§4.4 The Fundamental Theorem of Calculus (continued)

Summary

A piecewise function $f$ on $[0,5]$ is built from three parabolic or linear pieces, and direct computation of the pieces gives $\int_0^5 f(x)\,dx = \frac{5}{2}$. Its average value on $[0,5]$ is

An aircraft's climb rate table gives $c(h)$ in feet per minute. To estimate total time to climb to $10000$ feet, the right move is to

For an object with position $s$ and velocity $v = s'$, the average velocity $\frac{s(b)-s(a)}{b-a}$ and the average value of $v$ on $[a,b]$ satisfy

An object's velocity on $[12,24]$ integrates to $150$ meters. If velocity is increased everywhere by a constant $c$, the value of $c$ making the distance traveled on $[12,24]$ equal $210$ meters is

On $0 \lt t \lt 4$ the velocity obeys $v(t) = -\frac{1}{4}t^3 + \frac{3}{2}t^2 + 1$, so $a(t) = -\frac{3}{4}t^2 + 3t$, and on $12 \lt t \lt 16$ the velocity is linear with slope $\frac{3}{2}$. The greatest acceleration is

For the piecewise function with average value $\frac{1}{2}$ on $[0,5]$, compute $\int_0^5 f(x)\,dx$ as a decimal.

For the object with velocity graph on $[0,24]$ minutes, the accumulated displacement on $[12,24]$ is $150$ meters. Compute the average velocity on that window.

With $v(t) = -\frac{1}{4}t^3 + \frac{3}{2}t^2 + 1$ on $[0,4]$ and antiderivative $-\frac{1}{16}t^4 + \frac{1}{2}t^3 + t$, compute the exact distance traveled in meters.

For $x \lt 0$, the function $\ln(-x)$ is a valid antiderivative of $\frac{1}{x}$.