Connect accumulation and derivatives

• Interpret an integral with a variable upper limit as an accumulation function.
• Explain why a thin added interval makes the accumulation derivative equal to the current rate.
• Evaluate a definite integral using any antiderivative and endpoint subtraction.

Build the idea

Worked example

1. Step 1

   An antiderivative of two x plus one is x squared plus x because its derivative returns the integrand.

   $$F(x)=x^2+x$$

2. Step 2

   At three the antiderivative is twelve, and at one it is two. These values represent accumulation relative to a shared reference.

   $$F(3)=12,\quad F(1)=2$$

3. Step 3

   Subtract lower from upper to isolate the accumulation over the requested interval. The result is ten.

   $$\int_1^3(2x+1)\,dx=12-2=10$$

• Adding the antiderivative values at the two endpoints.

  Accumulation over an interval is the later stored total minus the earlier stored total, so use upper minus lower.

• Forgetting that a definite integral records signed rather than always positive area.

  Contributions below the horizontal axis are negative; geometric area requires absolute-value handling by region.

Evaluate the integral from zero to two of three x squared minus two and check the sign of the result.

This reviewed reading order covers the lesson explanation, equations, example, and checks without claiming that a video exists.

1. Part 1

   Lesson overview

   The Fundamental Theorem of Calculus connects local rate and accumulated change: differentiating an accumulated area recovers the integrand, and antiderivatives evaluate net accumulation. An accumulation function starts at a fixed input and totals signed contributions up to x. Moving x a small distance adds a thin slice whose height is approximately the current integrand value. Interpret an integral with a variable upper limit as an accumulation function.

2. Part 2

   Visual model and equations

   The added slice divided by its width approaches that height as the width shrinks. This makes the derivative of the accumulation function equal to the integrand under continuity conditions. Shade the region under a continuous curve from a fixed point a to a movable point x. Shifting x right adds a narrow rectangle-like slice whose area is approximately f of x times the shift. The derivative with respect to x of the integral from a to x of f of t d t equals f of x. The integral from a to b of f of x d x equals F of b minus F of a.

   • The derivative with respect to x of the integral from a to x of f of t d t equals f of x.
   • The integral from a to b of f of x d x equals F of b minus F of a.
3. Part 3

   Worked example

   Evaluate the integral from one to three of two x plus one, and interpret the result as accumulated change. Find an antiderivative, evaluate it at the upper and lower limits, subtract in the correct order, and compare with geometric area. An antiderivative of two x plus one is x squared plus x because its derivative returns the integrand. At three the antiderivative is twelve, and at one it is two. These values represent accumulation relative to a shared reference. Subtract lower from upper to isolate the accumulation over the requested interval. The result is ten. The accumulated change from x equals one to x equals three is ten. The graph is positive and forms a trapezoid with parallel heights three and seven over width two. Its area is one half times their sum times two, also ten.

   • F of x equals x squared plus x.
   • F of three is twelve and F of one is two.
   • The integral from one to three of two x plus one d x equals twelve minus two, which is ten.
4. Part 4

   Checks, practice, and scope

   Conversely, an antiderivative stores accumulated change. Subtracting its value at the lower endpoint from its value at the upper endpoint gives the definite integral. Adding the antiderivative values at the two endpoints. Accumulation over an interval is the later stored total minus the earlier stored total, so use upper minus lower. Forgetting that a definite integral records signed rather than always positive area. Contributions below the horizontal axis are negative; geometric area requires absolute-value handling by region. Evaluate the integral from zero to two of three x squared minus two and check the sign of the result. Use x cubed minus two x as an antiderivative and subtract the value at zero. The integral equals four, representing positive net accumulation after below-axis and above-axis contributions combine. A variable-upper-limit integral is an accumulation function. Differentiating accumulation recovers the current input rate under suitable continuity. Endpoint subtraction with an antiderivative evaluates net signed accumulation. The stated derivative form assumes continuity near the point; more general integrable functions require refined hypotheses. This lesson does not develop improper integrals, numerical quadrature, or measure-theoretic integration.

Scope and limitations

• The stated derivative form assumes continuity near the point; more general integrable functions require refined hypotheses.
• This lesson does not develop improper integrals, numerical quadrature, or measure-theoretic integration.