Why Equal Bounds Matter
A definite integral records accumulated quantity across an interval. If the interval has no length, there is no width over which any quantity can accumulate. This leads to a simple identity that is useful whenever integral expressions contain matching upper and lower bounds.
The bounds do not need to look identical for the interval to have zero length. For example, \(2+3\) and \(10/2\) are different expressions for the same number. The first task is therefore to compare the values of the bounds, not merely their written forms.
For an ordinary proper Riemann integral, the integrand must be defined on the interval in question. When the interval consists only of the point \(a\), it is enough that \(f(a)\) be defined. The identity does not say that \(f(a)=0\); it says that the integral over an interval of zero length is zero.
Why the Integral Is Zero
A Riemann sum adds contributions of the form “function value times subinterval width.” On an interval of positive length, each width is nonnegative and the widths together cover the interval. For equal endpoints, the total width is zero. In the degenerate-interval convention, every width in the corresponding sum is zero, so every contribution vanishes.
Consider the interval whose left and right endpoints are both \(a\).
Every subinterval width \(\Delta x_i\) is zero, because there is no distance between any points of the degenerate interval.
Each term has the form \(f(\xi_i)\Delta x_i=f(\xi_i)\cdot 0=0\), so the full sum is zero.
Every such sum is zero, so its limit, the definite integral, is zero as well.
This is a complete justification through the Riemann-sum meaning of the integral. More formally, the interval \([a,a]\) is degenerate, so its integral is assigned the value given by its zero-width Riemann sums. The result does not depend on the shape of the graph or the sign of the integrand: no width is available to contribute any signed area.
Worked Examples
Worked Example: A Polynomial on a Zero-Length Interval
Simplify \(\displaystyle \int_4^4 (3x^2-2x+7)\,dx\).
The lower bound is \(4\), and the upper bound is also \(4\). The integrand is defined at \(x=4\), so the zero-length interval identity applies:
As a check, the interval width is \(4-4=0\). Every Riemann-sum term is a function value multiplied by a width of zero, so the integral is indeed \(0\). The value of the integrand at the endpoint is \(3(4)^2-2(4)+7=48-8+7=47\), which does not change the result.
Worked Example: Bounds Written in Different Forms
Simplify \(\displaystyle 6+\int_{2+3}^{10/2}(x-1)^4\,dx\).
First evaluate each bound:
Thus the integral has equal bounds. The integrand is defined at \(5\), so
Substitute this into the full expression:
Check: \(5-5=0\), so the integral contributes zero; the remaining constant is \(6\). The simplified answer is therefore consistent with the original expression.
Worked Example: A Parameterized Zero-Length Integral
For a real number \(t\), simplify \(\displaystyle (t-2)\int_{t^2+1}^{1+t^2}\sqrt{x+1}\,dx\).
The two bounds have the same value for every real \(t\):
The radicand at this common bound is \(t^2+2\), which is positive because \(t^2\geq 0\). Thus the integrand is defined there, and the identity gives
Multiplying by the outside factor yields
Check: substituting the integral's value \(0\) makes the whole product \(0\) for every real \(t\), including \(t=2\). The identity applies for all \(t\), not just for one selected parameter value.
Worked Example: Simplifying a Difference of Integrals
Simplify \(\displaystyle \int_7^7 \cos x\,dx-\left(3\int_{-2}^{-2}(x^3+1)\,dx\right)\).
For the first integral, both bounds equal \(7\), and \(\cos x\) is defined at \(7\). Therefore,
For the second integral, both bounds equal \(-2\), and \(x^3+1\) is defined at \(-2\). Hence
Now substitute both values, keeping the outside coefficient and subtraction in place:
Check: each integral is zero because its own pair of bounds is equal. Multiplication by \(3\) and subtraction do not alter those evaluated values, so the final result is \(0\).
A Common Mistake: Using the Endpoint Value
A frequent error is to treat the integral as though it were simply the value of the integrand at the common endpoint. For example, an incorrect calculation might say
This substitutes \(x=2\) into the integrand but ignores the interval width. It calculates \(f(2)\), not the accumulated quantity over \([2,2]\). The correction is to apply the zero-length identity:
Indeed, the width is \(2-2=0\), while the endpoint function value is \(f(2)=5\). The Riemann-sum contribution is a function value multiplied by width, not the function value alone: \(5\cdot 0=0\). This distinction is especially useful when the integrand has a large, negative, or nonzero value at the common endpoint.
A Reliable Simplification Check
When an expression contains an integral with possibly matching bounds, use a short sequence of checks:
- Evaluate or simplify the lower and upper bounds.
- Confirm that they are equal as real numbers.
- Confirm that the integrand is defined at the common endpoint, as required for the ordinary integral being used.
- Replace that integral by \(0\), then simplify the remaining expression.
Do not infer that the bounds are equal merely because they look similar, and do not infer that they differ merely because they are written differently. For example, \(t+1\) and \(1+t\) are equal for every real \(t\), whereas \(t\) and \(t+1\) are not equal for any real \(t\). The equality check is part of applying the identity correctly.
The result also gives a useful interpretation: a zero-length interval carries no accumulated quantity. It therefore cannot contribute signed area, regardless of whether the integrand is positive, negative, or changes sign elsewhere. The value of the integrand alone is not enough to determine an integral; the interval matters too.
Check Your Understanding
For each question, first compare the bounds, then apply the identity only when they are equal.
- Simplify \(\displaystyle \int_{-3}^{-3}(2x-5)\,dx\). What is the value of the integrand at the common endpoint, and why is it different from the integral?
- Simplify \(\displaystyle 4+\int_{1+6}^{14/2} e^x\,dx\).
- For real \(s\), determine the value of \(\displaystyle \int_{s^2}^{s\cdot s}\frac{1}{x+2}\,dx\). Is the integrand defined at the common bound for every real \(s\)?
- A student claims \(\displaystyle \int_5^5 (x-5)\,dx=0\) because the integrand is zero at \(5\). Give the correct reason, and say whether the same integral identity would hold for an integrand whose value at \(5\) is nonzero.
