# What is the area of the region below the graph of f and the lines x=1, x=2?f(x)=1/(x^2 + 5x + 6)

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### 2 Answers

The function f(x) = 1/(x^2 + 5x + 6)

f(x) = 1 / (x^2 + 3x + 2x + 6)

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

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

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

Int[f(x)] = ln |x + 2| - ln | x + 3| + C

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

Between the limits x = 1 and x = 2, we have the area given as

ln (4/5) - ln (3/4)

ln [(4/5)/(3/4)]

=> ln 16/15

**The required area is ln 16/15**

To compute the area of the given region, we'll have to determine the definite integral of f(x).

We'll re-write the denominator of f(x) as a product of linear factors.

We know that: ax^2 + bx + c = a(x-x1)(x-x2), where x1 and x2 are the roots of the quadratic.

We notice that x1 = 2 and x2 = 3.

f(x) = 1/(x+2)(x+3)

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 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 region located under the graph of f(x), the lines x =1 and x = 2 and x axis is**: **A = ln 16/15 square units.**