Learning goals

What you should understand

  • State both parts of the theorem with their continuity and antiderivative conditions.
  • Differentiate an integral whose upper bound depends on x.
  • Evaluate a definite integral and interpret the sign of the result.
  • Recognize when a discontinuity or variable bound requires extra care.
Before you begin

You should know continuity, definite integrals as signed accumulation, antiderivatives, and the chain rule for a variable upper bound.

Two statements, one inverse relationship

Part 1 says the instantaneous change in accumulated signed area equals the current rate. Part 2 says total signed accumulation can be found by evaluating an antiderivative at the endpoints.

Part 1: differentiate accumulationddx∫axf(t) dt=f(x)\frac{d}{dx}\int_a^x f(t)\,dt=f(x)
Part 2: evaluate a definite integral∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx=F(b)-F(a)
  • For the standard statement, f is continuous on the relevant interval.
  • For Part 2, F prime equals f on the interval.
  • A composed upper bound requires the chain rule.

Reusable process

Match the part to the task

  1. 1
    Identify the output

    A derivative of an integral calls for Part 1. A numerical definite integral calls for Part 2.

  2. 2
    Check continuity and interval

    Make sure the integrand satisfies the theorem's hypotheses across the relevant interval.

  3. 3
    Handle variable bounds

    Replace the integrand variable with the upper-bound expression and multiply by its derivative.

    ddx∫ag(x)f(t) dt=f(g(x))g′(x)\frac{d}{dx}\int_a^{g(x)}f(t)\,dt=f(g(x))g'(x)
  4. 4
    Handle endpoint evaluation

    Find a verified antiderivative and subtract lower from upper.

    F(b)−F(a)F(b)-F(a)
  5. 5
    Interpret and check

    A negative definite integral is possible because the integral measures signed accumulation rather than unsigned area.

Accumulation thought experiment

What changes when the upper bound moves?

Imagine increasing x by a small amount h. The new accumulated strip has approximate area f(x) times h. Dividing by h suggests the derivative of the accumulation.

1h∫xx+hf(t) dt→f(x)\frac1h\int_x^{x+h}f(t)\,dt\to f(x)
2
A rate curve and its accumulated areaThe shaded area runs from a to x equals 2 and represents the accumulation function.

The shaded area from a to x = 2 is the value of the accumulation function. Moving x changes that value.

Check the reasoning

Continuity makes the average value of f on the shrinking interval approach f(x). This is the central intuition behind Part 1.

Worked examples

Build from recognition to verification

Differentiate an accumulation function

The upper bound is x squared, so apply Part 1 and then the chain rule.

G(x)=∫1x2cos⁡(t) dtG(x)=\int_1^{x^2}\cos(t)\,dt
  1. Evaluate the integrand at the bound: Replace t with x squared.
    cos⁡(x2)\cos(x^2)
  2. Differentiate the bound: The derivative of x squared is 2x.
    ddx(x2)=2x\frac{d}{dx}(x^2)=2x
  3. Multiply: The chain rule completes the derivative.
    G′(x)=2xcos⁡(x2)G'(x)=2x\cos(x^2)

Answer

G′(x)=2xcos⁡(x2)G'(x)=2x\cos(x^2)

Check: An explicit antiderivative is sine t, so G(x) = sine of x squared minus sine 1; differentiating gives the same result.

Evaluate a definite integral

The polynomial is continuous and has a simple antiderivative.

∫023x2 dx\int_0^2 3x^2\,dx
  1. Find an antiderivative: The derivative of x cubed is 3x squared.
    F(x)=x3F(x)=x^3
  2. Evaluate upper minus lower: Apply Part 2 at 2 and 0.
    F(2)−F(0)=8−0F(2)-F(0)=8-0
  3. State the accumulated value: The integrand is nonnegative on the interval, so the positive sign is consistent.
    88

Answer

∫023x2 dx=8\int_0^2 3x^2\,dx=8

Check: A Riemann-sum or numerical quadrature check approaches 8, independently supporting the antiderivative evaluation.

Common mistakes

Notice the decision before the algebra

Forgetting the chain rule

Replacing t with x squared but not multiplying by 2x.

Fix: Differentiate the upper bound after evaluating the integrand there.

Swapping endpoints

Computing F(a) minus F(b).

Fix: Write upper minus lower before inserting numbers.

Calling signed integral area

Assuming a definite integral must be positive.

Fix: Distinguish signed accumulation from total geometric area, which may require splitting and absolute values.

Ignoring discontinuities

Applying the standard theorem across a point where the integrand is not continuous.

Fix: Inspect the interval, split improper integrals where appropriate, and use the definition needed for that case.