AP Calculus AB/BC · Unit 6 · Topic 6.4
The Fundamental Theorem of Calculus and Accumulation Functions
Differentiate an integral with a variable bound using FTC and the chain rule.
1. Topic Focus
Interpret definite integrals as accumulated change, connect sums to integrals, apply both Fundamental Theorems, and select antiderivative techniques.
This topic: Differentiate an integral with a variable bound using FTC and the chain rule.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
The derivative of ∫₀^(x²) cos t dt is 2x cos(x²).
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
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.
7. AP Reasoning Routine
Identify the accumulating quantity and units, preserve bounds, choose a valid integration technique, and check answers by differentiation.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Differentiate or interpret each accumulation function.
(a) Find \(F'(x)\) if \(F(x)=\int_4^x\cos(t^2)\,dt\).
(b) Find \(G'(x)\) if \(G(x)=\int_1^{x^3}e^{t^2}\,dt\).
(c) Find \(H'(x)\) if \(H(x)=\int_{x^2}^{7}\sqrt{1+t}\,dt\).
(d) Find \(J'(x)\) if \(J(x)=\int_{2x}^{x^2+1}\frac{1}{1+t^2}\,dt\).
(e) For \(A(x)=\int_3^x(t-1)\,dt\), find \(A(3),A'(x),\) and \(A''(x)\).
(f) A quantity satisfies \(Q(2)=10\) and \(Q'(t)=3t^2\). Represent \(Q(x)\) with an accumulation function and verify its derivative.
(g) Let \(P(x)=\int_2^x f(t)\,dt\). If \(\int_2^5f(t)\,dt=7\) and \(\int_5^8f(t)\,dt=-3\), find \(P(5)\) and \(P(8)\).
(h) Explain why \(\frac{d}{dx}\int_4^x f(t)\,dt=f(x)\) still holds for \(x<4\), assuming continuity.
(i) If \(f\) jumps from 2 to 5 at \(x=0\), what can be said about the differentiability at 0 of \(A(x)=\int_{-1}^x f(t)\,dt\)?
(j) Explain the two separate chain-rule substitutions needed when differentiating an integral with both bounds variable.
Check the solution
(a) \(F'(x)=\cos(x^2)\).
(b) Substitute \(x^3\) into the integrand and multiply by \(3x^2\): \(G'(x)=3x^2e^{x^6}\).
(c) The lower bound contributes a minus sign, so \(H'(x)=-2x\sqrt{1+x^2}\).
(d) \(J'(x)=\frac{2x}{1+(x^2+1)^2}-\frac{2}{1+4x^2}\).
(e) \(A(3)=0\), \(A'(x)=x-1\), and \(A''(x)=1\).
(f) \(Q(x)=10+\int_2^x3t^2\,dt\), and FTC gives \(Q'(x)=3x^2\).
(g) \(P(5)=7\) and \(P(8)=7+(-3)=4\).
(h) For \(x<4\), reversing orientation changes the value of the accumulated integral, but increasing \(x\) still adds a thin strip of height \(f(x)\). The derivative remains \(f(x)\).
(i) The accumulation function remains continuous, but its one-sided derivatives are 2 and 5. Therefore \(A'(0)\) does not exist.
(j) Substitute the upper bound into the integrand and multiply by its derivative; then subtract the integrand evaluated at the lower bound multiplied by the lower bound's derivative.