Sol-162E3-S2004

# Sol-162E3-S2004 - MATH 162 – SPRING 2004 – THIRD EXAM...

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Unformatted text preview: MATH 162 – SPRING 2004 – THIRD EXAM SOLUTIONS Useful formulas: Arc length L = Z b a p ( x ( t )) 2 + ( y ( t )) 2 dt Area of a surface of revolution S = Z b a 2 πy ( t ) p ( x ( t )) 2 + ( y ( t )) 2 dt Some power series: sin x = ∞ X n =0 (- 1) n x 2 n +1 (2 n + 1)! cos x = ∞ X n =0 (- 1) n x 2 n (2 n )! 1 1- x = ∞ X n =0 x n , provided | x | < 1 1) Find which series equals the definite integral R 1 sin( x 2 ) dx A) ∑ ∞ n =0 (- 1) n 1 (2 n +2)! B) ∑ ∞ n =0 (- 1) n 1 (2 n +3)! C) ∑ ∞ n =0 (- 1) n 1 (2 n +1)!(4 n +3) D) ∑ ∞ n =0 (- 1) n- 1 1 (2 n +5! E) ∑ ∞ n =0 (- 1) n 1 (2 n +1)!(4 n +2) Solution: Using the formula given above for the Maclaurin series of sin x, but with x replaced by x 2 , we have sin( x 2 ) = ∞ X n =0 (- 1) n x 4 n +2 (2 n + 1)! 1 2 Therefore Z 1 sin( x 2 ) dx = ∞ X n =0 (- 1) n Z 1 x 4 n +2 (2 n + 1)! dx = ∞ X n =0 (- 1) n 1 (4 n + 3)(2 n + 1)! . The correct answer is C. 2) The power series expansion of 1 (1+ x ) 2 is A) ∑ ∞ n =0 (- 1) n x n B) ∑ ∞ n =0 (- 1) n nx n- 1 C) ∑ ∞ n =0 (- 1) n- 1 nx n- 1 D) ∑ ∞ n =0 (- 1) n- 1 x n E) ∑ ∞ n =0 x n Solution: We know that 1 (1 + x ) 2 =- d dx 1 1 + x and by the formula given above we have 1 1 + x = 1 1- (- x )...
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Sol-162E3-S2004 - MATH 162 – SPRING 2004 – THIRD EXAM...

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