Page 359, #18. Find the total area enclosed by the functions \(f(x)=-x^3+5x^2+2x+1\) and \(g(x)=3x^2+x+3\).
First, we find the points of intersection. \[ -x^3+5x^2+2x+1=3x^2+x+3\\ 0=x^3-2x^2-x+2\\ 0=x^2(x-2)-1(x-2)\\ 0=(x^2-1)(x-2)\\ 0=(x+1)(x-1)(x-2)\\ x=-1,1,2 \]
On [-1,1], \(g(x)>f(x)\), so we integrate \(g(x)-f(x)\). \[ \int_{-1}^1 (3x^2+x+3)-(-x^3+5x^2+2x+1) \,dx\\ \int_{-1}^1 x^3-2x^2-x+2 \,dx\\ =\frac{1}{4}(1)^4-\frac{2}{3}(1)^3-\frac{1}{2}(1)^2+2(1)-(\frac{1}{4}(-1)^4-\frac{2}{3}(-1)^3-\frac{1}{2}(-1)^2+2(-1))\\ =\frac{1}{4}-\frac{2}{3}-\frac{1}{2}+2-(\frac{1}{4}+\frac{2}{3}-\frac{1}{2}-2)\\ =\frac{-4}{3}+4\\ =\frac{-4}{3}+\frac{12}{3}\\ =\frac{8}{3}\\ \]
On [1,2], \(f(x)>g(x)\), so we integrate \(f(x)-g(x)\). \[ \int_{1}^2 (-x^3+5x^2+2x+1)-(3x^2+x+3) \,dx\\ \int_{1}^2 -x^3+2x^2+x-2 \,dx\\ =-\frac{1}{4}(2)^4+\frac{2}{3}(2)^3+\frac{1}{2}(2)^2-2(2)-(-\frac{1}{4}(1)^4+\frac{2}{3}(1)^3+\frac{1}{2}(1)^2-2(1))\\ =-4+\frac{16}{3}+2-4-(-\frac{1}{4}+\frac{2}{3}+\frac{1}{2}-2)\\ =-6+\frac{16}{3}+\frac{1}{4}-\frac{2}{3}-\frac{1}{2}+2\\ =-4-\frac{1}{2}+\frac{14}{3}+\frac{1}{4}\\ =\frac{-48}{12}-\frac{6}{12}+\frac{56}{12}+\frac{3}{12}\\ =\frac{5}{12}\\ \]
Therefore, the total area bounded by the two curves is: \[ A=\frac{8}{3}+\frac{5}{12}\\ A=\frac{32}{12}+\frac{5}{12}\\ A=\frac{37}{12}\approx3.083\\ \]