§5.1 Constructing accurate graphs of antiderivatives (continued)

Summary

A particle with $s(0) = 0.5$ has velocity graph enclosing $A_1 = \frac{7}{6}$ on $[0,1]$. Then $s(1)$ equals

The same particle's velocity is a downward parabola with vertex at $t = 3$, positive between its zeros at $1$ and $5$, negative outside. The position $s$ is

For a piecewise linear $f$, the accumulations $A(x) = \int_{-1}^{x}f$, $B(x) = \int_{0}^{x}f$, and $C(x) = \int_{1}^{x}f$ satisfy $\int_{-1}^{0}f = 1$ and $\int_{0}^{1}f = \frac{1}{2}$. The relationship among them is

Compute $s(1)$ as a fraction, given $s(0) = 0.5$ and the signed area $\frac{7}{6}$ accumulating over the first second.

A workout burns calories at piecewise constant rates, and the accumulation function $C$ satisfies $C(5) = 12.5$, $C(10) = 50$, $C(15) = 125$, $C(20) = 187.5$. Compute the calories burned between minute $15$ and minute $20$.

With $B(1) = \frac{1}{2}$ and $B(2) = 0$, compute the signed area of $f$ on $[1,2]$ as a decimal.

If $A(x) = \int_{-1}^{x}f(t)\,dt$ and $B(x) = \int_{0}^{x}f(t)\,dt$ with $\int_{-1}^{0}f(t)\,dt = 1$, then $A(x) = B(x) + 1$.