8.4-PFD-part1

# 8.4-PFD-part1 - D^ ^ > Z x2 8x 3 dx x 3 3x 2 W W 1 Evaluate...

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MthSc 108-Learning Activity (Part 1) Section 8.4: Integration of Rational Functions by Partial Fractions 1. Evaluate 2 3 2 8 3 3 x x dx x x + - + . Solution: x 2 + 8 x - 3 x 3 + 3 x 2 = x 2 + 8 x - 3 x 2 ( x + 3) = A x + B x 2 + C x + 3 x 2 + 8 x - 3 = Ax ( x + 3) + B ( x + 3) + Cx 2 = Ax 2 + 3 Ax + Bx + 3 B + Cx 2 = ( A + C ) x 2 + (3 A + B ) x + (3 B ) A + C = 1 3 A = 8 3 B = - 3 B = - 1 8 = 3 A - 1 A = 3 1 = 3 + C C = - 2 Then, x 2 + 8 x - 3 x 3 + 3 x 2 dx = 3 x + - 1 x 2 + - 2 x + 3 dx = 3ln x + 1 x - 2ln x + 3 + C 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. 3 2 3 3 2 2 2 1 ( 1) ( 1) x x x x x - + + + Solution: 2 x 3 - x 2 + 1 x 3 ( x + 1) 3 ( x 2 + 1) 2 = A x + B x 2 + C x 3 + D x + 1 + E x + 1 ( ) 2 + F x + 1 ( ) 3 + Gx + H x 2 + 1 + Ix + J x 2 + 1 ( ) 2 3. Evaluate x 3 - x + 3 x 2 + x - 2 dx Solution : Long Division: x 2 + x - 2 x 3 - x + 3 x - 1 Remainder: 2 x + 1 x 3 - x + 3 x 2 + x - 2 = x - 1 + 2 x + 1 x 2 + x - 2 x 3 - x + 3 x 2 + x - 2 dx = x - 1 + 2 x + 1 x 2 + x - 2 dx = ( x - 1) dx + 2 x + 1 x 2 + x - 2 dx Consider 2 x + 1 x 2 + x - 2 dx : u = x 2 + x - 2 du = (2 x + 1) dx 2 x + 1 x 2 + x - 2 dx = 1 u du = ln u = ln x 2 + x - 2 x 3 - x + 3 dx = ( x - 1) dx + 2 x + 1 dx = x 2 - x + ln x 2 + x - 2 + C

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4. Find the area of the region bounded below by y = 7 16 - x 2 and above by y = 1 . Solution: A = 2 1 - 7 16 - x 2 dx 0 3 = 2 dx - 0 3 14 1 16 - x 2 dx 0 3 = 2 x 0 3 - 14 1 16 - x 2 dx 0 3 = 6 - 14 8 1 4 + x dx 0 3 + 1 4 - x dx 0 3 = 6 - 7 4 ln 4 + x - ln 4 - x 0 3 = 6 - 7 4 ln 4 + x 4 - x
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