8.4-PFD-part1

8.4-PFD-part1 - D^ ^ > / Z x2 + 8x − 3 dx . x 3 + 3x 2...

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Unformatted text preview: D^ ^ > / Z x2 + 8x − 3 dx . x 3 + 3x 2 W & W & 1. Evaluate Solution: ∫ C x2 + 8x − 3 x2 + 8x − 3 A B =2 = + 2+ x ( x + 3) xx x+3 x 3 + 3x 2 → x 2 + 8 x − 3 = Ax ( x + 3) + B( x + 3) + Cx 2 = Ax 2 + 3Ax + Bx + 3B + Cx 2 = ( A + C ) x 2 + (3A + B) x + (3B ) → A+C = 1 B = −1 3A = 8 → 8 = 3A − 1 → A = 3 3B = −3 1 = 3 + C → C = −2 −2 1 x2 + 8x − 3 3 −1 ∫ x 3 + 3x 2 dx = ∫ x + x 2 + x + 3 dx = 3ln x + x − 2 ln x + 3 + C Then, 2. Write out the just the form of the partial fraction decomposition of the rational expression. Do not determine the numerical values of the numerator coefficients. 2 x3 − x2 + 1 x 3 ( x + 1)3 ( x 2 + 1) 2 Solution: 2x3 − x2 + 1 ABC D E F Gx + H Ix + J = + 2+ 3+ + + +2 + 3 3 2 2 2 3 2 xx x +1 x +1 x ( x + 1) ( x + 1) x x +1 x +1 x2 + 1 ( )( ) ( ) x3 − x + 3 dx 3. Evaluate ∫ 2 x +x−2 Solution: Long Division: x3 − x + 3 2x + 1 = x − 1+ 2 2 x +x−2 x +x−2 3 x − x+3 2x + 1 2x + 1 →∫ 2 dx = ∫ x − 1 + 2 dx = ∫ ( x − 1) dx + ∫ x 2 + x − 2 dx x +x−2 x + x − 2 2 3 x −1 x + x − 2 x − x + 3 Remainder: 2 x + 1 → Consider 2x + 1 dx : 2 +x−2 u = x2 + x − 2 du = (2 x + 1) dx ∫x 1 2x + 1 dx = ∫ du = ln u = ln x 2 + x − 2 u +x−2 3 x − x+3 2x + 1 x2 →∫ dx = ∫ ( x − 1) dx + ∫ dx = − x + ln x 2 + x − 2 + C ∫x 2 4. Find the area of the region bounded below by y = 7 and above by y = 1 . 16 − x 2 Solution: 3 3 3 3 7 1 1 dx = 2 ∫ dx −14 ∫ A = 2 ∫ 1 − dx = 2 x 0 − 14 ∫ dx 2 2 2 16 − x 0 0 0 16 − x 0 16 − x 3 3 3 3 1 14 1 7 7 4+ x = 6 − ∫ dx + ∫ dx = 6 − ln 4 + x − ln 4 − x = 6 − ln 0 8 0 4+ x 4− x 4 4 4 − x 0 0 7 7 = 6 − ln 7 − ln1 = 6 − ln 7 4 4 3 1 : 16 − x 2 1 1 A B = = + 16 − x 2 (4 − x )(4 + x ) 4 − x 4 + x → 1 = A(4 + x ) = B(4 − x ) = ( A − B ) x + (4 A + 4 B ) PFD for 1 1 when x = 4, A = 8 ; when x = −4, B = 8 → 1 1 1 1 1 1 =8+8= + 2 4 − x 4 + x 8 4 − x 4 + x 16 − x x 2 ( x 2 + 1) values of the coefficients. Solution: 5. Rewrite x 3 − 3x 2 + 2 x − 3 in terms of its partial fraction decomposition. Determine the numerical x 3 − 3x 2 + 2 x − 3 A B Cx + D = + 2+ 2 xx x 2 ( x 2 + 1) x +1 → x 3 − 3x 2 + 2 x − 3 = Ax ( x 2 + 1) + B( x 2 + 1) + (Cx + D ) x 2 = ( A + C ) x 3 + ( B + D ) x 2 + Ax + B A+C = 1 → B + D = −3 → A = 2; B = −3; C = −1; D = 0 ...
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This note was uploaded on 01/30/2011 for the course MTHSC 108 taught by Professor Any during the Spring '08 term at Clemson.

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