What Is the Fundamental Theorem of Calculus?
The Fundamental Theorem of Calculus (FTC) is the bridge between the two main branches of calculus — differentiation and integration. It says that these operations are inverses of each other, and it gives you a practical way to evaluate definite integrals without ever computing a Riemann sum.
The theorem comes in two parts that work together.
F(x) = ∫ₐˣ f(t) dt, then F'(x) = f(x). In words, the derivative of the accumulated area under f from a to x is simply f(x).F is an antiderivative of f on [a, b], then ∫ₐᵇ f(x) dx = F(b) − F(a).Together these two statements unify decades of geometry and algebra into one clean idea: integration undoes differentiation, and differentiation undoes integration.
Before FTC, computing the area under a curve required summing thousands of thin rectangles and taking a limit — a process so brutal that even the area of a parabola took ancient Greek mathematicians like Archimedes years to nail down. FTC reduces that calculation to two evaluations and a subtraction.
Discovered independently by Newton and Leibniz in the late 1600s, FTC is rightly called the theorem of calculus, not merely a theorem of calculus. Every integration technique you'll learn — substitution, integration by parts, partial fractions — is ultimately a tool for finding the antiderivative F that FTC then plugs in.
FTC Part 1 — The Derivative of an Accumulation Function
FTC Part 1 deals with functions defined by integrals with a variable upper limit.
f is continuous on [a, b] and F(x) = ∫ₐˣ f(t) dt for x ∈ [a, b], then F is differentiable on (a, b) and F'(x) = f(x).The key idea is the accumulation function: F(x) measures the net signed area under f from a fixed starting point a up to a moving endpoint x. As x increases, area accumulates; the rate at which it accumulates at the instant x is exactly the height of the curve at x, which is f(x).
This is profoundly intuitive once you see it: if you are filling a pool and the water level at time t is f(t), then the total water added from time a to time x is F(x), and the instantaneous rate of water flowing in right now is F'(x) = f(x).
Watch the variable
Watch the variable. The dummy variable inside the integral (we used t) must be different from the variable in the limit (x). Writing ∫ₐˣ f(x) dx is a notation error — the same symbol is doing two different jobs.
Chain rule extension
Chain rule extension. When the upper limit is a function of x rather than just x, you apply the chain rule. If G(x) = ∫ₐ^{u(x)} f(t) dt, then G'(x) = f(u(x)) · u'(x).
If G(x) = ∫₁^{x²} sin(t) dt, then G'(x) = sin(x²) · 2x.
G'(x) = sin(x²) · 2xWhen both limits are functions of x
When both limits are functions of x: split the integral at any convenient constant. If H(x) = ∫_{u(x)}^{v(x)} f(t) dt, write H(x) = ∫_a^{v(x)} f(t) dt − ∫_a^{u(x)} f(t) dt and apply FTC1 with the chain rule to each piece: H'(x) = f(v(x))·v'(x) − f(u(x))·u'(x).
Why is this useful? Many physics and probability quantities are defined as integrals — the cumulative distribution function in statistics, the position of a particle from its velocity, the total charge from current density. FTC1 says: differentiating one of these accumulation functions just hands you back the integrand. That recovery is the workhorse behind every "rate-of-change of an integral" problem on AP Calc free-response.
FTC Part 2 — Evaluating Definite Integrals With Antiderivatives
FTC Part 2 is the part you'll use most often as a working tool.
f is continuous on [a, b] and F is any antiderivative of f (meaning F'(x) = f(x)), then ∫ₐᵇ f(x) dx = F(b) − F(a). This is usually written with the evaluation bracket: ∫ₐᵇ f(x) dx = [F(x)]ₐᵇ = F(b) − F(a).The bombshell: the impossible-looking Riemann-sum limit that defines the definite integral collapses to a difference of two numbers. You no longer need to slice [a, b] into n rectangles, sum f(xᵢ)Δx for each, and take n → ∞. You just need to find any antiderivative of f and subtract.
Any antiderivative works
Any antiderivative works. If F is an antiderivative, so is F + C for any constant C. The constants cancel: (F(b) + C) − (F(a) + C) = F(b) − F(a). So you don't need a + C when computing definite integrals — only when computing indefinite ones.
Evaluate ∫₁³ x² dx.
The antiderivative of x² is x³/3.
By FTC2, ∫₁³ x² dx = [x³/3]₁³ = 27/3 − 1/3 = 26/3.
