Why Reflect an Interval?
A definite integral accumulates the values of a function across an interval. Sometimes those values have a useful relationship when the input is reflected from one endpoint toward the other. The substitution \(x=a+b-t\) makes that relationship explicit: as \(t\) runs from \(a\) to \(b\), \(x\) runs from \(b\) back to \(a\).
The key point is that reflection does not change the interval of integration as a set, but it reverses its orientation. Accounting for that reversal gives an identity between the original integrand and its reflected version. The identity can simplify an integral even when the original integrand does not look symmetric.
The Reflection Substitution
Let \(f\) be continuous on \([a,b]\), where \(a<b\). Set \(x=a+b-t\). Then
When \(x=a\), the new variable is \(t=b\); when \(x=b\), it is \(t=a\). Thus the endpoints reverse:
The final integral has the original bounds, so the variable \(t\) can be renamed \(x\). The result is the reflection identity.
The sign and the bounds must be handled together. Reversing the bounds contributes a minus sign, and the differential contributes another minus sign; these cancel. Keeping the original bounds while retaining \(dx=-dt\) would give the wrong sign.
Pairing the Original and Reflected Values
Because the original integral equals the integral of the reflected function, we can add the two equal expressions and divide by two:
This form is useful when the sum of the two reflected values is simpler than either value alone. In particular, if that sum is a constant \(C\) throughout the interval, then
This conclusion follows directly from the reflection identity and the integral of a constant. No assumption that \(f\) itself is constant is needed.
Worked Example: A Rational Integrand with a Simple Reflected Pair
Worked Example: A Rational Integrand with a Simple Reflected Pair
Evaluate
Here \(a+b=4\), so reflection replaces \(x\) by \(4-x\). Denote the integrand by \(f(x)\). Its reflected value is
The denominators agree, so adding the two values gives
The paired-integral identity therefore gives
Complete the square in the denominator:
Thus
Evaluating the endpoints,
To check the value independently, put \(u=x-2\) in the original integrand. Its denominator is \(2(u^2+4)\), and its numerator is \(u+2\), so
An antiderivative is
since differentiating gives \(\frac{u}{2(u^2+4)}+\frac{1}{u^2+4}\). At \(u=2\) and \(u=-2\), the logarithmic terms are equal and cancel in the difference. The arctangent terms give \(\frac12(\frac{\pi}{4}-(-\frac{\pi}{4}))=\frac{\pi}{4}\), confirming the answer.
When the Reflected Values Add to One
A particularly direct application occurs when the two reflected values sum to a constant. The next example uses an exponential expression for which the reflected pair simplifies exactly.
Worked Example: An Exponential Integrand on an Interval with Sum \(\ln 4\)
Evaluate
The endpoints sum to \(\ln 4\). Reflect \(x\) to \(\ln 4-x\). Since \(e^{\ln 4-x}=4e^{-x}\), the reflected value is
Adding this to the original value gives
The paired-integral identity now yields
For a check, an antiderivative of the original integrand is \(\ln(e^x+2)\), because
Evaluating it at the endpoints gives
This agrees with the result from reflection.
Reflection Symmetry Can Make the Pair Constant
For another example, consider an integrand whose numerator and denominator transform together under reflection. On \([0,4]\), define
Its reflected value is
The denominators match, and the numerators add to that denominator. Therefore,
Worked Example: A Rational Function Whose Reflected Values Sum to One
Evaluate
Using the reflected-pair identity and the sum just found,
To check, set \(u=x-2\). Then \(x=u+2\), and the denominator becomes \(2(u^2+4)\). The integrand becomes
An antiderivative is \(\frac{u}{2}+\ln(u^2+4)\), since its derivative is \(\frac12+\frac{2u}{u^2+4}\). The bounds \(x=0\) and \(x=4\) correspond to \(u=-2\) and \(u=2\), so
The direct antiderivative check agrees with the reflection calculation.
A Common Bounds Error
A frequent mistake is to make the substitution \(x=4-t\) but leave the bounds unchanged while keeping the minus sign in \(dx=-dt\). For example, that incorrect procedure gives
The original integral is positive, so the negative result is a warning. The error is that \(x=0\) corresponds to \(t=4\), and \(x=4\) corresponds to \(t=0\). The correct substitution keeps the reversed bounds:
The answer also matches the direct evaluation \(\left[\frac{x^2}{2}\right]_0^4=8\). When changing variables in a definite integral, always transform both the differential and the bounds.
When Reflection Helps
Reflection is most useful when it reveals a simple expression for \(f(x)+f(a+b-x)\). That sum might be constant, as in the exponential and rational examples, or it might have a simpler integral than the original function. If the reflected pair is no simpler, the substitution still gives a valid identity, but it may not shorten the calculation.
The method also gives a precise test for two familiar situations. If \(f(a+b-x)=f(x)\), the function has equal values at reflected inputs, and the paired identity reduces to the original integral unchanged. If \(f(a+b-x)=-f(x)\), the paired sum is zero, so the integral is zero. These conclusions require the stated relation to hold throughout the interval.
Check Your Understanding
Use reflection substitutions and the paired-integral identity to answer these questions.
- Under the substitution \(x=a+b-t\), what are the new bounds when \(x\) goes from \(a\) to \(b\), and what is \(dx\) in terms of \(dt\)?
- If \(f(x)+f(a+b-x)=6\) throughout \([a,b]\), express \(\int_a^b f(x)\,dx\) in terms of \(a\) and \(b\).
- For the interval \([0,2]\), find the reflection of \(x\). What is the reflected value of \(f(x)=e^x/(e^x+1)\)?
- Explain why retaining the original bounds while using \(dx=-dt\) leads to an incorrect result.
- If \(f(a+b-x)=-f(x)\) throughout \([a,b]\), what does the paired-integral identity imply about the definite integral?
