Calculus I › Integrals › full formula sheet
FTC Part 2
Evaluate areas without Riemann sums — find an antiderivative, plug in the top, subtract the bottom.
Notation on this page: F is any antiderivative of f (so F′ = f). No +C appears — it cancels.
Before this lesson: FTC Part 1 · Integral power rule
Where it comes from
The problem: the definite integral is defined as a limit of Riemann sums — infinitely many rectangles, then a limit. Computing even ∫02 x³ dx that way is a long, painful algebra exercise. FTC Part 2 says: skip all of it. Find any antiderivative F, and the answer is F(b) − F(a).
The tempting shortcut:
Before reading on: the wrong guess says ∫ab f(x) dx = F(b), ignoring the lower limit. But the integral measures area between a and b. Using that area picture, say in one sentence why the lower limit must matter.
Kill it with f(x) = x on [1, 2], using F(x) = x²/2. The guess says ∫12 x dx = F(2) = 2. But the region is a trapezoid with heights 1 and 2 over width 1 — area (1+2)/2 = 3/2, not 2. The guess forgot where the region starts:
And it does not matter which antiderivative you pick: if G = F + C, then G(b) − G(a) = (F(b)+C) − (F(a)+C) = F(b) − F(a). The constant cancels — which is why definite integrals never carry +C.
Derivation
Let G(x) = ∫ax f(t) dt, the accumulation function. By FTC Part 1, G′ = f — so G is an antiderivative, and any other antiderivative F differs from G by a constant. The rest is algebra.
What this really says: Part 1 tells you accumulation differentiates to f; Part 2 tells you any antiderivative is accumulation (up to a constant). Together they weld differentiation and integration into inverse operations.
How to use it
Before reading on: the derivation will show every antiderivative is F(x) = G(x) + C, yet the +C never affects the answer. Before reading: why does the subtraction F(b) − F(a) make the choice of C irrelevant?
The procedure, every time:
- Find an antiderivative F of the integrand (power rule, trig rules, whatever it takes). Use C = 0 — the simplest one.
- Evaluate F(b) − F(a). Top minus bottom, in that order. Write both substitutions explicitly — do not do it in your head.
- Watch the signs. Negative lower limits are the #1 error source: F(−1) means plug −1 into F, parentheses and all.
- No +C. It cancels — writing it is harmless but pointless; forgetting F(a) is fatal.
- Plausibility-check the sign: if f ≥ 0 on [a, b], the answer must be ≥ 0. A negative answer for a positive integrand means an arithmetic slip.
Worked examples
Four problems, easiest first. In each one, read every step — the why of each move is the lesson.
Example 1 — the basic move: ∫02 x³ dx
- Antiderivative: F(x) = x⁴/4 (power rule; C = 0).
- Top minus bottom: F(2) − F(0) = 16/4 − 0 = 4.
- Sanity check: x³ on [0,2] runs from 0 to 8, curving upward, so the area should be well under the 2·8 = 16 rectangle and above the triangle 8. 4 fits. ✓
Your turn: Compute ∫03 x2 dx.
Answer: 9
Antiderivative F(x) = x3/3 (C = 0). Top minus bottom: 27/3 − 0 = 9. Sanity: x2 on [0,3] runs 0→9 curving upward — well under the 27 rectangle, above the 13.5 triangle. ✓
Example 2 — nonzero limits: ∫13 (2x + 1) dx
- Antiderivative: F(x) = x² + x.
- Evaluate both ends: F(3) = 9 + 3 = 12; F(1) = 1 + 1 = 2.
- Subtract: 12 − 2 = 10.
- Geometry check: trapezoid, heights f(1) = 3 and f(3) = 7, width 2: area (3+7)/2·2 = 10. Matches ✓
Your turn: Compute ∫24 (3x − 1) dx.
Answer: 16
Antiderivative F(x) = 3x2/2 − x: F(4) = 24 − 4 = 20; F(2) = 6 − 2 = 4. Subtract: 20 − 4 = 16. Geometry check: trapezoid with heights f(2) = 5, f(4) = 11, width 2: (5+11)/2·2 = 16. ✓
Example 3 — negative limit: ∫−11 x² dx
- Antiderivative: F(x) = x³/3.
- Evaluate carefully: F(1) = 1/3; F(−1) = (−1)³/3 = −1/3.
- Subtract: 1/3 − (−1/3) = 2/3. (Why not zero? Subtracting a negative adds — the two minus signs are doing work.)
- Sanity check: x² ≥ 0 everywhere, so the area must be positive. 2/3 > 0 ✓. (An answer of 0 here always signals a dropped sign.)
Your turn: Compute ∫−22 x2 dx.
Answer: 16/3
Antiderivative x3/3: F(2) = 8/3; F(−2) = −8/3. Subtract: 8/3 − (−8/3) = 16/3. Sanity: x2 ≥ 0 everywhere, so the area must be positive. ✓ (Equivalently: even function, so 2·∫02 x2 dx = 16/3.)
Example 4 — signed area: ∫−10 x³ dx
- Antiderivative: F(x) = x⁴/4.
- Evaluate: F(0) − F(−1) = 0 − 1/4 = −1/4.
- Interpret: negative — correct, because x³ < 0 on [−1, 0]. FTC Part 2 computes signed area; below the axis counts negative.
Your turn: Compute ∫01 x3 dx.
Answer: 1/4
Antiderivative x4/4: F(1) − F(0) = 1/4 − 0 = 1/4. Sanity: positive, since x3 ≥ 0 on [0,1]. ✓
Memorization tips
- Say it aloud: “top minus bottom of any antiderivative.” Six words, the whole theorem.
- The C-cancel: (F(b)+C) − (F(a)+C) = F(b) − F(a). Say it once and you will never write +C in a definite integral again.
- Parentheses for negatives: F(−1) = (−1)³/3, written with the parentheses every time. Half of all FTC2 errors are sign slips on negative limits.
- The sign check: positive integrand → positive answer. If the signs disagree, recheck the arithmetic before anything else.
- Pair the parts: Part 1: d/dx ∫ax = f(x) (differentiate an integral). Part 2: ∫ab = F(b)−F(a) (evaluate an integral). One undoes, one computes.
- Geometry as backup: for lines and parabolas, a trapezoid/triangle area formula double-checks your antiderivative arithmetic in seconds.
Final challenge
Five mixed questions — sign traps, a backwards solve, and the order trap. Score 5/5 and FTC Part 2 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 does FTC Part 2 say?
If F′ = f, then ∫ab f(x) dx = F(b) − F(a). Find any antiderivative, evaluate at the top limit, subtract the value at the bottom limit.
Why can’t I just compute F(b)?
Because the integral measures area from a to b, not from 0. Counterexample: ∫12 x dx = 3/2, but F(2) = 2. The lower limit must be subtracted.
Does it matter which antiderivative I use?
No — any two antiderivatives differ by a constant C, and (F(b)+C) − (F(a)+C) = F(b) − F(a). The C cancels, so pick the simplest one (C = 0).
Do I need +C in a definite integral?
No. The constant cancels in F(b) − F(a), so definite integrals never carry +C.
How is FTC Part 2 different from Part 1?
Part 1 differentiates an integral (d/dx ∫ax f = f(x)); Part 2 evaluates an integral using an antiderivative (∫ab f = F(b) − F(a)). Part 1 undoes ∫ with d/dx; Part 2 computes ∫ with F.
More from the codex
Support the codex
This page is free, with no account and no ads. If it helped you learn, consider supporting the indie dev behind it.
Questions or a bug to report? Email [email protected].