Why the Starting Point Matters
An accumulation function records the signed integral of a function from a chosen starting point to a variable endpoint. Choosing a different starting point changes the amount accumulated by the time we reach a given endpoint. Yet, under the continuity condition in the Fundamental Theorem of Calculus, that choice does not change the derivative of the accumulation function.
This distinction is useful: the starting point determines the function's values, while the integrand determines its rate of change. We will make the relationship precise and use it to compare fixed-start integrals.
Fixed-Start Accumulation Functions
Let \(f\) be continuous on an interval \(I\), and choose a fixed point \(a\) in \(I\). Define the accumulation function with starting point \(a\) by integrating from \(a\) to \(x\). The first part of the Fundamental Theorem of Calculus says that this function is differentiable at every interior point \(x\) of \(I\), and its derivative is \(f(x)\).
Now choose another fixed starting point \(b\in I\), and define \(A_b(x)=\int_b^x f(t)\,dt\). Both functions use the same integrand; only the fixed lower limit differs. The question is how their values compare.
Changing the Start Adds a Constant
Definite integrals are additive across adjacent intervals. In particular, for \(a,b,x\in I\), the integral from \(a\) to \(x\) can be split at \(b\): the signed accumulation from \(a\) to \(x\) is the accumulation from \(a\) to \(b\), followed by the accumulation from \(b\) to \(x\). This gives the identity
The identity holds regardless of the order of \(a\), \(b\), and \(x\), provided the integrals are defined. Definite integrals are oriented: reversing the endpoints reverses the sign. Rearranging the identity gives a direct comparison of the two accumulation functions:
The integral \(\int_a^b f(t)\,dt\) does not depend on \(x\), because both \(a\) and \(b\) are fixed. Thus \(A_b\) differs from \(A_a\) by a constant. The value of that constant depends on the direction of the comparison: when we express \(A_b\) in terms of \(A_a\), we subtract the accumulation from \(a\) to \(b\).
Why the Derivatives Agree
The relationship above gives a short proof of why the starting point affects accumulation values but not the derivative formula. Both fixed-start functions satisfy the hypotheses of the first part of the Fundamental Theorem of Calculus: the integrand is continuous on the interval, and each lower limit is fixed. Therefore each derivative is the integrand evaluated at the variable endpoint.
The same conclusion also follows directly from the constant-difference identity. Differentiating that identity with respect to \(x\), the term \(\int_a^b f(t)\,dt\) has derivative zero because it is constant with respect to \(x\). The remaining derivative is \(A_a'(x)\), which equals \(f(x)\) by the Fundamental Theorem.
Let \(a\) and \(b\) be points in an interval on which \(f\) is continuous.
Use \(\int_a^x f(t)\,dt=\int_a^b f(t)\,dt+\int_b^x f(t)\,dt\).
Rearranging shows that the two accumulation functions differ by \(\int_a^b f(t)\,dt\), which is independent of \(x\).
Continuity of \(f\) gives \(A_a'(x)=f(x)\) and \(A_b'(x)=f(x)\) at interior points.
This proof explains both parts of the result. The integral between the two starting points shifts the values by a constant, and a constant shift does not affect the derivative. The continuity condition matters: it is what permits the derivative formula from the Fundamental Theorem of Calculus to be used at the point in question.
Worked Examples
Worked Example: Polynomial Integrand With Two Starting Points
Let \(f(t)=3t^2+2t+1\). Compare the accumulation functions starting at \(0\) and \(2\), and check that their derivatives agree.
First, for the starting point \(0\), evaluate the integral using the antiderivative \(t^3+t^2+t\):
Next compute the fixed accumulation from \(0\) to \(2\):
The relationship between starting points gives
Check the starting-point value: \(A_2(2)=8+4+2-14=0\), as required by \(\int_2^2 f(t)\,dt=0\). Differentiate both expressions:
The functions differ by \(-14\), but their derivatives are identical.
Worked Example: Trigonometric Integrand
Let \(f(t)=\cos t\). Compare the accumulation functions with starting points \(0\) and \(\frac{\pi}{2}\).
Starting at \(0\), integrate \(\cos t\):
The integral between the two starting points is
Therefore the accumulation function starting at \(\frac{\pi}{2}\) is
At its own starting point, it has value zero: \(A_{\pi/2}(\frac{\pi}{2})=\sin(\frac{\pi}{2})-1=1-1=0\). Differentiation checks the formula:
Thus the change of start shifts the function down by \(1\) without changing its derivative.
Worked Example: Comparing Starts in the Reverse Order
Let \(f(t)=2t+3\), and compare starts \(1\) and \(4\). The integral from \(1\) to \(4\) is
The function starting at \(1\) is
Using the comparison formula, the function starting at \(4\) is
Check its value at \(x=4\): \(A_4(4)=16+12-28=0\). The derivative check is
Both derivatives equal \(f(x)\), while the functions differ by \(24\).
A Common Sign Error
A frequent mistake is to reverse the order of the starting points without reversing the sign of the integral. In the last example, the correct comparison is \(A_4(x)=A_1(x)-\int_1^4 f(t)\,dt=A_1(x)-24\). A mistaken calculation might write \(A_4(x)=A_1(x)+24\), treating the integral from \(1\) to \(4\) as though it were subtracted in the opposite direction.
That wrong formula fails a direct check at the new starting point. Since \(A_1(4)=4^2+3\cdot4-4=24\), the mistaken formula would give \(A_4(4)=24+24=48\). But the definition requires \(A_4(4)=\int_4^4 f(t)\,dt=0\). The correct formula gives \(A_4(4)=24-24=0\).
Interpreting the Result
A fixed-start accumulation function is one particular function whose derivative is \(f\). Changing the fixed start selects a different such function, with a different value at each endpoint by the same constant amount. For instance, the start point \(a\) ensures that \(A_a(a)=0\), while the start point \(b\) ensures that \(A_b(b)=0\). Those two normalizations need not give the same function values elsewhere.
The derivative describes local change, not the accumulated amount since a particular origin. If two functions differ by a constant, their changes over small increments are the same, so their derivatives agree. The Fundamental Theorem makes this precise for these accumulation functions: with a continuous integrand, either fixed starting point gives derivative \(f(x)\) at an interior endpoint.
This result concerns changing one fixed starting point to another fixed starting point. It does not assert a derivative rule for an endpoint that itself varies with \(x\). Here the lower limit remains constant while the upper endpoint is \(x\).
Check Your Understanding
Use the fixed-start comparison formula and the Fundamental Theorem of Calculus to answer these questions.
- If \(A_a(x)=\int_a^x f(t)\,dt\) and \(A_b(x)=\int_b^x f(t)\,dt\), what constant relates \(A_b(x)\) to \(A_a(x)\)?
- What continuity condition allows the derivative formula for a fixed-start accumulation function to apply?
- For \(f(t)=t\), compare the functions starting at \(0\) and \(3\). Give both functions and verify their derivatives.
- Why does changing a fixed starting point not change the derivative?
- If \(\int_2^5 f(t)\,dt=-6\), express \(A_5(x)\) in terms of \(A_2(x)\).
