§4.3 The definite integral (continued)

Summary

An object's piecewise linear velocity encloses signed areas on $[0,4]$ with net change $P = -2.625$ feet and total distance $D = 3.375$ feet. The object

For $v(t) = t(t-1)(t-3)$ on $[0,4]$, the correct expression for total distance traveled is

For the piecewise linear velocity on $[0,4]$ with net change $-2.625$ feet, the average velocity on the interval is

On $[0,1]$, a function $f$ bounds a triangle of area $1$ above the axis while $g$ bounds a quarter circle of radius $1$ below it. The value of $\int_0^1 [f(x) + g(x)]\,dx$ is

With $f(x) = 3 - x^2$ sitting above $g(x) = 2x^2$ on $[-1,1]$, the exact area between the curves is written as

With net change $-2.625$ feet over four seconds, compute the average velocity in feet per second, rounded to four decimals.

For $v(t) = t(t-1)(t-3)$ on $[0,4]$, the total change in position is approximately $2.665$ feet. Report that value to three decimals.

An object whose net change in position on an interval is negative ends the interval to the left of where it began.