Net Change, Distance and a Calculus I Checkpoint

Accumulation retains direction

If $q'(t)=r(t)$ , then $q(b)=q(a)+\int_a^b r(t)dt$ under the usual continuity assumptions on the rate. The initial value supplies the starting amount; the integral supplies the net change. Neither alone determines the final state.

For velocity, positive and negative contributions cancel in displacement. Distance traveled adds the magnitudes, so it integrates $|v|$ . Find the zeros and sign changes of velocity before splitting the interval.

Visual guide

VISUAL GUIDEVelocity sign separates distance and displacement

v(t) = t − 1

For v(t) = t − 1 on [0, 3], motion reverses at t = 1. The signed areas give displacement −1/2 + 2 = 3/2; adding their magnitudes gives total distance 5/2.

Worked example: a turning particle

Let $v(t)=t-2$ m/s for $0\le t\le4$ , with $s(0)=3$ m. An antiderivative is $t^2/2-2t$ , so net displacement is zero and $s(4)=3$ m. This does not mean the particle stayed still.

Velocity is negative on $[0,2)$ and positive on $(2,4]$ . Distance is

-\int_0^2(t-2)dt+\int_2^4(t-2)dt=2+2=4\text{ m}.

At $t=2$ , position is $1$ m and the particle reverses direction. Its speed is zero at that instant, while acceleration is the constant $1$ m/s².

Worked example: an inventory balance

A tank initially holds $10$ liters. Its net inflow over $0\le t\le2$ is $r(t)=3-t$ liters/minute. The accumulated change is $[3t-t^2/2]_0^2=4$ liters, so the final amount is $14$ liters. The average net rate is $4/2=2$ liters/minute.

A negative net rate would reduce the inventory. The model is physically valid only while constraints such as capacity and nonnegative volume are respected. Integration does not remove those constraints.

Mixed checkpoint

Work these before opening the solutions. Each question asks you to choose a method, not just imitate the immediately preceding example.

Evaluate $\lim_{x\to2}(x^2-4)/(x-2)$ and explain whether the original formula is continuous at $2$ .
Differentiate $x^2e^{3x}$ and find its derivative at zero.
Find the absolute extrema of $x^3-3x$ on $[0,2]$ .
Differentiate $A(x)=\int_0^{x^2}(1+t^3)dt$ .
Evaluate $\int_0^1 2x/(1+x^2)dx$ .
A particle has velocity $v(t)=2t-2$ on $[0,2]$ . Find displacement and distance.

Show worked solutions

Cancel to obtain $x+2$ for nearby non-target inputs; the limit is $4$ . The original expression is undefined at $2$ , so it is not continuous there.
Product and chain rules give $e^{3x}(2x+3x^2)$ , which is zero at the origin.
The interior critical point is $1$ . Values at $0,1,2$ are $0,-2,2$ , so the minimum is $-2$ at $1$ and maximum is $2$ at $2$ .
FTC plus the chain rule gives $2x(1+x^6)$ .
Substitution $u=1+x^2$ changes the bounds to $1,2$ , giving $\ln2$ .
Displacement is $[t^2-2t]_0^2=0$ . Split at the zero $t=1$ and integrate absolute velocity to obtain distance $2$ .

What to review next

A limit error points back to one-sided behavior or algebraic simplification. A missing factor in a derivative or variable-bound integral points to the chain rule. A wrong extrema or distance answer often indicates a missed endpoint, domain restriction or sign change. Review that specific step before moving into integration techniques and infinite processes in Calculus II.