Question
Find the area bounded by
x-9/ x^2 - 12x +35
y=0
x=8
x=10
enter an exact answer. Area =?
x-9/ x^2 - 12x +35
y=0
x=8
x=10
enter an exact answer. Area =?
Answers
GPT-4o mini
To find the area bounded by the curve \( y = \frac{x - 9}{x^2 - 12x + 35} \), the lines \( y = 0 \), \( x = 8 \), and \( x = 10 \), we need to calculate the definite integral of the function from \( x = 8 \) to \( x = 10 \).
1. **First, simplify the function \( y = \frac{x - 9}{x^2 - 12x + 35} \)**:
We start by factoring the denominator:
\[
x^2 - 12x + 35 = (x - 5)(x - 7)
\]
Thus, we rewrite \( y \) as:
\[
y = \frac{x - 9}{(x - 5)(x - 7)}
\]
2. **Next, set up the definite integral to find the area**:
The area \( A \) bounded by the curves can be calculated as:
\[
A = \int_{8}^{10} \left(\frac{x - 9}{(x - 5)(x - 7)}\right) \, dx
\]
3. **Now we need to compute \( \int \frac{x - 9}{(x - 5)(x - 7)} \, dx \)**:
We can use partial fractions to simplify the integral:
\[
\frac{x - 9}{(x - 5)(x - 7)} = \frac{A}{x - 5} + \frac{B}{x - 7}
\]
Multiplying through by the denominator, we have:
\[
x - 9 = A(x - 7) + B(x - 5)
\]
Expanding this gives:
\[
x - 9 = Ax - 7A + Bx - 5B = (A + B)x - (7A + 5B)
\]
By equating coefficients, we get the system of equations:
\[
A + B = 1
\]
\[
-7A - 5B = -9
\]
From \( A + B = 1 \), we can express \( B = 1 - A \). Substituting into the second equation:
\[
-7A - 5(1 - A) = -9
\]
\[
-7A - 5 + 5A = -9
\]
\[
-2A - 5 = -9
\]
\[
-2A = -4 \Rightarrow A = 2
\]
\[
B = 1 - 2 = -1
\]
Thus we have:
\[
\frac{x - 9}{(x - 5)(x - 7)} = \frac{2}{x - 5} - \frac{1}{x - 7}
\]
4. **Now we can integrate**:
\[
A = \int_{8}^{10} \left( \frac{2}{x - 5} - \frac{1}{x - 7} \right) dx
\]
This splits into two integrals:
\[
A = \int_{8}^{10} \frac{2}{x - 5} \, dx - \int_{8}^{10} \frac{1}{x - 7} \, dx
\]
Calculating these integrals:
\[
= 2 \left[ \ln |x - 5| \right]_{8}^{10} - \left[ \ln |x - 7| \right]_{8}^{10}
\]
Now substituting the limits:
\[
= 2 \left( \ln |10 - 5| - \ln |8 - 5| \right) - \left( \ln |10 - 7| - \ln |8 - 7| \right)
\]
\[
= 2 \left( \ln 5 - \ln 3 \right) - \left( \ln 3 - \ln 1 \right)
\]
\[
= 2 \ln \frac{5}{3} - \ln 3
\]
5. **Combining terms using logarithm properties**:
\[
= \ln \left( \frac{(5/3)^2}{3} \right) = \ln \left( \frac{25}{9} \right)
\]
Thus, the exact area bounded by the curves is:
\[
\text{Area} = \ln \left( \frac{25}{9} \right)
\]
1. **First, simplify the function \( y = \frac{x - 9}{x^2 - 12x + 35} \)**:
We start by factoring the denominator:
\[
x^2 - 12x + 35 = (x - 5)(x - 7)
\]
Thus, we rewrite \( y \) as:
\[
y = \frac{x - 9}{(x - 5)(x - 7)}
\]
2. **Next, set up the definite integral to find the area**:
The area \( A \) bounded by the curves can be calculated as:
\[
A = \int_{8}^{10} \left(\frac{x - 9}{(x - 5)(x - 7)}\right) \, dx
\]
3. **Now we need to compute \( \int \frac{x - 9}{(x - 5)(x - 7)} \, dx \)**:
We can use partial fractions to simplify the integral:
\[
\frac{x - 9}{(x - 5)(x - 7)} = \frac{A}{x - 5} + \frac{B}{x - 7}
\]
Multiplying through by the denominator, we have:
\[
x - 9 = A(x - 7) + B(x - 5)
\]
Expanding this gives:
\[
x - 9 = Ax - 7A + Bx - 5B = (A + B)x - (7A + 5B)
\]
By equating coefficients, we get the system of equations:
\[
A + B = 1
\]
\[
-7A - 5B = -9
\]
From \( A + B = 1 \), we can express \( B = 1 - A \). Substituting into the second equation:
\[
-7A - 5(1 - A) = -9
\]
\[
-7A - 5 + 5A = -9
\]
\[
-2A - 5 = -9
\]
\[
-2A = -4 \Rightarrow A = 2
\]
\[
B = 1 - 2 = -1
\]
Thus we have:
\[
\frac{x - 9}{(x - 5)(x - 7)} = \frac{2}{x - 5} - \frac{1}{x - 7}
\]
4. **Now we can integrate**:
\[
A = \int_{8}^{10} \left( \frac{2}{x - 5} - \frac{1}{x - 7} \right) dx
\]
This splits into two integrals:
\[
A = \int_{8}^{10} \frac{2}{x - 5} \, dx - \int_{8}^{10} \frac{1}{x - 7} \, dx
\]
Calculating these integrals:
\[
= 2 \left[ \ln |x - 5| \right]_{8}^{10} - \left[ \ln |x - 7| \right]_{8}^{10}
\]
Now substituting the limits:
\[
= 2 \left( \ln |10 - 5| - \ln |8 - 5| \right) - \left( \ln |10 - 7| - \ln |8 - 7| \right)
\]
\[
= 2 \left( \ln 5 - \ln 3 \right) - \left( \ln 3 - \ln 1 \right)
\]
\[
= 2 \ln \frac{5}{3} - \ln 3
\]
5. **Combining terms using logarithm properties**:
\[
= \ln \left( \frac{(5/3)^2}{3} \right) = \ln \left( \frac{25}{9} \right)
\]
Thus, the exact area bounded by the curves is:
\[
\text{Area} = \ln \left( \frac{25}{9} \right)
\]