$f_{avg} = \frac{1}{b-a}\int_a^b f(x)dx$

$s(t) = \int v(t)dt$

$v(t) = \int a(t)dt$

• Net change: $\int_a^b v(t)dt = s(b) - s(a)$ (signed)
• Total distance: $\int_a^b |v(t)|dt$ (always positive)

Gives total distance traveled, regardless of direction.

$A = \int_a^b [f(x) - g(x)]dx$

where f(x) ≥ g(x) on [a, b].

$A = \int_c^d [h(y) - k(y)]dy$

Divide at intersection points, apply formula on each subinterval.

$A = \int_a^c |f(x) - g(x)|dx$

$V = \int_a^b A(x)dx$

For square cross-sections: $A(x) = [s(x)]^2$ where s(x) is side length.

• Triangle: $A = \frac{1}{2}bh$
• Semicircle: $A = \frac{1}{2}\pi r^2$

$V = \pi \int_a^b [R(x)]^2 dx$

R(x) = radius (distance from curve to axis of rotation).

When rotating around line other than x- or y-axis, adjust radius accordingly.

$V = \pi \int_a^b ([R(x)]^2 - [r(x)]^2)dx$

R(x) = outer radius, r(x) = inner radius.

Account for distance from rotation axis.