Everything in calculus until now ran from position to velocity to acceleration by differentiating. Integration is the reverse journey: rebuilding the accumulated total from the rate. This chapter builds the intuition for that reversal — adding slices, reading signs, and turning a sum into an area — so the definite integral arrives as an old friend rather than a new boss.
1 · Sigma notation: Σ
Σ is compact notation for 'add a pattern of terms'. Σ_{k=1}^{n} a_k means a_1 + a_2 + … + a_n, where k is the index and n the upper limit. Core rules worth memorizing:
- Σ_{k=1}^{n} c = n·c (constant terms)
- Σ_{k=1}^{n} (a_k ± b_k) = Σ a_k ± Σ b_k (linearity)
- Σ_{k=1}^{n} k·c = c·n(n + 1)/2 (constants factor out)
- Σ_{k=1}^{n} k = n(n + 1)/2 and Σ_{k=1}^{n} k² = n(n + 1)(2n + 1)/6
Sigma notation is not intellectual decoration — it is what the limit of a sum machinery needs in order to talk about 'infinitely many tiny pieces'.
2 · A nice telescoping example
Σ_{k=1}^{n} 1/(k(k + 1))
- Decompose: 1/(k(k + 1)) = 1/k − 1/(k + 1) (partial fractions — check by recombining).
- Write terms: (1/1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) + … + (1/n − 1/(n + 1)).
- Cancel neighbours: every intermediate −1/k kills the next +1/k. Only 1 and −1/(n + 1) survive.
- Result: Σ_{k=1}^{n} 1/(k(k + 1)) = 1 − 1/(n + 1) = n/(n + 1). For n = 10: 10/11 ≈ 0.909.
Telescoping is the same trick behind product rules for sums: explode each term into a difference, and watch the middle collapse. Many definite-integral computations in later chapters rest on exactly this idea.
3 · Displacement vs distance travelled
Velocity is signed: driving forward at 60 km/h for 30 minutes covers +30 km. Reversing at 60 km/h for 15 minutes covers −15 km. Displacement is the net change in position: 30 − 15 = 15 km. Distance travelled ignores sign: 30 + 15 = 45 km. Whenever velocity changes sign, the two answers diverge — and the odometer, unlike the sat-nav, only counts the absolute value.
4 · Three cases of motion
- Constant velocity v: distance = v·t. Simple multiplication.
- Constant acceleration a from rest: v = a·t grows linearly, and distance = ½·a·t² — the average speed (a·t)/2 times t.
- Varying velocity v(t) with a known formula: no single multiplication works. Chop [0, T] into tiny intervals where v is nearly constant; each slice contributes ≈ v(t_i)·Δt; add them all up.
5 · Signed area vs unsigned area
Area between y = x and the x-axis on [−1, 2]
- Below the axis (x from −1 to 0): triangle with legs 1 → area ½, sitting under the axis, so its signed value is −½.
- Above the axis (x from 0 to 2): triangle with legs 2 → signed area +2.
- Signed total: −½ + 2 = 3/2. Unsigned total (geometric area): ½ + 2 = 5/2.
Signed area treats below-axis regions as negative — matching displacement. Unsigned area sums absolute values — matching odometer distance. Which one you want is a question about the problem, not about the curve.
6 · The journey: position from velocity
The fundamental loop of motion analysis: position s(t) → velocity v(t) = s′(t) → acceleration a(t) = v′(t). To run it backwards, accumulate the signed-area slices of v from a start time to t. A continuous velocity, approximated by rectangles on [0, T], gives
distance ≈ Σ v(t_i)·Δt, and displacement = lim (Σ v(t_i)·Δt) as the slices shrink
Two natural approximations sandwich the truth: left-endpoint rectangles (use the velocity at the left edge of each slice) underestimate when v is increasing, and right-endpoint rectangles overestimate in that case. Midpoint rectangles split the difference and usually land closest. The narrower the slices, the tighter the sandwich — squeezing Δt → 0 pins down one exact number: the definite integral.
Letting the slices shrink without bound defines the definite integral:
∫ from a to b of f(x)dx = lim n→∞ Σ_{i=1}^{n} f(x_i)·Δx, Δx = (b − a)/n
This is a preview, not yet a computational engine: the real machine (the Fundamental Theorem of Calculus — differentiate an antiderivative and subtract) arrives in the next chapter. For now: Σ is how we add; the integral sign ∫ is what the add becomes in the limit; f(x) is the height curve; dx is the slice width. Whenever you see ∫, picture an infinitely fine trilogy of rectangles nesting under the curve.
Signed and unsigned answers differ exactly on stretches where f dips below the x-axis. If a problem asks 'total distance covered' and your velocity function goes negative, you must split the interval at the zero of v and add |area| on each piece — simply integrating once gives only the net displacement.
Σ_{k=1}^{n} k² = n(n + 1)(2n + 1)/6
Σ_{k=1}^{n} (a_k ± b_k) = Σ a_k ± Σ b_k
Telescoping: 1/(k(k + 1)) = 1/k − 1/(k + 1)
displacement = Σ v·Δt (signed); distance = Σ |v|·Δt
∫_a^b f(x)dx = lim Σ f(x_i)·Δx
distance under a: from rest, s = ½at²
Common mistakes
- Confusing displacement (signed) with distance travelled (unsigned) when velocity changes sign.
- Forgetting the correction −1/(n+1) in telescoping sums and reporting 1 instead of n/(n + 1).
- Using right-endpoint rectangles and expecting an underestimate for increasing v (they overestimate).
- Applying Σ rules to non-linear patterns like Σ 1/k² with the linear formula for Σ k.
- Reading ∫ as 'area' without splitting at zeros of f — it is signed area, and signs matter.
- Assuming Σ v(t_i)·Δt stops being approximate the moment any formula is available — only the limit Δt → 0 makes it exact.