Unit 6 · Topic 6.4

6.4 The Fundamental Theorem of Calculus and Accumulation Functions

You can define a new function with an integral, like g(x) = ∫ₐˣ f(t) dt, which measures the area built up from a to x. The Fundamental Theorem of Calculus says the derivative of that function is just f(x), so differentiation and integration undo each other.

Key terms

accumulation function
Fundamental Theorem of Calculus
integrand
chain rule
dummy variable

Accumulation functions

An accumulation function has a variable upper limit: g(x) = ∫ₐˣ f(t) dt. For each input x, g(x) is the signed area under f from a to x. Picture a vertical line sliding to the right from a; g(x) is the signed area that has piled up behind it.

The letter t inside is a dummy variable. It's only a placeholder for the horizontal axis, so ∫ₐˣ f(t) dt and ∫ₐˣ f(s) ds are the same function of x. We use t so it doesn't clash with the x in the upper limit.

Every accumulation function has g(a) = ∫ₐᵃ f(t) dt = 0, because no area has built up yet. That makes a handy point for tangent lines and initial conditions.

The Fundamental Theorem of Calculus (derivative form)

If f is continuous on an interval containing a, then for every x in that interval,

d/dx ∫ₐˣ f(t) dt = f(x).

Why: g(x + h) − g(x) is the area of a thin strip from x to x + h. That strip is about f(x) tall and h wide, so its area is about f(x)·h. Divide by h and let h → 0: the rate at which the area grows is the height of the curve at x.

Notice the lower limit a doesn't matter. Changing a only shifts g up or down by a constant, which doesn't change the derivative.

Variable limits and the chain rule

If the upper limit is a function u(x), use the chain rule: d/dx ∫ from a to u(x) of f(t) dt = f(u(x))·u′(x). Plug the upper limit into f, then multiply by the upper limit's derivative.

If x is in the lower limit, reverse the limits first, which flips the sign: ∫ from x to b of f(t) dt = −∫ from b to x of f(t) dt. So d/dx ∫ from x to b of f(t) dt = −f(x).

If both limits involve x, split the integral at any constant c and handle each part: d/dx ∫ from v(x) to u(x) of f(t) dt = f(u(x))·u′(x) − f(v(x))·v′(x).

Function	Derivative
∫ₐˣ f(t) dt	f(x)
∫ from x to b of f(t) dt	−f(x)
∫ from a to u(x) of f(t) dt	f(u(x))·u′(x)
∫ from v(x) to u(x) of f(t) dt	f(u(x))·u′(x) − f(v(x))·v′(x)

Why this is a big deal

The theorem says every continuous function has an antiderivative: g(x) = ∫ₐˣ f(t) dt is one. That's true even for functions like e^(−t²) whose antiderivatives can't be written with ordinary formulas. You can still differentiate g(x) = ∫₀ˣ e^(−t²) dt (the answer is e^(−x²)), find where it increases, and estimate its values with a calculator.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Basic derivative of an accumulation function
Let g(x) = ∫₁ˣ √(t³ + 1) dt. Find g′(2), and write the tangent line to g at x = 1.
Show the solutionHide the solution
1. Step 1: √(t³ + 1) is continuous for t ≥ −1, which includes the interval from 1 to 2, so the Fundamental Theorem applies: g′(x) = √(x³ + 1).
2. Step 2: g′(2) = √(8 + 1) = √9 = 3.
3. Step 3: For the tangent line at x = 1: g(1) = ∫₁¹ √(t³ + 1) dt = 0, and g′(1) = √2.
4. Step 4: Tangent line: y = 0 + √2(x − 1).
Answer: g′(2) = 3; tangent line at x = 1: y = √2(x − 1).
Example 2
Exam level: both limits depend on x
Let F(x) = ∫ from 3x to x² of √(1 + t⁴) dt. Find F′(x).
Show the solutionHide the solution
1. Step 1: Split at 0: F(x) = ∫ from 0 to x² of √(1 + t⁴) dt − ∫ from 0 to 3x of √(1 + t⁴) dt.
2. Step 2: First piece: plug x² into the integrand and multiply by (x²)′ = 2x: √(1 + (x²)⁴)·2x = 2x√(1 + x⁸).
3. Step 3: Second piece: plug 3x in and multiply by (3x)′ = 3: √(1 + (3x)⁴)·3 = 3√(1 + 81x⁴).
4. Step 4: Subtract.
Answer: F′(x) = 2x√(1 + x⁸) − 3√(1 + 81x⁴)
Example 3
Trap: x in the lower limit
Find d/dx ∫ from x to 5 of sin(t²) dt.
Show the solutionHide the solution
1. Step 1: Reverse the limits to put x on top: ∫ from x to 5 of sin(t²) dt = −∫ from 5 to x of sin(t²) dt.
2. Step 2: Apply the theorem: the derivative is −sin(x²).
3. Step 3: Common wrong answer: sin(x²), with the sign missing.
Answer: −sin(x²)

Common mistakes

Forgetting the chain-rule factor when the upper limit isn't just x. d/dx ∫ from 0 to x² of cos t dt is 2x·cos(x²), not cos(x²).
Hunting for an antiderivative first. You don't need one: the theorem lets you substitute the upper limit straight into the integrand.
Thinking the lower limit appears in the answer. For d/dx ∫ₐˣ f(t) dt, the constant a disappears.
Missing the negative sign when x is the lower limit.

On the exam

Expect multiple-choice questions on derivatives of integrals with x², sin x or 3x as a limit. Plug in the limit, multiply by its derivative.
Free-response questions often define g(x) = ∫ₐˣ f(t) dt and ask for g′(x) or g″(x) at a point. Write g′(x) = f(x) and g″(x) = f′(x), then read values from the given graph or table.

Connected topics

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