How can we recover a cart’s position from velocity 2t+1 m/s? Velocity alone is insufficient: two carts can move identically from different starting places. An antiderivative recovers a quantity from its derivative up to a constant; an initial condition selects that constant. Use derivative rules.

What we recover

An antiderivative F of f on an interval has derivative f at every point there. The integral notation is an indefinite integral and denotes the entire family:

F′(x) = f(x), ∫f(x) dx = F(x) + C.

The equation F′=f, with the function F unknown, is a simple differential equation: it specifies its derivative. Differentiating removes C, so velocity cannot recover it. If F and G have equal derivatives, their difference has derivative zero; on one interval that difference is constant. This accounts for every antiderivative, not just some examples. OpenStax

For our hypothetical model t counts seconds and s metres, 0≤t≤3:

v(t) = 2t + 1, s(t) = (t)^(2) + t + C; s(0) = 5⇒C = 5.

At 2 seconds the position is 11 m. Displacement from the start is 11−5=6 m, not 11 m. Changing the initial condition shifts the whole position graph without changing velocity.

A family with the same velocity

s=t²+t+C; C=0,3,5

0123051015t, ss, m
Hypothetical s=t²+t+C, seconds and metres. Dashed C=0, dotted C=3, orange C=5. Initial points differ; slopes at the same time agree. Uniform scales.

A short table you can check

Differentiate each right-hand side to check it. For real r choose x>0; nonnegative integer powers extend as polynomials.

∫(x)^(r) dx = ((x)^(r + 1))/(r + 1) + C (r ≠ − 1, x > 0).

r=−1 needs a separate rule, since the previous denominator would vanish:

∫(1)/(x) dx = ln⁡∣x∣ + C (x ≠ 0).

Use an interval not crossing zero. The negative and positive half-lines may have different constants. For b>0, b≠1:

∫(e)^(x) dx = (e)^(x) + C; ∫(b)^(x) dx = ((b)^(x))/(ln⁡b) + C.

Trigonometric arguments are radians; denominators in the last line must be nonzero throughout the chosen interval:

∫cos⁡x dx = sin⁡x + C; ∫sin⁡x dx = − cos⁡x + C.

∫(dx)/((cos⁡)^(2)x) = tan⁡x + C; ∫(dx)/((sin⁡)^(2)x) = − cot⁡x + C.

Simple rules

If F′=f, G′=g and a,b are constants, linearity gives:

∫(af + bg) dx = aF + bG + C.

For a linear inner argument, reverse the chain rule:

∫f(kx + b) dx = (F(kx + b))/(k) + C (k ≠ 0).

∫cos⁡(3x + 1) dx = (1)/(3)sin⁡(3x + 1) + C.

Differentiating the sine produces an extra factor 3, hence 1/3 in the antiderivative. This does not permit dividing by an arbitrary variable inner derivative. Check the interval, differentiate your answer and apply the initial condition: the constant C then becomes known.