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areaCalculate the area of the surface between the graph of f and the lines x = 1, x =...
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Int f(x)dx = Int dx/(x+2)(x+3)
We'll apply Leibniz-Newton rule to calculate the definite integral.
We'll decompose the fraction into partial fractions.
1/(x+2)(x+3) = A/(x+2) + B/(x+3)
1 = A(x+3) + B(x+2)
We'll remove the brackets:
1 = Ax + 3A + Bx + 2B
We'll combine like terms:
1 = x(A+B) + 3A + 2B
A + B = 0
A = -B
-3B + 2B = 1
-B = 1
B = -1
A = 1
1/(x+2)(x+3) = 1/(x+2) - 1/(x+3)
Int dx/(x+2)(x+3) = Int dx/(x+2) - Int dx/(x+3)
Int dx/(x+2)(x+3) = ln |x+2| - ln|x+3| + C
Int dx/(x+2)(x+3) = ln |(x+2)/(x+3)|
But, Int dx/(x+2)(x+3) = F(2) - F(1)
F(2) = ln |(2+2)/(2+3)|
F(2) = ln 4/5
F(1) = ln |(1+2)/(1+3)|
F(1) = ln 3/4
F(2) - F(1) = ln 4/5 - ln 3/4
F(2) - F(1) = ln (4/5)*(4/3)
F(2) - F(1) = ln 16/15
The area of the surface bounded by the graph of f(x), the lines x =1 and x = 2 and x axis is: A = ln 16/15 square units
Posted by giorgiana1976 on May 24, 2011 at 10:16 PM (Answer #2)
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