Question

Find the area of the region bounded by the graphs of the given equations. $$ y=\frac{12}{x^{2}+5 x+6}, y=0, x=0, x=1 $$

Short answer

The area of the region bounded by the graphs of the given equations is \(12\ln{(\frac{2}{3})}\)

Step-by-step solution

Understand the problem

The problem gives four bounding equations: \(y=\frac{12}{x^{2}+5 x+6}\), \(y=0\), \(x=0\), and \(x=1\). These equations set up the boundaries of the area to find.

Set up the integral

From the equation \(y=\frac{12}{x^{2}+5 x+6}\) the integral can be set up as: \(\int_{0}^{1} \frac{12}{x^{2}+5 x+6} dx\).

Simplify the integrand

The integrand can be simplified by doing a partial fractions decomposition to get: \(\int_{0}^{1} (\frac{6}{x+2} - \frac{6}{x+3}) dx\).

Evaluate the integral

The integral can now be evaluated using the fundamental theorem of calculus: \([\frac{6 \ln|x+2|}{1} - \frac{6 \ln|x+3|}{1} ]_{0}^{1} = 6(\ln2-\ln3)-6(\ln3-\ln2) = 12\ln{(\frac{2}{3})}\).