Question

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

Answer 1

The area is 42 area units.

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) = (x^(3))/(3) - 4x^(2) + 17x`

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

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

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

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

The area is 42 area units.