Two Roles in One Integral
An accumulation function is written using an integral with a variable endpoint, such as \(A(x)=\int_a^x f(t)\,dt\). The symbol \(x\) tells us which endpoint determines the value of the function. The symbol \(t\) is the variable that runs through the interval while the integral is formed.
These symbols have different roles. The endpoint \(x\) is the input to \(A\); the integration variable \(t\) is a temporary name used inside the integral. Confusing them can change the expression, or make it difficult to tell what the expression means.
The word “dummy” does not mean that the integration variable is unimportant. The integrand and differential must still use the same variable. It means only that the particular name chosen for that variable does not determine the value of the integral.
Renaming the Dummy Variable
For example, the following expressions mean the same thing, provided \(u\) is just a new name for the integration variable:
The limits remain \(a\) and \(x\). Only the dummy-variable name changes. The integrand and differential change together: \(f(t)\,dt\) becomes \(f(u)\,du\). Renaming \(t\) to \(u\) does not rename the endpoint \(x\).
The same rule applies when the integrand has more than one symbol in it. If \(x\) is a parameter as well as the endpoint, it remains fixed during the integration and must not be changed when the dummy variable is renamed. For instance, \(\int_0^x (t+x)\,dt\) becomes \(\int_0^x (u+x)\,du\), not an expression in which the parameter \(x\) has been replaced by \(u\).
Why Renaming Does Not Change the Integral
The equality is built into the definition of a definite integral. For an integrable function on an interval from \(a\) to \(x\), a Riemann sum uses a partition of that interval and a sample point in each subinterval. Its terms have the form \(f(\xi_i)\Delta t_i\). The letter used to describe the interval’s variable does not affect the partition, the sample points, or the resulting sum.
Here is a complete justification that covers all possible positions of the endpoint.
Choose any partition \(a=t_0<t_1<\cdots<t_n=x\). For each subinterval, choose a sample point \(\xi_i\). The Riemann sum for either integral is \(\sum_{i=1}^n f(\xi_i)(t_i-t_{i-1})\). The notation used for the integration variable does not appear in this sum. Taking the same limit of such sums gives equal integrals.
Both integrals have equal endpoints, so each is zero. Therefore they are equal.
Use the reversed-endpoint identity \(\int_a^x f(t)\,dt=-\int_x^a f(t)\,dt\). The integral on the right has its endpoints in increasing order. By the argument in Step 1, renaming its dummy variable does not change its value. Applying the reversed-endpoint identity again gives \(\int_a^x f(t)\,dt=\int_a^x f(u)\,du\).
This proof concerns a change of notation, not a change to the function being integrated or to the endpoints. In the reversed-endpoint case, using the identity explicitly matters: a partition that starts at \(a\) and ends at \(x\) in increasing order would not exist when \(x<a\).
Worked Example: Rewriting and Evaluating an Accumulation Function
Worked Example: Rewriting and Evaluating an Accumulation Function
Let \(A(x)=\int_1^x (3t^2-2t+4)\,dt\). Rewrite the integral using \(u\) as the dummy variable, then evaluate it.
Renaming \(t\) to \(u\) in the integrand and differential, while leaving both endpoints unchanged, gives
An antiderivative of \(3u^2-2u+4\) is \(u^3-u^2+4u\), since its derivative is \(3u^2-2u+4\). Thus
Check the calculation: the antiderivative differentiates to the stated integrand, and substituting \(x=1\) gives \(A(1)=1-1+4-4=0\), as required by the equal endpoints. The variable \(u\) is integrated out; \(x\) remains the input of \(A\).
Changing the Input Is Not Renaming the Dummy Variable
In a definition such as \(A(x)=\int_a^x f(t)\,dt\), the input \(x\) can itself be given a different name, but that is a separate change. If the input is renamed to \(y\), it must be changed consistently in the function notation and at the endpoint: \(A(y)=\int_a^y f(t)\,dt\). The dummy variable may stay \(t\), or it may be renamed separately. Either way, the expression describes the same function with its input labelled \(y\).
