Test 1 Solution

# 33 1 solution 2 1 x d x 1 x 2 2 2 0 1 2 2 2 4 2 x

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Unformatted text preview: 2 1 2 x3 ∫ dx = ∫ 0 x2 + 1 = 1 dt 2x 2 2 2 x3 1 1 x2 1 t −1 1 1 1 2 1 2 dt = ∫ dt = ∫ dt = ∫ t − dt = [ t 3 / 2 − t1 / 2 ]1 = 2 − 2 − ( − 1) 21 t 21 t 21 3 3 3 t 2x t 2 2. − 33 1 Solution 2: 1 ∫ x d ( x + 1) = x 2 2 2 0 = 1 2 2 2 4 2 x + 1 | − ∫ x + 1 ⋅ d ( x ) = 2 − ∫ t dt = 2 − t 3 / 2 | = 2 − ( 2 − ) 3 3 3 0 1 0 1 2 1 2 2 2 2 − 33 1 (e) x ∫ x + 1 dx Let t = x ⇒ dt = 0 1 2x dx ⇒ dx = 2 x dt 1 x t t2 π π dx = ∫ 2 2 x dt = 2 ∫ 2 dt = 2[t − tan −1 t ] = 2(1 − ) = 2 − ∫ x +1 0 t +1 t +1 4 2 0 0 0 1 1 1 3 [6 pts]. Find the area of the region between the x axis and the curve y = sin −1 x over [0, 1]. Solution 1: use integral of the inverse (see our lecture notes) π Solution 2: use the geometry Area = ⋅ 1 − 2 π /2 ∫ sin ydy = 0 4. Area of an ellipse. (a) [3 pts] Graph the curve: π /2 π π + cos y | = − 1 2 2 0 x2 y 2 + = 1. 4 9 See page 358 (b) [6pts] Find the area of the region bounded by this curve. x2 y 2 y x x + = 1 ⇒ ( )2 = 1 − ( )2 ⇒ y = 3 1 − ( )2 4 9 3 2 2 2 1 1 x 1 Area = 2 ∫ 3 1 − ( ) 2 dx =6 ∫ 1 − t 2 ⋅ 2dt =12 ∫ 1 − t 2 dt =12 ⋅ π ⋅ 12 = 6π 2 2 −2 −1 −1...
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## This note was uploaded on 01/22/2014 for the course MTH 153 taught by Professor Wu during the Fall '10 term at Michigan State University.

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