The fundamental theorem of calculus, explained

By Ashraf Bhura, Co-Founder, STEM-Tutor · Updated October 2026

The fundamental theorem of calculus (FTC) connects the two halves of calculus: derivatives and integrals. It says that integration and differentiation undo each other, and it is behind almost every integral you evaluate.

Two ideas that turn out to be one

Differentiation measures rates of change: the slope of a curve. Integration measures accumulation: the area under a curve, built up from thin slices. At first these look like unrelated problems. The fundamental theorem shows they are inverse processes. That discovery, made by Newton and Leibniz, is what turned calculus into a practical tool, because it means you can find areas by reversing derivatives instead of adding up infinitely many slices.

Part 1: the derivative of an accumulation function

Define a function that measures the area under f from a fixed starting point a up to a variable point x:

F(x) = ∫_a^x f(t) dt

Part 1 says: F'(x) = f(x). In words, the rate at which the area grows is the height of the curve at the right-hand edge. If you move the edge a tiny distance h, the extra area is a thin strip of width h and height about f(x), so the area changes at rate f(x).

The variable t inside the integral is a dummy variable. Using t avoids confusing it with the x in the upper limit.

Example 1: Part 1 directly

Find d/dx ∫₂^x (t³ + 1) dt.

• By Part 1, the derivative is the integrand evaluated at x.
• x³ + 1. No integration needed.

Example 2: with the chain rule

Find d/dx ∫₀^x² cos t dt.

• The upper limit is x², not x, so apply the chain rule: f(upper limit) × derivative of the upper limit.
• cos(x²) · 2x.

Example 3: a variable lower limit

Find d/dx ∫_x⁵ √(1 + t²) dt.

• Swap the limits to put x on top, which introduces a minus sign: −∫₅^x √(1 + t²) dt.
• Derivative: −√(1 + x²).

Part 2: evaluating definite integrals

If F is any antiderivative of f, meaning F' = f, then

∫_a^b f(x) dx = F(b) − F(a)

This is the rule you use every time you evaluate a definite integral: find an antiderivative, substitute the upper limit, substitute the lower limit, and subtract. The constant of integration cancels, so you can leave it out. Write the antiderivative in square brackets with the limits, so the examiner can follow each step and award method marks.

Example 4: Part 2

Evaluate ∫₁³ (2x + 1) dx.

• Antiderivative: x² + x.
• (9 + 3) − (1 + 1) = 10.

Example 5: a trig integral

Evaluate ∫₀^π sin x dx.

• Antiderivative: −cos x.
• (−cos π) − (−cos 0) = 1 + 1 = 2.

Net change

Part 2 has a powerful consequence. If F'(x) is a rate of change, then the integral of the rate gives the total change in the quantity:

F(b) = F(a) + ∫_a^b F'(x) dx

This is the net change theorem, and AP free-response questions use it constantly: water in a tank, people in a queue, distance from velocity. Start with the initial amount and add the integral of the rate.

Example 6: water in a tank

A tank holds 50 litres at t = 0. Water flows in at r(t) = 4t litres per minute. How much water is in the tank at t = 5?

• Change = ∫₀⁵ 4t dt = [2t²]₀⁵ = 50 litres.
• Amount at t = 5: 50 + 50 = 100 litres.

Example 7: position from velocity

A particle has velocity v(t) = t² − 4 and is at x = 3 when t = 0. Find its position at t = 3.

• Displacement = ∫₀³ (t² − 4) dt = [t³/3 − 4t]₀³ = 9 − 12 = −3.
• Position: 3 + (−3) = 0. Note that displacement is not distance; for total distance travelled, integrate |v(t)|.

AP-style graph questions

A classic AP question gives the graph of f and defines g(x) = ∫₀^x f(t) dt. Use these facts:

• g(x) is the signed area under f from 0 to x. Area above the axis counts as positive, below as negative.
• g'(x) = f(x), so g is increasing where f is positive and decreasing where f is negative.
• g has a relative maximum where f changes from positive to negative, and a relative minimum where f changes from negative to positive.
• g''(x) = f'(x), so g is concave up where f is increasing, with points of inflection where f has a turning point.

Example 8: from a graph

f is a straight line from (0, 2) to (4, −2), and g(x) = ∫₀^x f(t) dt. Find g(2), g(4) and where g has a maximum.

• f crosses zero at x = 2. Area from 0 to 2 is a triangle: ½ × 2 × 2 = 2, so g(2) = 2.
• From 2 to 4 the triangle is below the axis, area −2, so g(4) = 0.
• f changes from positive to negative at x = 2, so g has its maximum at x = 2.

Common mistakes

• Forgetting the chain rule when the upper limit is a function of x.
• Forgetting the minus sign when x is in the lower limit.
• Subtracting in the wrong order: it is F(upper) − F(lower).
• Treating area below the axis as positive when finding g(x).
• Leaving out the initial value in net change problems.

Practice questions

Try each one before opening the answer.

Frequently asked questions

Is the fundamental theorem of calculus on AP Calculus AB?

Yes. It is in Unit 6 of both AB and BC, and accumulation function questions based on it appear on almost every free-response section.

Is it in A level Maths?

A level Maths uses Part 2 every time you evaluate a definite integral, though it is not usually named or tested as a theorem.

Key takeaways

• Part 1: d/dx ∫_a^x f(t) dt = f(x).
• With a function in the limit, multiply by its derivative (chain rule).
• Part 2: ∫_a^b f(x) dx = F(b) − F(a).
• Net change: final = initial + integral of the rate.
• If g = ∫ f, then g' = f: read g's behaviour from f's graph.

Related guides

• Riemann sums
• U-substitution
• The chain rule