Answer to Question #97981 in Calculus for Paul

Question

Find the area enclosed be these three functions: f(x) = 1, g(x) = x + 1, h(x) = 9 − x

Answer 1

$g(x)=h(x) \to x+1=9-x\to x=4.$

$g(x)=f(x)\to x+1=1\to x=0.$

$h(x)=f(x)\to 9-x=1\to x=8.$

$A=\int_0^4(g(x)-1)dx+\int_4^8(h(x)-1)dx=\int_0^4xdx+\int_4^8(8-x)dx=\\ =\frac{x^2}{2}|_{x=0}^{x=4}+(8x-\frac{x^2}{2})|_{x=4}^{x=8}=8.$