Question

Find the area of the region under the graph of f on [a, b]. f(x) = x2 − 8x + 17; [−1, 2]

Answer 1

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:

$A = \int\limits^a_b {f(x)} \, dx$

Applying to this question:

$A = \int\limits^-1_2 {x^2 - 8x + 17} \, dx$

Then

$F(x) = \frac{x^{3}}{3} - 4x^{2} + 17x$

$A = F(2) - F(-1)$

$F(2) = frac{2^{3}}{3} - 4*2^{2} + 17*2 = \frac{8}{3} + 18 = \frac{8 + 3*18}{3} = \frac{62}[3}$

$F(-1) = frac{(-1)^{3}}{3} - 4*(-1)^{2} + 17*(-1) = -\frac{1}{3} - 21 = \frac{-1 - 21*3}{3} = -\frac{64}[3}$

$A = F(2) - F(-1) = \frac{62}[3} - (-\frac{64}[3}) = \frac{62+64}{3} = 42$

The area is 42 area units.