Why endpoint order matters
A change of variable in a definite integral changes more than the expression being integrated: it also changes the endpoints. When the inner function decreases, the value at the original lower endpoint becomes the larger new endpoint. Keeping track of that order is essential, because reversing the bounds changes the sign of an integral.
Suppose \(u=g(x)\) and \(g\) decreases from \(x=a\) to \(x=b\), where \(a<b\). Then \(g(a)>g(b)\). The transformed integral therefore runs from the larger value \(g(a)\) to the smaller value \(g(b)\). That order is not an error to fix automatically; it is part of the substitution formula.
The substitution rule for a decreasing function
The theorem uses the differential relation \(du=g'(x)\,dx\) together with the endpoint substitutions \(x=a\mapsto u=g(a)\) and \(x=b\mapsto u=g(b)\). If the integrand contains \(-g'(x)\) instead of \(g'(x)\), a minus sign appears: $$ \int_a^b f(g(x))\bigl(-g'(x)\bigr)\,dx = -\int_{g(a)}^{g(b)}f(u)\,du = \int_{g(b)}^{g(a)}f(u)\,du. $$
The last equality uses the orientation rule \(\int_A^B f(u)\,du=-\int_B^A f(u)\,du\). Thus there are two consistent ways to handle decreasing bounds: leave them in the order produced by substitution, or swap them and include the corresponding minus sign. The two forms give the same value.
Why the rule works
The endpoint order follows directly from the chain rule and the Fundamental Theorem of Calculus. Because \(f\) is continuous, it has an antiderivative \(F\) on the interval containing the range of \(g\). The chain rule gives the derivative of the composite function \(F(g(x))\). Applying the Fundamental Theorem then proves the substitution formula.
Therefore \(F(g(x))\) is an antiderivative of the integrand in the \(x\)-integral. Evaluating at the original endpoints gives $$ \int_a^b f(g(x))g'(x)\,dx = F(g(b))-F(g(a)). $$ By the same theorem, this difference is exactly the integral of \(f\) from \(g(a)\) to \(g(b)\): $$ F(g(b))-F(g(a)) = \int_{g(a)}^{g(b)}f(u)\,du. $$
This argument explains why decreasing \(g\) does not require a new substitution rule. The bounds record where the original endpoints go; the orientation of the transformed integral records the sign correctly.
Worked examples
Worked Example: Keep the decreasing endpoint order
Evaluate $$ \int_0^1 -2(3-2x)^2\,dx. $$
Set \(u=3-2x\). Then \(du=-2\,dx\), and the inner function decreases because its derivative is \(-2<0\). Transform both endpoints: $$ x=0\Rightarrow u=3,\qquad x=1\Rightarrow u=1. $$ Thus $$ \int_0^1 -2(3-2x)^2\,dx = \int_3^1 u^2\,du. $$ An antiderivative of \(u^2\) is \(u^3/3\), so $$ \int_3^1 u^2\,du = \left[\frac{u^3}{3}\right]_3^1 = \frac{1^3}{3}-\frac{3^3}{3} = \frac{1}{3}-9 = -\frac{26}{3}. $$ Check using an antiderivative in \(x\): $$ \frac{d}{dx}\left(\frac{(3-2x)^3}{3}\right) = \frac{1}{3}\cdot 3(3-2x)^2(-2) = -2(3-2x)^2. $$ Its endpoint difference is $$ \frac{(3-2)^3}{3}-\frac{(3-0)^3}{3} = \frac{1}{3}-9 = -\frac{26}{3}, $$ as required.
Worked Example: Swap the bounds and include a minus sign
Evaluate $$ \int_0^2 2(5-2x)^3\,dx. $$
Let \(u=5-2x\). Then \(du=-2\,dx\), so \(2\,dx=-du\). The inner function decreases on the interval, and the transformed endpoints are $$ x=0\Rightarrow u=5,\qquad x=2\Rightarrow u=1. $$ Substituting gives $$ \int_0^2 2(5-2x)^3\,dx = -\int_5^1 u^3\,du. $$ Swap the bounds and keep the minus sign: $$ -\int_5^1 u^3\,du = \int_1^5 u^3\,du = \left[\frac{u^4}{4}\right]_1^5 = \frac{5^4}{4}-\frac{1^4}{4} = \frac{625-1}{4} = 156. $$ Check with an antiderivative in \(x\): $$ \frac{d}{dx}\left(-\frac{(5-2x)^4}{4}\right) = -\frac{1}{4}\cdot4(5-2x)^3(-2) = 2(5-2x)^3. $$ The endpoint difference is $$ -\frac{(5-4)^4}{4}-\left(-\frac{(5-0)^4}{4}\right) = -\frac14+\frac{625}{4} = 156. $$ The original integrand is positive throughout the interval, and the positive result is consistent with that fact.
