1 - Solution set for MATH 155, of Assignment 1 Prob. 1...

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Solution set for MATH 155, of Assignment 1 Prob. 1 Given the function as, f ( x ) = x - 2 x 2 + 3 x 3 - 4 x 4 Integrating, we get Z f ( x ) = F ( x ) = x 2 - 2 x 3 3 + 3 x 4 4 - 4 x 5 5 + C . where C is a constant. Prob. 2 Given the function as, f ( x ) = x 1 + x We can rewrite it as, f ( x ) = 1 + x - 1 1 + x = 1 - 1 1 + x Integrating, we get Z f ( x ) = F ( x ) = x - ln ( 1 + x ) + C . where C is a constant. Prob. 3 Given the function as, f ( x ) = x 7 + 1 x 7 Integrating, we get Z f ( x ) = F ( x ) = x 8 8 - 1 6 x 6 + C . where C is a constant. Prob. 4 Given the function as, f ( x ) = e x 2 + e - x 2 Integrating, we get Z f ( x ) = F ( x ) = 2 e x 2 - 2 e - x 2 = 2 ( e x 2 - e - x 2 ). + C where C is a constant. 1
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Prob. 5 Given the equation as, dy dx = 1 + sec x 4 Thus, we have the general solution by finding the antiderivative of the above function. Hence, y = x + 4 tan x 4 + C where C is a constant. Prob. 6
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1 - Solution set for MATH 155, of Assignment 1 Prob. 1...

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