Why the Bounds Must Change
A substitution changes the variable used to describe the integrand. In a definite integral, the limits describe the interval of values of that variable, so they must change as well. If \(u=g(x)\), the original endpoints \(x=a\) and \(x=b\) correspond to \(u=g(a)\) and \(u=g(b)\).
This endpoint transformation is part of the substitution, not an optional final adjustment. It also preserves orientation: if \(g(a)>g(b)\), the new integral runs from the larger value to the smaller one. That reversed order carries a minus sign in the value of the integral.
The Definite-Integral Substitution Rule
The standard rule applies when the integral is an ordinary proper integral and the functions satisfy conditions that guarantee its integrand is continuous. In particular, it is enough for \(g\) to be continuously differentiable on the closed interval and for \(f\) to be continuous on an interval containing all values \(g(x)\) takes there.
The differential notation \(du=g'(x)\,dx\) records how the change in \(u\) relates to a change in \(x\). The bounds, however, are found directly from the endpoints: substitute \(x=a\) and \(x=b\) into \(u=g(x)\). Do not choose the new bounds by comparing which values seem smaller or by keeping the old numerical bounds automatically.
Why the Rule Works
The chain rule explains the transformed integrand. Because \(f\) is continuous on an interval containing the range of \(g\), it has an antiderivative \(F\) there, so \(F'(u)=f(u)\). The chain rule then gives
Thus \(F(g(x))\) is an antiderivative of the original integrand. Applying the Fundamental Theorem of Calculus to the original integral, and then using \(F'(u)=f(u)\) for the transformed one, gives
This also explains why no monotonicity assumption is needed. The chain rule holds whether \(g\) increases, decreases, or changes direction. The transformed limits retain the direction from the image of the starting endpoint to the image of the finishing endpoint.
Worked Examples
Worked Example: An Increasing Substitution
Evaluate \(\displaystyle\int_0^2 2x\cos(x^2)\,dx\) by changing variables, and check the result.
Choose \(u=x^2\). Then \(du=2x\,dx\). Transform the endpoints: when \(x=0\), \(u=0^2=0\); when \(x=2\), \(u=2^2=4\). Therefore,
To check, an antiderivative of the original integrand is \(\sin(x^2)\), since \(\frac{d}{dx}\sin(x^2)=2x\cos(x^2)\). Evaluating at the original endpoints gives \(\sin(2^2)-\sin(0^2)=\sin(4)-0=\sin(4)\), as required.
Worked Example: Reversed Bounds and a Negative Differential
Evaluate \(\displaystyle\int_0^1 2x e^{1-x^2}\,dx\). Here \(u=1-x^2\) decreases as \(x\) goes from \(0\) to \(1\), so the transformed bounds will be reversed.
Set \(u=1-x^2\). Then \(du=-2x\,dx\), or \(2x\,dx=-du\). Transform the endpoints separately: at \(x=0\), \(u=1-0^2=1\); at \(x=1\), \(u=1-1^2=0\). Thus
Check in the original variable: an antiderivative is \(-e^{1-x^2}\), because its derivative is \(-e^{1-x^2}(-2x)=2x e^{1-x^2}\). The endpoint difference is \((-e^{1-1^2})-(-e^{1-0^2})=-1-(-e)=e-1\), matching the substitution result.
Worked Example: A Substitution That Is Not One-to-One
Evaluate \(\displaystyle\int_{-1}^1 \frac{2x}{1+x^2}\,dx\) using \(u=1+x^2\). This \(g(x)=1+x^2\) is not one-to-one on \([-1,1]\): different \(x\)-values can give the same \(u\). The substitution rule still applies because \(g\) is continuously differentiable and \(f(u)=1/u\) is continuous on an interval containing the range \([1,2]\).
Set \(u=1+x^2\), so \(du=2x\,dx\). At \(x=-1\), \(u=1+(-1)^2=2\); at \(x=1\), \(u=1+1^2=2\). Therefore,
Check using an antiderivative in \(x\): \(\ln(1+x^2)\) differentiates to \(\frac{2x}{1+x^2}\). Its endpoint difference is \(\ln(1+1^2)-\ln(1+(-1)^2)=\ln(2)-\ln(2)=0\). Equal transformed endpoints give zero here, in agreement with the original integral.
How to Apply the Rule Reliably
A dependable substitution keeps three transformations together: the expression in the integrand, the differential, and the limits. Use this sequence before evaluating the new integral.
Identify a substitution that makes the integrand simpler and whose derivative factor matches the differential.
Compute \(du=g'(x)\,dx\), including any constant or negative factor.
Calculate \(g(a)\) for the lower endpoint and \(g(b)\) for the upper endpoint. Keep them in that order.
Express the full integral in \(u\), with no \(x\)-terms left, and integrate between the transformed limits.
When useful, differentiate an antiderivative in the original variable and evaluate it at the original endpoints.
A Common Mistake: Keeping the Original Bounds
In the reversed-bounds example, \(u=1-x^2\) maps \(x=0\) to \(u=1\) and \(x=1\) to \(u=0\). A common error is to change the integrand and differential but leave the limits at \(0\) and \(1\). That gives the wrong calculation
This is the negative of the correct value \(e-1\). The mistake is not in integrating \(-e^u\); it is in assigning the limits. The transformed interval runs from \(u=1\) to \(u=0\), not from \(u=0\) to \(u=1\). Using the correct limits gives
There are two equivalent ways to handle this situation: retain the reversed limits and the negative sign in the differential, or reverse the limits and change the overall sign. If both adjustments are made, they cancel; if neither is made, the answer may also have the wrong sign.
Why This Method Matters
Changing the bounds allows the entire definite integral to be evaluated in the new variable. There is no need to return to \(x\) merely to insert the endpoints: the transformed endpoints already specify the interval of integration. This is especially useful when the substitution simplifies the integrand substantially.
The endpoint order also records information about how the original interval is mapped. Increasing substitutions often preserve the order of the bounds; decreasing substitutions reverse it; and a substitution that changes direction may map both endpoints to the same value. In every case, the rule uses the actual images of the original endpoints, together with the transformed differential. The conditions in the theorem matter: continuous differentiability of \(g\) and continuity of \(f\) ensure that the integrand and the ordinary definite integrals in the rule are well defined.
Check Your Understanding
For each question, identify the transformed limits before evaluating.
- Under \(u=3-x\), what do the limits \(x=1\) and \(x=4\) become, in that order?
- For \(u=x^2+2\), what are the transformed limits when \(x\) runs from \(0\) to \(3\)?
- In \(\displaystyle\int_0^1 2x\cos(1-x^2)\,dx\), what are \(du\) and the transformed bounds for \(u=1-x^2\)?
- Why does the substitution rule not require \(g\) to be one-to-one?
- What error is made by keeping the original limits after changing from \(x\) to \(u\)?
