Calculus I › Integrals › full formula sheet
Splitting intervals
Cut the interval anywhere — the total is the sum of the pieces. The key that unlocks absolute values and piecewise functions.
Notation on this page: c is the split point — any number you choose, inside [a, b] or not.
Before this lesson: FTC Part 2
Where it comes from
The problem: some integrands change their personality mid-interval. |x−1| is two different lines glued at x = 1; a piecewise function is literally defined in pieces. No single antiderivative formula covers the whole [a, b] — so cut the interval where the behavior changes, integrate each well-behaved piece, and add.
The intuition is pure area: the area from a to b is the area from a to c plus the area from c to b. Nobody would dispute it for floor tiles; integrals just say it in symbols.
The tempting double-count:
Before reading on: the wrong guess adds the whole interval twice: ∫ab f = ∫ac f + ∫ab f. Sketch the areas: which piece gets double-counted, and what should the correct split look like?
Kill it with f(x) = 1 on [0, 2], split at c = 1. The left side is 2 (a 1×2 rectangle). The guess gives ∫01 1 dx + ∫02 1 dx = 1 + 2 = 3 ≠ 2 — the [0, 1] region got counted twice. The rule chains the pieces end to end: the second piece must start where the first ended:
Derivation
Via FTC Part 2: write the integral as F(b) − F(a) and watch the middle terms telescope. The split point c can be anywhere — even outside [a, b].
c outside [a, b]? Still works. Try ∫02 x dx with c = 5: ∫05 x dx + ∫52 x dx = 25/2 + (2 − 25/2) = 2. The second piece is negative (backwards bounds), exactly canceling the overshoot. Signed areas keep the books balanced.
How to use it
Before reading on: ∫03 |x−1| dx has an absolute value with a corner somewhere inside [0,3]. Before reading: where exactly does the inside hit zero — and why must the split happen there?
The procedure, every time:
- Find where the integrand changes behavior: kinks of absolute values (split |x−1| at x = 1), joints of piecewise definitions, points where the formula would need a different antiderivative.
- Split the interval there. One split point → two integrals; two kinks → three integrals.
- Write each piece without the absolute value / with the correct branch. On each sub-interval the integrand is a single well-behaved formula.
- Integrate each piece and add.
Splitting the interval vs. splitting the integrand
Do not confuse this rule with the sum rule ∫(f+g) = ∫f + ∫g. The sum rule splits the integrand (two functions, same interval); this rule splits the interval (one function, two ranges). Different cuts, different rules.
Worked examples
Four problems, easiest first. In each one, read every step — the why of each move is the lesson.
Example 1 — absolute value: ∫03 |x−1| dx
- Find the kink. |x−1| = 0 at x = 1: it equals 1−x for x < 1 and x−1 for x > 1. Split at x = 1.
- Write the pieces: ∫01 (1−x) dx + ∫13 (x−1) dx.
- Integrate: [x − x²/2]01 = 1/2; [x²/2 − x]13 = (9/2 − 3) − (1/2 − 1) = 3/2 + 1/2 = 2.
- Add: 1/2 + 2 = 5/2.
- Geometry check: two triangles — base 1 × height 1 (area 1/2) and base 2 × height 2 (area 2). Total 5/2. Matches ✓
Your turn: Compute ∫04 |x−2| dx.
Answer: 4
Kink where x − 2 = 0, at x = 2: ∫02 (2−x) dx + ∫24 (x−2) dx = [2x − x2/2]02 + [x2/2 − 2x]24 = 2 + 2 = 4. Geometry: two triangles, base 2 × height 2, area 2 each. ✓
Example 2 — piecewise: f(x) = x for x < 0, x² for x ≥ 0; ∫−22 f(x) dx
- Split at the joint x = 0: ∫−20 x dx + ∫02 x² dx.
- First piece: [x²/2]−20 = 0 − 2 = −2. (Negative — correct: x < 0 there.)
- Second piece: [x³/3]02 = 8/3.
- Add: −2 + 8/3 = 2/3.
Your turn: Let f(x) = x2 for x < 0 and f(x) = x for x ≥ 0. Compute ∫−22 f(x) dx.
Answer: 14/3
Split at the joint x = 0: ∫−20 x2 dx + ∫02 x dx = [x3/3]−20 + [x2/2]02 = 8/3 + 2 = 14/3. ✓
Example 3 — split for convenience: ∫04 (2x+1) dx, split at 2
- Split (no kink here — just demonstrating): ∫02 (2x+1) dx + ∫24 (2x+1) dx.
- Each piece: [x²+x]02 = 6; [x²+x]24 = 20 − 6 = 14.
- Add: 6 + 14 = 20. Direct: [x²+x]04 = 20. Matches ✓
Your turn: Compute ∫06 (2x+1) dx by splitting at 3.
Answer: 42
Split: ∫03 (2x+1) dx + ∫36 (2x+1) dx = [x2+x]03 + [x2+x]36 = 12 + (42 − 12) = 42. Direct: [x2+x]06 = 42. ✓
Example 4 — even shortcut: ∫−22 |x| dx
- Split at the kink x = 0: ∫−20 (−x) dx + ∫02 x dx.
- Each piece: [−x²/2]−20 = 2; [x²/2]02 = 2.
- Add: 4. (Shortcut: |x| is even, so 2·∫02 x dx = 4 — same answer, half the work.)
Your turn: Compute ∫−33 |x| dx.
Answer: 9
Split at the kink x = 0: ∫−30 (−x) dx + ∫03 x dx = 9/2 + 9/2 = 9. Shortcut: |x| is even, so 2·∫03 x dx = 9. ✓
Memorization tips
- Say it aloud: “cut anywhere, add the pieces.” The split point is your choice — the sum does not care.
- The kink rule: absolute value |x−a| → split at a. Piecewise with joint at c → split at c. Find where the formula changes, cut there.
- End to end: a→c, then c→b. If your second piece does not start where the first ended, you double-counted or left a gap.
- Unsigned vs signed: |·| forces a split because signed area would cancel. Whenever an absolute value appears, your first reflex is “where is the kink?”
- Even/odd shortcut: symmetric interval + even integrand → double the right half. It is splitting with the kink at 0, done smartly.
- Splitting never changes the value — it only reorganizes the work. Use it freely whenever a piece looks friendlier.
Final challenge
Five mixed questions — kinks, signed areas, and c outside the interval. Score 5/5 and splitting is yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
What is the interval-splitting rule?
∫ab f(x) dx = ∫ac f(x) dx + ∫cb f(x) dx: cut the interval at any point c, integrate each piece, and add. Areas add.
Where should I split the interval?
Wherever the integrand’s behavior changes: at kinks of absolute values (split |x−1| at x = 1), at the joints of piecewise functions, or where the function crosses the axis.
Can c be outside [a, b]?
Yes — the algebra F(b)−F(a) = [F(c)−F(a)] + [F(b)−F(c)] telescopes for any c. Signed areas keep the books balanced even when c is outside.
Why split ∫03|x−1| dx at x = 1?
Because |x−1| changes formula there: it’s 1−x for x < 1 and x−1 for x > 1. Splitting lets each piece use its simple form: ∫01(1−x) dx + ∫13(x−1) dx = 1/2 + 2 = 5/2.
Is splitting the interval the same as splitting the integrand?
No — splitting the integrand ∫(f+g) = ∫f + ∫g is the sum rule (different rule). Splitting the interval cuts the x-range while keeping the same f.
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