From Accumulated Change to a Slope
An integral can describe the total accumulated effect of a function over an interval. If the endpoint of that interval changes, the accumulated total changes too. This creates a natural question: how quickly is the total changing as the endpoint moves?
The Fundamental Theorem of Calculus, Part 1 answers this question when the integrand is continuous. It connects the value of a function at the moving endpoint to the instantaneous rate of change of the accumulated integral.
In this tutorial, the lower endpoint stays fixed. Only the upper endpoint \(x\) varies. This distinction matters: the result below concerns a fixed starting point and does not require differentiating a variable bound.
The integral is signed: if \(x<a\), then \(\int_a^x f(t)\,dt=-\int_x^a f(t)\,dt\). The definition therefore applies whether \(x\) is to the right or the left of \(a\), as long as the interval between them lies in the domain where \(f\) is continuous.
The Fixed-Endpoint Theorem
The fixed number \(a\) determines where accumulation begins. It does not appear in the derivative formula. The slope at \(x\) is the value of the integrand at \(x\), not the total accumulated amount from \(a\) to \(x\).
Continuity is the key condition. The theorem does not require \(f\) itself to be differentiable. It does require \(f\) to be continuous at the point where the derivative of \(A\) is being found; the stated theorem assumes continuity throughout the interval under consideration.
Why the Derivative Equals the Integrand
The reasoning comes from the definition of a derivative. To find \(A'(x)\), compare the accumulated total at \(x+h\) with the total at \(x\). Additivity of definite integrals gives $$ A(x+h)-A(x) = \int_a^{x+h}f(t)\,dt-\int_a^x f(t)\,dt = \int_x^{x+h}f(t)\,dt. $$ The change in accumulation is exactly the integral over the short interval added at the endpoint.
For \(h\ne 0\), the difference quotient is therefore $$ \frac{A(x+h)-A(x)}{h} = \frac{1}{h}\int_x^{x+h}f(t)\,dt. $$ As \(h\) approaches zero, this quotient is the average value of \(f\) over an interval shrinking toward \(x\). Continuity ensures that values of \(f(t)\) on that short interval are close to \(f(x)\).
This argument works for positive and negative \(h\). When \(h<0\), the oriented integral and the negative denominator are both accounted for in the difference quotient; the absolute-value estimate uses the interval length \(|h|\).
Applying the Theorem
In a typical problem, the integral may look complicated or may not have an elementary expression in terms of \(x\). If the task is to differentiate the accumulation function, finding a formula for the integral first is unnecessary. Check the continuity condition, identify the integrand, and evaluate it at the input \(x\).
Write the function as \(A(x)=\int_a^x f(t)\,dt\). The letter \(t\) is only the integration variable.
Confirm that \(f\) is continuous on the interval between \(a\) and the inputs being considered.
Set \(A'(x)=f(x)\), then substitute any requested value of \(x\).
Worked Example: A Polynomial Integrand
Let $$ A(x)=\int_0^x (3t^2-4t+1)\,dt. $$ Find \(A'(x)\) and the slope of \(A\) at \(x=2\).
Step 1: Identify the integrand. The fixed lower endpoint is \(a=0\), and \(f(t)=3t^2-4t+1\).
Step 2: Check the condition. A polynomial is continuous for every real \(t\), so the theorem applies for every real \(x\).
Step 3: Differentiate the accumulation function. By Part 1, $$ A'(x)=f(x)=3x^2-4x+1. $$
Step 4: Evaluate at \(x=2\). $$ A'(2)=3(2)^2-4(2)+1=3(4)-8+1=12-8+1=5. $$ The slope of \(A\) at \(x=2\) is \(5\).
Check: Substituting \(x=2\) into the derivative formula gives \(12-8+1=5\), matching the evaluated slope. No antiderivative of the integrand is needed to find this derivative.
Worked Example: A Trigonometric Integrand
Define $$ B(x)=\int_{\pi/2}^x \cos t\,dt. $$ Find \(B'(x)\) and \(B'(\pi)\).
Step 1: Identify the fixed endpoint and integrand. The fixed lower endpoint is \(\pi/2\), and \(f(t)=\cos t\).
Step 2: Check continuity. The cosine function is continuous for all real inputs, so the theorem applies.
Step 3: Apply Part 1. $$ B'(x)=\cos x. $$
Step 4: Evaluate at \(x=\pi\). $$ B'(\pi)=\cos(\pi)=-1. $$ Thus \(B\) has slope \(-1\) at \(x=\pi\).
Check: The integrand evaluated at the endpoint is \(\cos(\pi)=-1\), exactly the value given by the derivative formula. The fixed lower endpoint \(\pi/2\) affects the accumulated values of \(B\), but it does not alter this slope.
Worked Example: A Continuous Integrand with a Corner
Let $$ C(x)=\int_{-1}^x |t|\,dt. $$ Find \(C'(x)\) for all real \(x\), including \(x=0\).
Step 1: Identify the integrand. Here \(f(t)=|t|\), and the fixed lower endpoint is \(-1\).
Step 2: Check continuity. The absolute-value function is continuous everywhere, including at \(t=0\). It has a corner there, but that does not prevent the theorem from applying.
Step 3: Apply Part 1. $$ C'(x)=|x|. $$
Step 4: Check representative inputs. $$ C'(-2)=|-2|=2,\qquad C'(0)=|0|=0,\qquad C'(1)=|1|=1. $$ In particular, the accumulation function is differentiable at \(x=0\), even though the integrand \(|t|\) is not differentiable there.
Check: The theorem requires continuity of \(f\), not differentiability of \(f\). Since \(|t|\) is continuous at every input, the derivative formula \(C'(x)=|x|\) applies at all three checked values, including zero.
What Continuity Does—and Does Not—Require
Continuity is what makes the short-interval average of \(f\) approach the endpoint value \(f(x)\). If the integrand is discontinuous at a point, the theorem cannot be applied there on the basis of this result. A derivative might still exist in a particular situation, but Part 1 alone does not guarantee it.
It is also useful to distinguish the requirements on \(f\) from conclusions about \(A\). The integrand needs to be continuous; it need not have its own derivative. The theorem then guarantees that \(A\) is differentiable at interior points and identifies its derivative.
Common Mistake: Differentiating the Integrand
A common error is to treat the integral sign as if differentiating the accumulation function meant differentiating the expression inside the integral. For example, with $$ A(x)=\int_0^x (3t^2-4t+1)\,dt, $$ one might incorrectly write $$ A'(x)=6x-4. $$ That expression is the derivative of \(3x^2-4x+1\), not the derivative of the accumulation function.
The correction is to identify the integrand itself and evaluate it at the upper endpoint: $$ A'(x)=3x^2-4x+1. $$ The fixed-endpoint theorem says \(A'(x)=f(x)\), not \(f'(x)\). In the example, the incorrect expression would give \(A'(2)=8\), whereas the correct slope is \(3(2)^2-4(2)+1=5\).
Check Your Understanding
For each question, identify the integrand and check whether the continuity condition is satisfied before applying the theorem.
- If \(P(x)=\int_2^x (t^3+1)\,dt\), what is \(P'(x)\)?
- If \(Q(x)=\int_{-1}^x \sin t\,dt\), find \(Q'(0)\).
- If \(R(x)=\int_0^x |t-3|\,dt\), is Part 1 applicable at \(x=3\), and what is \(R'(3)\)?
- For \(S(x)=\int_1^x (5t^2)\,dt\), explain why \(S'(x)\) is not \(10x\).
- State the continuity condition used by the fixed-endpoint version of the theorem.
