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Worksheet2_1B_Solutions

# Worksheet2_1B_Solutions - Math 1B Worksheet 1 Monday...

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Math 1B Worksheet 1 Monday, January 24 1. Graph the ellipse given by the equation ( x 2 /a 2 )+( y 2 /b 2 ) = 1 and compute its area, where a and b are constants. What happens when a = b ? The equation of the ellipse in the first quadrant is y = b p 1 - ( x/a ) 2 = ( b/a ) a 2 - x 2 , so we can compute the area as follows, using the substitution x = a sin θ , dx = a cos θ dθ : A = 4 Z a 0 b a a 2 - x 2 dx = 4 b a Z π/ 2 0 p a 2 - a 2 sin 2 θ ( a cos θ dθ ) = 4 b a Z π/ 2 0 a 2 cos 2 θ dθ = 2 ab Z π/ 2 0 (1 + cos 2 θ ) = 2 abθ + ab sin 2 θ | π/ 2 0 = πab. 2. Evaluate Z dx ( x - 2) 2 - x 2 + 4 x . First, we complete the square and write - x 2 + 4 x = 4 - ( x 2 - 4 x + 4) = 4 - ( x - 2) 2 . Then, making the substitution x - 2 = 2 sin θ , dx = 2 cos θ dθ , we get Z dx ( x - 2) 2 - x 2 + 4 x = Z dx ( x - 2) 2 p 4 - ( x - 2) 2 = Z 2 cos θ dθ 4 sin 2 θ p 4 - 4 sin 2 θ = Z 2 cos θ dθ (4 sin 2 θ )(2 cos θ ) = 1 4 Z sin 2 θ = 1 4 Z csc 2 θ dθ = - cot θ 4 + C = - 4 x - x 2 4( x - 2) + C. 3. Evaluate Z 16 - e 2 x dx . 1

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Making the substitution e x = 4 sin θ , we have e x dx = 4 cos θ dθ , or dx = cot θ dθ since e x = 4 sin θ . Then, we have Z 16 - e 2 x dx = Z p 16 - 16 sin 2 θ (cot θ dθ ) = Z 4 cos
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Worksheet2_1B_Solutions - Math 1B Worksheet 1 Monday...

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