## anonymous 5 years ago (a) integrate the function over the interval (b) find the area of the region between the graph and the x-axis. y=x^2-6x+8, [0,2]

1. anonymous

(a)$\int\limits_{0}^{2} \ \ x^2 - 6x+8 \ dx = \left[ x^3/3 - 3x^2+8x \right] \ from \ 0 \to 2$$= 2^3/3 - 3(2)^2+8(2) = 20/3$

2. anonymous

(b)$\int\limits_{2}^{4} \ \ x^2 - 6x+8 \ dx = 4/3$ (below the x-axis).

3. anonymous

On part b...where did the 4 and 2 come from?

4. anonymous

The only definite area between the curve and the x-axis is on the interval [2,4].