## Calculus: Early Transcendentals (2nd Edition)

$$\int_0^1{[-x+2 - (x^2-1)]dx}=\frac{13}{6}$$
Find region bounded by $y = x^2 -1, y= -x+2, x=0, x=1$ The formula we will use here is the following: $$\int_a^b {(top - bottom)}dx$$ We must identify which functions is on top versus on bottom for $x$ in $[0 , 1]$. By graphing both function on between $0$ and $1$. If we do so, we observer $y = -x+2$ is above $y = x^2 -1$, Hence: $$\int_0^1{[-x+2 - (x^2-1)]dx}$$ Simplify $$\int_0^1{(-x^2-x+3)dx}$$ We use a power rule to integrate the quadratic polynomial $$\frac{-x^3}{3} -\frac{x^2}{2}+3x |_0^1$$ Then evaluate the result $$\left(\frac{-(1)^3}{3} -\frac{(1)^2}{2}+3(1)\right) - \left( \frac{-(0)^3}{3} -\frac{(0)^2}{2}+3(0) \right) = \frac{13}{6}-0=\frac{13}{6}$$