By contrast, changing the endpoint from \(x\) to a different expression, such as \(x^2\), generally changes the value. The rule for dummy variables permits a new name for the variable of integration; it does not permit replacing an endpoint with a new expression.
Worked Example: Renaming the Input Consistently
Let \(B(z)=\int_{-2}^z (u+1)\,du\). Write the same function with input \(w\), then calculate its value at \(w=0\).
Renaming the input means replacing \(z\) by \(w\) in the function’s input and endpoint. The dummy variable \(u\) can remain unchanged:
An antiderivative is \(\frac{u^2}{2}+u\), because its derivative is \(u+1\). Therefore
Check: substituting the endpoint \(w=0\) into the evaluated expression \(B(w)=\frac{w^2}{2}+w\) also gives \(B(0)=0\). The input name changed from \(z\) to \(w\); the dummy variable \(u\) did not need to change.
When the Integrand Depends on the Input
An integral can contain both an integration variable and the accumulation input in its integrand. In that case, the input acts as a parameter during integration. A safe way to read the notation is to ask which variable is paired with the differential. In \(\int_0^x (t+x)\,dt\), the differential \(dt\) identifies \(t\) as the integration variable; \(x\) remains the input and is held fixed while integrating with respect to \(t\).
Worked Example: Keeping a Parameter Fixed
Define \(C(x)=\int_0^x (t+x)\,dt\). Rewrite the integral using \(u\) as the dummy variable, and evaluate \(C(2)\).
Only the integration variable changes name. The input \(x\) remains in the integrand and at the upper endpoint:
For the calculation, keep \(x\) fixed while integrating with respect to \(u\). An antiderivative in \(u\) is \(\frac{u^2}{2}+xu\), because its derivative with respect to \(u\) is \(u+x\). It follows that
Thus \(C(2)=\frac{3(2)^2}{2}=\frac{12}{2}=6\). Check by direct substitution in the integral: \(C(2)=\int_0^2(u+2)\,du=[\frac{u^2}{2}+2u]_0^2=(2+4)-0=6\).
Common Pitfall: Replacing the Variable of Integration with the Input
A common error is to treat the dummy variable and the input as interchangeable inside the integrand. They are not interchangeable. In the previous example, a mistaken rewrite might be \(\int_0^x (x+x)\,dt\). This is a different integrand: it has replaced \(t\) by \(x\), rather than renaming \(t\).
At \(x=2\), the correct value is \(6\), while the altered expression gives \(2(2)^2=8\). The difference is not a matter of notation. In the correct integral, the integrand varies with the integration variable; in the altered expression, \(2x\) is constant with respect to \(t\). To avoid this error, first identify the differential, then rename only the variable paired with it.
A Reliable Reading Procedure
When you encounter a variable-limit integral, separate its roles before manipulating it. The following checks help keep the meaning fixed:
- Locate the differential, such as \(dt\), to identify the integration variable.
- Identify the endpoint that serves as the function’s input.
- When renaming the dummy variable, change it consistently in the integrand and differential.
- Keep the limits and any separate parameter unchanged unless you intend to change the integral.
- If the endpoint lies below the basepoint, use the reversed-endpoint identity when reasoning with partitions.
These checks make it easier to compare formulas and to read accumulation functions without confusing the temporary variable of integration with the input that selects their value.
Check Your Understanding
Answer each question by identifying which symbols are integration variables, inputs, endpoints, or parameters.
- Rewrite \(\int_2^x (s^2+1)\,ds\) using \(v\) as the dummy variable. Which symbol remains the input?
- In \(\int_0^x (t+x)\,dt\), which symbol is held fixed during the integration?
- Why does renaming \(t\) to \(u\) require changing \(dt\) to \(du\) as well?
- What identity lets you handle \(\int_a^x f(t)\,dt\) when \(x<a\), and why can’t an increasing partition from \(a\) to \(x\) be used?
- Explain why \(\int_0^x (t+1)\,dt\) and \(\int_0^x (x+1)\,dt\) are generally different expressions.
