Answer:
The area is 42 area units.
Step-by-step explanation:
The area of a function f(x) on an interval [a,b] is given by:
[tex]A = \int\limits^a_b {f(x)} \, dx[/tex]
Applying to this question:
[tex]A = \int\limits^-1_2 {x^2 - 8x + 17} \, dx[/tex]
Then
[tex]F(x) = \frac{x^{3}}{3} - 4x^{2} + 17x[/tex]
[tex]A = F(2) - F(-1)[/tex]
[tex]F(2) = frac{2^{3}}{3} - 4*2^{2} + 17*2 = \frac{8}{3} + 18 = \frac{8 + 3*18}{3} = \frac{62}[3}[/tex]
[tex]F(-1) = frac{(-1)^{3}}{3} - 4*(-1)^{2} + 17*(-1) = -\frac{1}{3} - 21 = \frac{-1 - 21*3}{3} = -\frac{64}[3}[/tex]
[tex]A = F(2) - F(-1) = \frac{62}[3} - (-\frac{64}[3}) = \frac{62+64}{3} = 42[/tex]
The area is 42 area units.
