Calculate the area of the surface between the graph of f and the lines x = 1, x = 2. f(x) = 1 / ( x + 2 )( x + 3 )

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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

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