Calculus I › Integrals › full formula sheet

d/dx ∫ax f(t) dt = f(x)
Say it: the derivative with respect to x of the integral from a to x of f of t equals f of x

FTC Part 1

Differentiation undoes integration — the upper limit is the star, the lower limit vanishes, and the dummy variable stays inside.

Notation on this page: t is a dummy variable (it lives only inside the integral); a is a constant lower limit; f is assumed continuous.

Before this lesson: Definition of the derivative · Continuity at a point

Where it comes from

Define the accumulation function A(x) = ∫ax f(t) dt: the area swept out from a fixed starting point a up to a moving endpoint x. The question FTC Part 1 answers: how fast is that accumulated area growing right now?

Intuition before formalism. Nudge x by a tiny h. The extra area is a thin sliver between x and x+h:

Before reading on: A(x) = ∫ax f(t) dt is the area swept out so far. Nudge x a tiny bit right: what shape is the new sliver of area — and what are its height and width?

A(x+h) − A(x)
=
∫xx+h f(t) dt ≈ f(x)·h
The sliver is nearly a rectangle: height ≈ f(x) (f barely changes over a tiny width), width = h.
[A(x+h) − A(x)]/h
≈
f(x)
Divide by h: the growth rate of the accumulated area is just the function’s current height, f(x). Letting h → 0 makes “≈” exact.

The tempting confusions, killed with one example. Take f(t) = t², a = 0: A(x) = ∫0x t² dt = x³/3, so A′(x) = x² = f(x). Notice what happened: the a = 0 vanished (it only set where accumulation started — a constant, killed by differentiation), and the t became x (t was always a dummy variable; x is the variable you differentiate with respect to). Guesses like “f(t)” or “something with a in it” die right here.

Derivation

Let A(x) = ∫ax f(t) dt with f continuous. We compute A′(x) from the limit definition. The key move is splitting the integral at x.

A(x+h) − A(x)
=
∫ax+h f − ∫ax f = ∫xx+h f(t) dt
Step 1 — the sliver. Splitting intervals: area a→x+h minus area a→x leaves area x→x+h. The starting point a cancels before we even differentiate.
∫xx+h f(t) dt
=
f(ch)·h,   ch between x and x+h
Step 2 — continuity squeezes. On the tiny interval [x, x+h], continuous f attains values near f(x); precisely, the integral equals f(ch)·h for some ch in the sliver (Extreme Value Theorem + squeeze).
[A(x+h) − A(x)]/h
=
f(ch) → f(x)
Step 3 — divide and limit. The h cancels. As h → 0, ch → x, and continuity gives f(ch) → f(x).
A′(x)
=
f(x)
Step 4 — conclude. The limit of the difference quotient is f(x). Differentiation undid the integration. ∎

Why continuity matters: Step 2 needs f(t) ≈ f(x) throughout the sliver. A jump discontinuity at x would break the squeeze — which is why the theorem assumes f is continuous.

How to use it

Before reading on: the procedure below lists five cases, and one of them is the classic exam trap. Before reading: for d/dx ∫0x2 et dt, a careless student answers ex2. Which differentiation rule must ride along when the upper limit is not bare x?

The procedure, every time:

  1. Check the shape: d/dx of ∫(constant)(x-stuff) f(t) dt. The variable x must be in the upper limit (possibly inside a function g(x)).
  2. Plain case (upper limit exactly x): replace t with x — answer f(x). Ignore the lower limit entirely.
  3. Chain twist (upper limit g(x)): answer f(g(x))·g′(x). The g′(x) factor is the classic exam trap — never drop it.
  4. x in the lower limit: flip first — ∫xb = −∫bx, so d/dx ∫xb f(t) dt = −f(x).
  5. Never integrate first unless the problem demands it — FTC Part 1 is the shortcut.

The chain twist, slowly

For d/dx ∫0x² et dt, think of A(u) = ∫0u et dt with u = x². Then:

d/dx ∫0x² et dt
=
A′(u)·u′
Chain rule with u = x².
=
eu·2x
A′(u) = eu by FTC Part 1; u′ = 2x.
=
2x·ex²
Two moves: plug g(x) into f, then multiply by g′(x).
Common mistake: answering f(t) — leaving the dummy variable in the answer. The derivative is with respect to x, so the answer must be a function of x. If you see a t in your final answer, something went wrong.

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: d/dx ∫0x t³ dt

  1. Check the shape. d/dx of an integral with constant lower limit 0 and upper limit exactly x — plain FTC Part 1.
  2. Replace t with x: f(t) = t³ becomes x³. (The 0 vanishes — constants die under differentiation.)
  3. Verify the slow way: ∫0x t³ dt = x⁴/4, and d/dx [x⁴/4] = x³. Matches ✓
Your turn: Compute d/dx ∫0x t5 dt.