26/3Evaluate ∫₀^π sin(x) dx.
The antiderivative of sin(x) is −cos(x).
So ∫₀^π sin(x) dx = [−cos(x)]₀^π = −cos(π) − (−cos(0)) = −(−1) − (−1) = 1 + 1 = 2.
The sine curve over [0, π] sits entirely above the x-axis, so the answer is the geometric area there, which is 2.
2What about discontinuities? FTC2 requires f to be continuous on the closed interval [a, b]. If f has a vertical asymptote or jump inside [a, b], blindly plugging in endpoints gives the wrong answer — you have to break the integral into pieces or treat it as an improper integral. We will see this trap below.
The Intuition — Why FTC Connects Slopes and Areas
FTC can look like magic — area calculation collapsing to a slope calculation. But the geometric reason is illuminating.
Think of the accumulation function F(x) = ∫ₐˣ f(t) dt as the area swept out by f from a to x. To find F'(x₀), ask: how fast is that area growing at x = x₀?
- Push x by a tiny amount
Push
xfromx₀tox₀ + hwithhtiny. The extra area added is a thin sliver of widthhand height approximatelyf(x₀), so the extra area is approximatelyf(x₀) · h. - Form the difference quotient
Therefore
F(x₀ + h) − F(x₀) ≈ f(x₀) · h, and dividing byhgives(F(x₀ + h) − F(x₀)) / h ≈ f(x₀). - Take the limit
Letting
h → 0turns the approximation into equality:F'(x₀) = f(x₀). That's FTC1 in three lines.
And FTC2 follows from FTC1
And FTC2 follows from FTC1. If F is an antiderivative of f, define A(x) = ∫ₐˣ f(t) dt. By FTC1, A'(x) = f(x) = F'(x), so A and F differ by a constant: A(x) = F(x) + C. Plug in x = a: A(a) = 0 (the area from a to a is zero), so 0 = F(a) + C, giving C = −F(a). Then ∫ₐᵇ f(t) dt = A(b) = F(b) − F(a).
The deepest takeaway: derivatives and integrals are mirror operations. Differentiating an accumulation function recovers the integrand. Integrating a derivative recovers the original function (up to its starting value). This duality structures all of calculus — the chain rule has substitution as its integral counterpart, the product rule has integration by parts as its counterpart, partial-fraction differentiation in reverse becomes partial-fraction integration. Once you internalize FTC, you stop seeing differential and integral calculus as two subjects and start seeing them as one.
Step-by-Step — Evaluating Definite Integrals With FTC2
Most integration problems on an exam reduce to a clean procedure.
- Verify continuity on [a, b]
Confirm that
fhas no vertical asymptotes, removable discontinuities, or jumps inside the interval. If it does, the integral is improper and you must split it up or take limits. - Find an antiderivative F
Use the standard rules: power rule for integration (
∫xⁿ dx = xⁿ⁺¹/(n+1) + Cforn ≠ −1),∫1/x dx = ln|x| + C,∫eˣ dx = eˣ + C,∫sin(x) dx = −cos(x) + C,∫cos(x) dx = sin(x) + C. Use u-substitution for composites and integration by parts for products. - Drop the + C
When evaluating a definite integral, the constant cancels in the subtraction, so omit it.
- Plug in the limits and subtract
F(b) − F(a). Work top-minus-bottom carefully; sign errors here are the most common source of wrong answers. - Simplify the result
Combine fractions, evaluate radicals, and reduce exact forms.
Evaluate ∫₀² (3x² − 2x + 1) dx.
Step 1: the integrand is a polynomial, continuous everywhere ✓.
Step 2: antiderivative is F(x) = x³ − x² + x.
Step 3: drop the + C.
Step 4: F(2) − F(0) = (8 − 4 + 2) − (0 − 0 + 0) = 6.
Step 5: already simplified — answer is 6.
6Evaluate ∫₀^{π/2} cos(x)·esin(x) dx.
Let u = sin(x), so du = cos(x) dx.
Change limits: when x = 0, u = 0; when x = π/2, u = 1.
The integral becomes ∫₀¹ eᵘ du = [eᵘ]₀¹ = e − 1.
e − 1Note on changing limits. When you u-substitute inside a definite integral, you have two choices: (a) change the limits to u-values and stay in u all the way to the end, or (b) keep the limits in x and undo the substitution before evaluating. Choice (a) is cleaner and less error-prone. Don't mix the two — that's how students get wrong answers from correct setups.
