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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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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f(x) = 1/(x+2)(x+3)   and x=1 and x=2

Let F(x) = intg f(x)

Then the area between f , x=1, and x= 2 is:

A = F(2) - F(1)

First we will integrate f(x):

F(x) = ing (1/(x+2)(x+3)

We will rewrite using partial fractions:

1/(x+2)(x+3) = A/(x+2) + B/(x+3)

==> A(x+3) + B(x+2) = 1

==> (A+B)x + 3A+2B = 1

==> A+b = 0

==> A = -B

==> 3A+ 2B = 1

==> 3(-B) + 2B = 1

==> -3B + 2B = 1

==> -B = 1

==> B =1 ==> A = -1

==> 1/(x+2)(x+3) = 1/(x+3) - 1/(x+2)

==> F(x) = intg (1/(x+3)  dx  - intg 1/(x+2) dx

               = ln (x+3) - ln (x+2)

==> F(x) = ln (x+3)/(x+2)

==> F(2) = ln 5/4 

==> F(1) = ln 4/3

==> A = f(2) - F(1)

           = ln 5/4 - ln 4/3

           = ln (5/4 * 3/4)

           = ln 15/16

==> A = ln 15/16  square units

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The area to be computed is the definite integral of f(x).

Int f(x)dx = Int dx/(x+2)(x+3)

We'll apply Leibniz-Newton to calculate the definite integral.

First, we'll calculate the indefinite integral of f(x). To determine the indefinite integral, we'll write the function as a sum or difference of elementary ratios.

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 located between 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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