From a graph to its antiderivative

What is it about?

$F$ is an antiderivative of $f$ if $F'(x) = f(x)$ : the height of the graph of $f$ is the slope of the graph of $F$ . So without calculating:

where $f > 0$ , $F$ increases; where $f < 0$ , $F$ decreases;
a zero of $f$ with a sign change is an extremum of $F$ ( $+ \to -$ : maximum, $- \to +$ : minimum), without a sign change a saddle point;
an extremum of $f$ is an inflection point of $F$ ;
the area between $f$ and the $x$ -axis between two zeros is the change in height of $F$ .

All antiderivatives $F_0 + C$ are vertical shifts of each other (a family of curves; in the slope field the strokes are the same in every column). An initial value $F(x_0) = y_0$ picks one: $C = y_0 - F_0(x_0)$ . Example: $f(x) = 3x^2 - 2$ , $F(1) = 4$ gives $F(x) = x^3 - 2x + 5$ .

Try it

Transfer the slope: press “Let P run along”.
Slope field: drag the initial point, then “Try different C”.
Features step by step, and a zero without sign change.
Area pieces of sin and reverse: F → f.
Everyday life: savings account, lift, CO₂ in the classroom.

Tasks

Task 1: Sketch it yourself

Sketch the antiderivative of $x^2 - 1$ with $F(0) = 0$ , then reveal the solution. Load task

Show solution

Maximum at $-1$ , minimum at $1$ , inflection at $0$ : $F(x) = \frac13 x^3 - x$ . Show

Task 2: Initial value problem

Find the antiderivative of $3x^2 - 2$ with $F(1) = 4$ and compute $F(2)$ . Load task

Show solution

$F(x) = x^3 - 2x + C$ , $1 - 2 + C = 4$ , so $C = 5$ and $F(2) = 9$ . Show

Task 3: Exponential function

Which antiderivative of $e^x$ passes through $(0 \mid 3)$ ? Load task

Show solution

$e^0 + C = 3$ , so $C = 2$ and $F(x) = e^x + 2$ – not $C = 3$ , because $F_0(0) = 1$ . Show

Task 4: Storage basin

Net inflow $q(t) = -0.5t^2 + 4t - 3.5$ m³/h, $V(0) = 20$ m³. When is the basin fullest, and when does it hold the least water, for $0 \leq t \leq 10$ h? Load task

Show solution

$q$ changes sign from $+$ to $-$ at $t = 7$ : $V(7) = \frac{109}{3} \approx 36.33$ m³. Least water at $t = 1$ h and at the end point $t = 10$ h: $V(1) = V(10) = \frac{55}{3} \approx 18.33$ m³. Fastest increase at $t = 4$ . Show

Ideas for the classroom

Use the views as a sequence: transfer the slope → slope field → features → area pieces → reverse → own sketch. In the task view the card only gives hints until the solution is revealed.
Combine with the derivative and fundamental theorem simulations.
Common stumbling blocks: confusing maxima of $f$ and $F$ ; forgetting the constant of integration; reading a zero without sign change as an extremum; setting $C = y_0$ although $F_0(x_0) \neq 0$ .