PostClass 7.4-solutions

Eakin ace669 postclass 74 morales 56430 consequently

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Unformatted text preview: ven by the integral 3 2 1− 0 1 1 7 + 8 4−x 4+x = 2 x− 7 4+x ln 8 4−x dx 3 0 . eakin (ace669) – PostClass 7.4 – morales – (56430) Consequently, Area = 6− 008 7 ln 7 sq.units . 4 10.0 points Evaluate the definite integral 1 I= 0 1. I = ln 7 3 3. I = ln 3 8 4. I = ln 8 7 5. I = ln 7 8 6. I = ln 5 dx . +x−6 8 3 2. I = ln x2 3 7 correct Explanation: By partial fractions, x2 5 1 1 = − . +x−6 x−2 x+3 Thus 1 I= 0 1 dx − x−2 1 0 1 dx . x+3 But 1 0 1 dx = x−2 l n ( | x − 2| ) 1 dx = x+3 l n ( | x + 3| ) 1 0 , while 1 0 1 0 . Consequently, I= ln x−2 x+3 1 = ln 0 3 8 . 5...
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