The Fundamental Theorem of Calculus and Accumulation Functions

3. Visual Connection

5. Concept Development

A definite integral can define a new function

When the upper bound varies, each input \(x\) determines a different signed area:

For every permitted \(x\), the integral returns one number, so \(A\) is a function even though the integrand is written using a different variable.

The integration variable is internal

In \(A(x)=\int_a^x f(t)\,dt\), \(t\) is a dummy variable used only inside the integral. Using \(t\) prevents confusion between the moving endpoint \(x\) and the values being accumulated.

The starting value is zero accumulation

At the base point, the interval has zero width:

This does not imply \(f(a)=0\); it says only that nothing has accumulated between identical endpoints.

FTC Part 1 connects accumulation and rate

If \(f\) is continuous on an interval containing \(a\) and \(x\), then

Differentiation recovers the instantaneous rate from its accumulated change.

Why the theorem is reasonable

Increasing the upper bound from \(x\) to \(x+h\) adds a thin signed strip whose area is approximately \(f(x)h\). Therefore

Continuity makes the average height on the shrinking interval approach \(f(x)\).

A composite upper bound requires the chain rule

If \(u(x)\) is differentiable,

Substitute the entire upper-bound expression into the integrand, then multiply by its derivative.

A variable lower bound contributes a minus sign

Reversing bounds gives \(\int_{u(x)}^a f=-\int_a^{u(x)}f\), so

Both bounds may vary

Separate the integral at any fixed reference point, then differentiate both pieces:

The upper contribution is added and the lower contribution is subtracted.

No antiderivative formula may be needed

FTC Part 1 differentiates an accumulation function even when the integrand has no elementary antiderivative. Expressions such as \(e^{t^2}\) or \(\sqrt{1+t^4}\) can be handled by endpoint substitution and the chain rule.

Orientation works when \(x<a\)

If the moving endpoint lies left of the base point, \(A(x)=\int_a^x f=-\int_x^a f\). The accumulation value changes sign appropriately, while the derivative formula \(A'(x)=f(x)\) remains valid wherever \(f\) is continuous.

Initial-value functions include a starting amount

If \(Q'(t)=r(t)\) and \(Q(a)=Q_0\), then the quantity itself can be represented as

The integral supplies net change; the constant supplies the known initial amount.

Higher derivatives come from the integrand

For \(A(x)=\int_a^x f(t)\,dt\), FTC gives \(A'(x)=f(x)\). If \(f\) is differentiable, then \(A''(x)=f'(x)\). This connection later allows the graph of \(f\) to determine the behavior of \(A\).

Continuity matters

If the integrand has a jump at \(x=c\), the accumulation function can remain continuous while failing to be differentiable at \(c\). FTC guarantees \(A'(c)=f(c)\) only under the theorem's continuity condition at that point.

FTC Part 1 and Part 2 have different jobs

Part 1 differentiates a function defined by an integral. Part 2 evaluates a definite integral using an antiderivative and endpoint subtraction. Identify whether the prompt asks for a derivative or an integral value before choosing the theorem.

Common errors

Frequent errors include omitting the chain-rule factor, forgetting the lower-bound minus sign, substituting \(x\) instead of the full bound into the integrand, adding an unnecessary constant of integration, and trying to find an antiderivative when FTC Part 1 already gives the derivative directly.

6. Detailed Worked Example and Error Check

Example 1: Basic accumulation function. Let

Then \(F(2)=0\), \(F'(x)=x^3-4x\), and \(F''(x)=3x^2-4\). The value of \(F\) is accumulated signed area, while \(F'\) is the current height of the integrand.

Example 2: Composite upper bound.

FTC followed by the chain rule gives

Example 3: Variable lower bound.

The minus sign comes from the moving lower bound. No elementary antiderivative of \(e^{-t^2}\) is required.

Example 4: Two variable bounds.

The upper and lower contributions give

Example 5: Endpoint left of the base point. Let \(A(x)=\int_3^x(t+1)\,dt\). Then

The negative accumulated value reflects reversed orientation; it does not alter the FTC derivative rule.

Example 6: Recover a quantity from its rate. A particle has velocity \(v(t)=t^2-2t\) and position \(s(0)=5\). Its position function is

FTC verifies \(s'(x)=x^2-2x=v(x)\). Also, \(s(3)=5\), because the positive and negative displacement contributions from 0 to 3 cancel.

Example 7: A jump in the integrand. Let \(f(t)=1\) for \(t<1\) and \(f(t)=3\) for \(t\ge1\), and define \(B(x)=\int_0^x f(t)\,dt\). Then

The two formulas meet at \(B(1)=1\), so \(B\) is continuous. Its left derivative is 1 and right derivative is 3, so \(B'(1)\) does not exist; this matches the failure of continuity of \(f\) at 1.