Problem 46 - College Algebra and Calculus: An Applied Approach

Question

Find the area of the region bounded by the graphs of the given equations. $$ y=\frac{-24}{x^{2}-16}, y=0, x=1, x=3 $$

Step-by-Step Solution

Answer

-12ln(7) + 12ln(15)

Step 1: Write down the integral

The area A of the region bounded by the graphs can be represented by the definite integral of y with respect to x from x=1 to x=3, that is \[A = \int_{1}^{3} \frac{-24}{x^{2}-16} dx\]

Step 2: Resolve the integral

The given integral can be resolved using the method of substitution. Let \(u = x^{2}-16\), then \(du = 2x\,dx\). After the substitution and dividing by 2 to balance the equation, we get the following integral \[A = -12 \int_{-15}^{-7} \frac{1}{u}\,du\]

Step 3: Evaluate the integral

The integral of \( \frac{1}{u}\) is \( ln|u|\), so we have \[A = -12[ln|-7| - ln|-15|] = -12ln|\frac{-7}{-15}| = -12ln| \frac{7}{15} |\]

Step 4: Simplify the answer

The natural logarithm can be simplified using properties of logarithms, so we get \[ A = -12ln(7) + 12ln(15) \]