Common Mistakes and Pitfalls
FTC problems look mechanical, but predictable traps account for most lost points.
Evaluating ∫₋₁¹ 1/x² dx by plugging into the antiderivative −1/x gives −1/1 − (−1/(−1)) = −1 − 1 = −2. But 1/x² is always positive, so a negative answer is impossible. The function has a vertical asymptote at x = 0 inside the interval; this is an improper integral and FTC2 does not apply directly. Always check continuity on the whole interval before computing.
Especially with trig and exponentials. Take your time and use parentheses around both F(b) and F(a).
When the upper limit is u(x), the derivative is f(u(x)) · u'(x), not just f(u(x)).
An indefinite integral ∫f(x) dx represents a family of antiderivatives and needs a + C. A definite integral ∫ₐᵇ f(x) dx represents a number and never has a + C.
The expression ∫ₐˣ f(x) dx is invalid because the same symbol means two different things. Use t (or any other letter) for the dummy: ∫ₐˣ f(t) dt.
The definite integral is signed area: regions below the x-axis contribute negatively. If you want the geometric area (unsigned) between the curve and the x-axis, integrate |f(x)| instead, or split the interval at the zeros and add the absolute values.
If you change the variable, change the limits. If you keep the limits in x, you must back-substitute before evaluating. Mixing the two gives nonsense.
FTC1 requires continuity of the integrand. For integrands with discontinuities, the accumulation function might not be differentiable everywhere, and the simple F'(x) = f(x) rule fails at those points.
Worked Examples and Applications
Find d/dx [∫₂^{x³} ln(1 + t²) dt].
By FTC1 with chain rule: ln(1 + (x³)²) · 3x² = 3x² · ln(1 + x⁶).
3x² · ln(1 + x⁶)Find d/dx [∫_{sin(x)}^{cos(x)} eᵗ² dt].
Split at any constant (say 0): the derivative is e^{cos²(x)} · (−sin(x)) − e^{sin²(x)} · cos(x).
e^{cos²(x)} · (−sin(x)) − e^{sin²(x)} · cos(x)Evaluate ∫₁⁴ (4x³ − 6x + 5) dx.
Antiderivative: F(x) = x⁴ − 3x² + 5x.
F(4) = 256 − 48 + 20 = 228.
F(1) = 1 − 3 + 5 = 3.
Answer: 228 − 3 = 225.
225Evaluate ∫₀^{π/4} sec²(x) dx.
Antiderivative is tan(x).
So [tan(x)]₀^{π/4} = tan(π/4) − tan(0) = 1 − 0 = 1.
1A particle's velocity is v(t) = t² − 4 m/s for 0 ≤ t ≤ 3.
Net displacement = ∫₀³ (t² − 4) dt = [t³/3 − 4t]₀³ = (9 − 12) − 0 = −3 m. The particle ends 3 m behind its start.
Total distance requires splitting at the velocity zero t = 2: ∫₀² |t² − 4| dt + ∫₂³ |t² − 4| dt.
On [0, 2], v < 0, so |v| = 4 − t² and the integral is [4t − t³/3]₀² = 8 − 8/3 = 16/3.
On [2, 3], v > 0, so the integral is [t³/3 − 4t]₂³ = (9 − 12) − (8/3 − 8) = −3 + 16/3 = 7/3.
Total distance = 16/3 + 7/3 = 23/3 ≈ 7.67 m.
Net displacement and total distance differ — a classic distinction that FTC2 makes explicit.
displacement = −3 m; distance = 23/3 mThe average value of f on [a, b] is (1/(b−a)) ∫ₐᵇ f(x) dx — an application of FTC2.
For f(x) = x² on [0, 3]: average = (1/3) · [x³/3]₀³ = (1/3)(9) = 3.
3Application — physics
Application — physics. Work done by a variable force F(x) moving an object from a to b is W = ∫ₐᵇ F(x) dx. FTC2 reduces this to evaluating any antiderivative of the force law at the endpoints — the foundational result that makes mechanical work computable in closed form for spring forces, gravity, and electric fields.
If you're working through FTC problems for AP Calc, Calc 1 homework, or a final exam, scan them with Solver AI for the full setup — antiderivative work, limit substitution, and verification of every sign.