Unit 4 · Lesson 2 of 5

Differentiation and accumulation undo one another.

Accumulated volume or energy changes at the rate entering at its current endpoint. See the accumulated curve and its slope together.

Change the input. Follow the mathematics.

What you'll be able to do

Set up the model and calculate its accumulated integral value.
Connect the equation to the changing visual.
Explain the assumptions and identify a misleading interpretation.

Predict before exploring

What will change if you increase upper endpoint x? Predict the direction, then explain the response.

A model you can changeMove one control at a time.

Read the model

F(x)=\int_0^x(1+t^2)\,dt=x+\frac{x^3}3,\quad F'(x)=1+x^2

A continuous integrand. The accumulated area is signed if endpoints are reversed; control range is nonnegative here.

Accumulation FSlope F′ · dashed

Upper endpoint x3

accumulated integral value

F(3)=3+\frac{3^3}{3}=12,\quad F'(3)=10

A tiny extra interval adds approximately integrand height times width. That is why the accumulation derivative recovers the integrand.

Assumptions, units, and limits

A continuous integrand. The accumulated area is signed if endpoints are reversed; control range is nonnegative here.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

From the situation to an equation

For a continuous integrand, the derivative of an accumulation integral is the integrand evaluated at the upper limit. The evaluation form uses any antiderivative to compute a definite integral. A varying upper limit also needs a chain-rule factor.

F(x)=\int_0^x(1+t^2)\,dt=x+\frac{x^3}3,\quad F'(x)=1+x^2

Keep the reasoning visible

\int_a^bf(t)\,dt=G(b)-G(a),\quad G'=f

A tiny extra interval adds approximately integrand height times width. That is why the accumulation derivative recovers the integrand.

What makes this model useful?

A continuous integrand. The accumulated area is signed if endpoints are reversed; control range is nonnegative here.

The accumulation function, integrand, and antiderivative play related but different roles. State which variable is the moving endpoint.

A worked example

Calculate the starting scenario

Use upper endpoint x = 3 .
$F(3)=3+\frac{3^3}{3}=12,\quad F'(3)=10$
A tiny extra interval adds approximately integrand height times width. That is why the accumulation derivative recovers the integrand. The checked quantity is accumulated integral value.

Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the accumulated integral value using upper endpoint x 3 . State the meaning of its sign and units.

Checked quantity: accumulated integral value · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Explore the visual solution

Read the model

F(x)=\int_0^x(1+t^2)\,dt=x+\frac{x^3}3,\quad F'(x)=1+x^2

A continuous integrand. The accumulated area is signed if endpoints are reversed; control range is nonnegative here.

Accumulation FSlope F′ · dashed

Upper endpoint x3

accumulated integral value

F(3)=3+\frac{3^3}{3}=12,\quad F'(3)=10

A tiny extra interval adds approximately integrand height times width. That is why the accumulation derivative recovers the integrand.

Assumptions, units, and limits

A continuous integrand. The accumulated area is signed if endpoints are reversed; control range is nonnegative here.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

F(x)=\int_0^x(1+t^2)\,dt=x+\frac{x^3}3,\quad F'(x)=1+x^2

Calculate and explain

Apply the idea to a new scenario

At $x=2$ find the accumulation of $1+t^2$ from 0 to x.

Checked quantity: accumulated integral value · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Explore the visual solution

Read the model

F(x)=\int_0^x(1+t^2)\,dt=x+\frac{x^3}3,\quad F'(x)=1+x^2

A continuous integrand. The accumulated area is signed if endpoints are reversed; control range is nonnegative here.

Accumulation FSlope F′ · dashed

Upper endpoint x2

accumulated integral value

F(2)=2+\frac{2^3}{3}=4.66667,\quad F'(2)=5

A tiny extra interval adds approximately integrand height times width. That is why the accumulation derivative recovers the integrand.

Assumptions, units, and limits

A continuous integrand. The accumulated area is signed if endpoints are reversed; control range is nonnegative here.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

F(x)=\int_0^x(1+t^2)\,dt=x+\frac{x^3}3,\quad F'(x)=1+x^2

Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual

Read the model

F(x)=\int_0^x(1+t^2)\,dt=x+\frac{x^3}3,\quad F'(x)=1+x^2

A continuous integrand. The accumulated area is signed if endpoints are reversed; control range is nonnegative here.

Accumulation FSlope F′ · dashed

Upper endpoint x3

accumulated integral value

F(3)=3+\frac{3^3}{3}=12,\quad F'(3)=10

A tiny extra interval adds approximately integrand height times width. That is why the accumulation derivative recovers the integrand.

Assumptions, units, and limits

A continuous integrand. The accumulated area is signed if endpoints are reversed; control range is nonnegative here.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. For a continuous integrand, the derivative of an accumulation integral is the integrand evaluated at the upper limit. The evaluation form uses any antiderivative to compute a definite integral. A varying upper limit also needs a chain-rule factor.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

Completion is saved on this browser. This is separate from account practice progress.