The function \(f(x)=\frac{16}{(x+1)(x-3)^2}\).

(a) Find the area between the graphs of \(f(x)\) and \(g(x)=4-2x\) on \([0,2]\) shown below.

(b) Find the area between the graphs of \(f(x)\) and \(h(x)=\frac{4}{x+1}\) on \([0,2]\) shown below.

Consider the function \(f(x)=\frac{16}{(x+1)(x-3)^2}\) and \(g(x)=4-2x\) on \([0,2]\).

The given graph is,

\(\frac{16}{(x+1)(x-3)^2}=4-2x\)

Solve for \(x\).

\(x=1\)

Set up the area of the region using integration.

\(\begin{align*}A&=\int_0^1[g(x)-f(x)]dx+\int_1^2[f(x)-g(x)]dx\\&=\int_0^1\left [ (4-2x)-\frac{16}{(x+1)(x-3)^2} \right ]dx+\int_1^2\left [ \frac{16}{(x+1)(x-3)^2}-(4-2x) \right ]dx\\&=1.23+2.09\\&=3.32\end{align*}\)

Consider the function \(f(x)=\frac{16}{(x+1)(x-3)^2}\) and \(g(x)=\frac{4}{x+1}\) on \([0,2]\).

The given graph is,

\(\frac{16}{(x+1)(x-3)^2}=\frac{4}{x+1}\)

Solve for \(x\).

\(x=1\)

Set up the area of the region using integration.

\(\begin{align*}A&=\int_0^1[g(x)-f(x)]dx+\int_1^2[f(x)-g(x)]dx\\&=\int_0^1\left [\frac{4}{x+1}-\frac{16}{(x+1)(x-3)^2} \right ]dx+\int_1^2\left [ \frac{16}{(x+1)(x-3)^2}-\frac{4}{x+1} \right ]dx\\&=1.01+1.47\\&=2.48\end{align*}\)