 t 4- 3 t- 6 + 1 t 2 parenrightbigg dt integraldisplay parenleftBig 6 t 4- 3 t- 1 2- 6 + t- 2 parenrightBig dt = 6 integraldisplay t 4 dt- 3 integraldisplay t- 1 2 dt- 6 integraldisplay dt + integraldisplay t- 2 dt = 6 1 5 t 5- 3 1 1 2 t 1 2- 6 t + 1- 1 t- 1 + c = 6 5 t 5- 6 t 1 2- 6 t- t- 1 + c IF we yet require that F (1) = 26 5 then we have to have F (1) = 6 5- 6- 6- 1 + c = 6 5- 13 + c = c- 59 5 = 26 5 c = 26 + 59 5 = 17 , i.e. F ( t ) = 6 5 t 5 + 6 t 1 2- 6 t- t- 1 + 17 ....
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This note was uploaded on 01/16/2011 for the course MAT 1330 taught by Professor Dumitriscu during the Fall '08 term at University of Ottawa.

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Calculus 1 - Assignment 3 Solutions - Pawesuppressl Lorek,...

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