Worked Example: A decreasing inner function with a radical
Evaluate $$ \int_1^2 2x\sqrt{4-x^2}\,dx. $$
Choose \(u=4-x^2\). Its derivative is \(du=-2x\,dx\), and \(g'(x)=-2x<0\) for every \(x\in[1,2]\), so the inner function decreases on this interval. The endpoints become $$ x=1\Rightarrow u=3,\qquad x=2\Rightarrow u=0. $$ Since \(2x\,dx=-du\), the integral is $$ \int_1^2 2x\sqrt{4-x^2}\,dx = -\int_3^0 \sqrt{u}\,du = \int_0^3 \sqrt{u}\,du. $$ Using \(\sqrt{u}=u^{1/2}\), an antiderivative is \(\frac{2}{3}u^{3/2}\). Hence $$ \int_0^3 \sqrt{u}\,du = \left[\frac{2}{3}u^{3/2}\right]_0^3 = \frac{2}{3}\cdot 3^{3/2}-0 = 2\sqrt{3}. $$ Check in the original variable: $$ \frac{d}{dx}\left(-\frac{2}{3}(4-x^2)^{3/2}\right) = -\frac{2}{3}\cdot\frac{3}{2}(4-x^2)^{1/2}(-2x) = 2x\sqrt{4-x^2}. $$ At the endpoints the antiderivative has values \(0\) at \(x=2\) and \(-2\sqrt{3}\) at \(x=1\). Their difference is \(0-(-2\sqrt{3})=2\sqrt{3}\), confirming the result.
A common sign mistake
A frequent error is to swap the transformed bounds but forget to account for the sign change. For the first example, the correct substitution gives $$ \int_0^1 -2(3-2x)^2\,dx = \int_3^1 u^2\,du = -\frac{26}{3}. $$ If we reverse the limits and write \(\int_1^3u^2\,du\) without a minus sign, we get the wrong value: $$ \int_1^3u^2\,du = \left[\frac{u^3}{3}\right]_1^3 = \frac{27-1}{3} = \frac{26}{3}. $$ The correction is to preserve the orientation identity: $$ \int_3^1u^2\,du = -\int_1^3u^2\,du = -\frac{26}{3}. $$
A reliable check is to ask whether the signs agree with the original integrand. In the first example the integrand is negative throughout, so its definite integral over an interval of positive length must be negative. In the second and third examples the integrands are positive throughout, and both results are positive.
A dependable procedure
Write \(u=g(x)\) and calculate \(du=g'(x)\,dx\), including any required constant or negative factor.
Verify that \(g'(x)<0\) on the interval, or otherwise establish that \(g\) is decreasing there.
Calculate \(g(a)\) for the lower \(x\)-endpoint and \(g(b)\) for the upper one. Do not sort the resulting values.
Keep the transformed limits as \(g(a)\) to \(g(b)\), or reverse them and include a minus sign.
Evaluate exactly and, when practical, differentiate an antiderivative in \(x\) and compare the signs.
When this approach applies
The decreasing-function case applies when one inner function \(g\) decreases across the entire integration interval. A sufficient condition is that \(g\) is differentiable and \(g'(x)<0\) throughout that interval. The substitution theorem also requires that the remaining function of \(u\) be continuous over the transformed range, so the definite integral is well-defined under the stated conditions.
If the inner function decreases only on part of the interval and then increases, it is not decreasing across the whole interval. One cannot treat the entire interval as a single decreasing change of variable under the condition used here. That situation requires separate consideration.
The central habit is simple: calculate the images of the original endpoints in their original order, then preserve that order or reverse it with a minus sign. Doing so keeps the substitution, the integral’s orientation, and its sign aligned.
Check Your Understanding
For each question, track the transformed endpoints before evaluating.
- If \(u=7-3x\) on \([1,2]\), what are the transformed lower and upper endpoints, in order?
- When \(u=g(x)\) decreases from \(g(a)\) to \(g(b)\), why must the transformed integral retain the order \(g(a)\) to \(g(b)\), unless a minus sign is introduced?
- Rewrite \(-\int_6^2 f(u)\,du\) with increasing bounds.
- In \(\int_0^1 -2(3-2x)^2\,dx\), what sign should the answer have, and what feature of the integrand determines it?
- What condition on \(g\) lets you use the single decreasing-interval procedure across the whole interval?
