Introduction
An accumulation function records the integral of a function up to a moving endpoint. When that endpoint is an affine expression, such as \(3x+1\) or \(2-4x\), its rate of change contributes a constant factor to the derivative.
This is an important combination of two familiar ideas: the Fundamental Theorem of Calculus describes the effect of moving the endpoint, and the Chain Rule accounts for how quickly that endpoint moves. The constant factor is easy to miss, especially when the endpoint decreases as \(x\) increases.
The affine upper-endpoint formula
Let \(f\) be continuous on an interval containing the fixed starting point \(a\) and the endpoint values under consideration. Let \(m\) and \(b\) be constants, and define the accumulation function
The variable \(t\) is the integration variable. The input \(x\) determines the upper endpoint \(mx+b\). The letter \(m\) denotes the endpoint’s constant slope, not a variable.
The formula has two parts. First evaluate the integrand at the current upper endpoint, giving \(f(mx+b)\). Then multiply by \(m\), the rate at which that endpoint changes with respect to \(x\). The fixed lower endpoint affects the value of \(A(x)\), but not this derivative formula.
The sign and size of \(m\) matter. If \(m\) is positive, the upper endpoint increases as \(x\) increases. If \(m\) is negative, the endpoint decreases, and the derivative formula includes that negative factor. If \(m=0\), the endpoint is fixed and the accumulation function is constant.
Why the chain factor appears
Define \(F(u)=\int_a^u f(t)\,dt\). By the Fundamental Theorem of Calculus, continuity of \(f\) gives \(F'(u)=f(u)\). The accumulation function can then be written as \(A(x)=F(mx+b)\). The inner function \(mx+b\) has derivative \(m\), so the Chain Rule gives \(A'(x)=F'(mx+b)m=m f(mx+b)\).
We can also see the factor directly from a change in the input. Increasing \(x\) by \(h\) changes the endpoint from \(mx+b\) to \(mx+b+mh\). The integral’s change is therefore over an interval of width \(mh\), not \(h\). For \(m\ne 0\), the difference quotient is
As \(h\) approaches zero, the average value in parentheses approaches \(f(mx+b)\), by continuity of \(f\). Thus the quotient approaches \(m f(mx+b)\). If \(m=0\), the endpoint does not move at all, so \(A(x)\) is constant and \(A'(x)=0\), also matching the formula.
Worked examples
Worked Example: A positive endpoint slope
Differentiate \(A(x)=\int_0^{3x+1}(t^2+1)\,dt\), and check the result by evaluating the integral first.
Here \(f(t)=t^2+1\) and the upper endpoint is \(3x+1\). Its slope is \(3\). Applying the formula gives
To check, an antiderivative of \(t^2+1\) is \(\frac{t^3}{3}+t\). Therefore
Differentiating this expression gives
This agrees with the formula. As a numerical check at \(x=0\), it gives \(3(1^2+1)=6\); the differentiated explicit expression gives \(3(1)^2+3=6\) as well.
Worked Example: A decreasing upper endpoint
Differentiate \(B(x)=\int_0^{2-4x}\cos t\,dt\), and verify the sign of the result.
The integrand is \(f(t)=\cos t\), and the upper endpoint \(2-4x\) has slope \(-4\). Hence
For a direct check, an antiderivative of \(\cos t\) is \(\sin t\), so
Differentiating explicitly, including the derivative of the inside expression, yields
The result matches the formula. At \(x=0\), both calculations give \(-4\cos 2\). The minus sign comes from the endpoint’s negative slope.
Worked Example: A fractional endpoint slope
Differentiate \(C(x)=\int_0^{x/3-1}(e^t+2)\,dt\), and check at \(x=3\).
Here \(f(t)=e^t+2\), and the endpoint \(x/3-1\) has slope \(1/3\). Thus
An antiderivative of \(e^t+2\) is \(e^t+2t\). Evaluating at the bounds gives
Differentiate this expression:
This agrees with the formula. At \(x=3\), the endpoint is \(3/3-1=0\), so both expressions give \(C'(3)=\frac{1}{3}(e^0+2)=1\).
Understanding increasing, decreasing, and fixed endpoints
The affine expression \(mx+b\) determines how the integral changes as \(x\) changes. The constant \(b\) shifts the endpoint but does not affect its rate of change. The constant \(m\) is the chain factor: it determines both the speed and direction of endpoint movement.
| Endpoint slope | Endpoint behavior as \(x\) increases | Derivative formula |
|---|---|---|
| \(m>0\) | The upper endpoint increases | \(m f(mx+b)\) |
| \(m<0\) | The upper endpoint decreases | \(m f(mx+b)\), with a negative factor |
| \(m=0\) | The upper endpoint is fixed | \(0\) |
For a decreasing endpoint, the negative factor is not an optional sign adjustment. It follows from the endpoint’s derivative. For example, an endpoint of \(5-2x\) has slope \(-2\), so its contribution to the derivative is \(-2 f(5-2x)\), not \(2 f(5-2x)\).
If \(m=0\), then \(A(x)=\int_a^b f(t)\,dt\) for every \(x\). This is a constant value, so its derivative is zero. The general formula gives the same answer because \(0\cdot f(b)=0\).
A common mistake: omitting the slope
A frequent error is to evaluate the integrand at the endpoint but forget to multiply by the endpoint’s derivative. For instance, with \(B(x)=\int_0^{2-4x}\cos t\,dt\), an incorrect calculation is
This treats the endpoint as if it changed at rate \(1\). But the endpoint is \(2-4x\), whose derivative is \(-4\). The corrected calculation is
Checking against the explicit form \(B(x)=\sin(2-4x)\) confirms that the corrected derivative is right. A reliable habit is to write down the endpoint slope before applying the formula.
A practical procedure
For an accumulation function with an affine upper endpoint, organize the calculation in three steps:
Write the integrand as \(f(t)\) and the upper endpoint as \(mx+b\).
Differentiate \(mx+b\) with respect to \(x\), obtaining \(m\), including its sign.
Substitute the endpoint into \(f\), then multiply by \(m\): \(A'(x)=m f(mx+b)\).
Before using the formula, check that \(f\) is continuous at the endpoint values in question. Continuity is what permits the accumulation derivative result. If \(f\) is continuous throughout an interval containing those values, the formula applies at each corresponding input.
Why this formula matters
Accumulation functions describe how a running total changes when its cutoff moves. The integrand gives the contribution at the current cutoff; the endpoint slope adjusts that contribution to the rate at which the cutoff moves with \(x\). When the cutoff moves twice as fast, the factor has magnitude \(2\). When it moves in the opposite direction, the factor is negative.
This interpretation is useful beyond examples that can be evaluated explicitly. The derivative formula can be applied directly even when the integral has no convenient elementary antiderivative. For affine upper endpoints, the essential information is the value of the integrand at the endpoint and the constant slope of that endpoint.
Check Your Understanding
For each question, identify the endpoint slope and apply the affine upper-endpoint formula.
- Find the derivative of \(\int_1^{5x-2} f(t)\,dt\), assuming \(f\) is continuous at the endpoint.
- What is the derivative of \(\int_0^{7-3x} \sin(t)\,dt\)? Explain why its sign includes a negative factor.
- If \(A(x)=\int_{-1}^{4} f(t)\,dt\), what is \(A'(x)\), and why?
- For \(C(x)=\int_0^{2x+1} t^2\,dt\), find \(C'(x)\) from the formula and verify it by evaluating the integral first.
- Which condition on \(f\) is used to justify the accumulation derivative formula?