Answer: x5

Plain case: upper limit exactly x, so replace t with x. Verify the slow way: ∫0x t5 dt = x6/6, and d/dx [x6/6] = x5. ✓

Example 2 — nonzero lower limit: d/dx ∫2x sin t dt

  1. Check the shape. Upper limit exactly x — plain case. The lower limit 2 is just a constant.
  2. Replace t with x: sin x. (Why ignore the 2? A(x) = ∫2x differs from ∫0x by the constant ∫02, which differentiation kills.)
  3. Verify: ∫2x sin t dt = −cos x + cos 2; d/dx = sin x. Matches ✓
Common mistake: answering sin x − sin 2 — “plugging the lower limit in.” FTC Part 1 never substitutes the lower limit; it only ever produces f(x).
Your turn: Compute d/dx ∫1x cos t dt.

Answer: cos x

Upper limit exactly x — the lower limit 1 is just a constant, ignore it. Verify: ∫1x cos t dt = sin x − sin 1; d/dx = cos x. ✓

Example 3 — the chain twist: d/dx ∫0x² et dt

  1. Check the shape. Upper limit is x² = g(x), not x — chain twist needed.
  2. Plug g(x) into f: f(g(x)) = ex².
  3. Multiply by g′(x) = 2x: 2x·ex².
  4. Verify the slow way: ∫0x² et dt = ex² − 1; d/dx = 2x·ex² (chain rule). Matches ✓
Common mistake: answering ex² — plugging in g(x) but forgetting ×g′(x). The chain rule always travels with FTC Part 1 when the limit is not bare x.
Your turn: Compute d/dx ∫0x3 et dt.

Answer: 3x2·ex3

Chain twist: g(x) = x3, g′(x) = 3x2. Plug in, then multiply: f(g(x))·g′(x) = ex3·3x2. Verify: ∫0x3 et dt = ex3 − 1; d/dx = 3x2ex3. ✓

Example 4 — x underneath: d/dx ∫x5 t² dt

  1. x is in the lower limit — FTC Part 1 wants x on top. Flip the bounds (which negates): ∫x5 = −∫5x.
  2. Now apply FTC Part 1: d/dx [−∫5x t² dt] = −x².
  3. Answer: −x².
  4. Verify: ∫x5 t² dt = 125/3 − x³/3; d/dx = −x². Matches ✓
Common mistake: answering +x² — forgetting the flip’s minus sign. Whenever x sits underneath, expect a negative.
Your turn: Compute d/dx ∫x2 t3 dt.

Answer: −x3

x is underneath: flip the bounds (negate): −∫2x t3 dt, then FTC Part 1 gives −x3. Verify: ∫x2 t3 dt = 4 − x4/4; d/dx = −x3. ✓

Memorization tips

  • Say it aloud: “dee-dee-ex of the integral from a to x is f of x.” The sentence names every piece: d/dx, the integral, a-to-x, f(x).
  • The dummy stays inside: t never leaves the integral. If your answer contains t, it is wrong — full stop.
  • The lower limit is decoration: a = 0, a = 2, a = 99 — the answer is f(x) regardless. Constants vanish under d/dx.
  • The chain tax: upper limit g(x) → multiply by g′(x). Write “×g′” as a reflex before you finish.
  • x underneath → minus: flip the bounds, take the negative. “Underneath is negative” rhymes on purpose.
  • The slow-way check: when in doubt, integrate first, then differentiate — the answers must match, and the slow way catches every dropped factor.

Final challenge

Five mixed questions — the chain twist, x underneath, and the dummy trap. Score 5/5 and FTC Part 1 is yours.

← Back to the Calculus I formula sheet

How to learn a formula here

  1. 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.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What does FTC Part 1 say?

If A(x) = ∫ax f(t) dt is the accumulated area from a to x, then A′(x) = f(x): differentiating the accumulation function returns the original integrand, with x in place of t.

Why does the lower limit a disappear?

a is a constant — it only shifts where accumulation starts. When you differentiate, additive constants vanish, so a leaves no trace in A′(x) = f(x).

Why is the answer f(x) and not f(t)?

t is a dummy variable — it only lives inside the integral. The derivative is with respect to x (the upper limit), so x is the variable that survives.

What if the upper limit is g(x) instead of x?

Use the chain rule: d/dx ∫ag(x) f(t) dt = f(g(x))·g′(x). The g′(x) factor is the classic exam twist — never drop it.

What is d/dx ∫xb f(t) dt?

−f(x). Flipping the limits negates the integral (∫xb = −∫bx), so differentiating gives −f(x). The minus is mandatory.

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