Two Consistent Routes
A substitution in a definite integral changes more than the name of the integration variable. If \(u=g(x)\), then the differential factor changes as well, and the limits must match whichever variable is being used at the point of evaluation.
There are two consistent routes. One is to change to \(u\), transform both bounds, and finish the calculation in \(u\). The other is to find an antiderivative using \(u\), express that antiderivative back in \(x\), and evaluate using the original \(x\)-bounds. The routes give the same value, but mixing their variables and bounds can give a wrong answer.
The Substitution Rule and Its Conditions
For the standard proper Riemann-integral substitution rule, the inner function needs enough regularity to make the integrand continuous. In the version used here, \(g\) is continuously differentiable on the closed interval \([a,b]\), and \(h\) is continuous on the range of \(g\) there. The product \(h(g(x))g'(x)\) is then continuous on \([a,b]\), so its proper definite integral is defined.
The notation \(g\in C^1([a,b])\) means that \(g\) has a continuous derivative on the interval, including the appropriate one-sided behavior at the endpoints. The theorem does not require \(g\) to be one-to-one for this matched integrand. The derivative factor \(g'(x)\) is essential to the formula; the theorem does not say that an arbitrary factor involving \(g(x)\) can be changed to \(du\).
Why the Two Routes Agree
Here is the reason the two approaches produce the same result. Since \(h\) is continuous, it has an antiderivative \(H\) on the interval containing the range of \(g\). The chain rule gives the derivative of \(H(g(x))\). Applying the standard definite-integral evaluation rule to that derivative proves the substitution identity.
Let \(H'(u)=h(u)\) on an interval containing the range of \(g\).
For \(F(x)=H(g(x))\), we have \(F'(x)=H'(g(x))g'(x)=h(g(x))g'(x)\).
The definite integral is \(F(b)-F(a)=H(g(b))-H(g(a))\).
Since \(H'=h\), the difference \(H(g(b))-H(g(a))\) equals \(\int_{g(a)}^{g(b)}h(u)\,du\).
This also explains the relationship between the two routes. If we integrate in \(u\), we evaluate \(H(u)\) at \(g(a)\) and \(g(b)\). If we write the antiderivative as \(H(g(x))\), we evaluate it at \(a\) and \(b\). These are the same endpoint values expressed in different variables.
Route One: Change the Bounds and Stay in \(u\)
To finish entirely in the new variable, identify \(u=g(x)\), replace the differential factor by \(du=g'(x)\,dx\), and calculate the new bounds by substituting the original endpoints into \(g\). Then integrate and evaluate using those new bounds.
Worked Example: Stay in the New Variable
Evaluate \(\int_1^2 2x\cos(x^2)\,dx\) by changing to \(u=x^2\) and completing the calculation in \(u\).
Set \(u=x^2\). Differentiating gives \(du=2x\,dx\). At the lower endpoint \(x=1\), \(u=1^2=1\). At the upper endpoint \(x=2\), \(u=2^2=4\). Therefore, both the integrand and the bounds change:
Check: the antiderivative in the original variable is \(\sin(x^2)\), since \(\frac{d}{dx}\sin(x^2)=2x\cos(x^2)\). Evaluating it at the original bounds gives \(\sin(2^2)-\sin(1^2)=\sin(4)-\sin(1)\), agreeing with the transformed calculation.
Route Two: Return to \(x\) Before Evaluating
Alternatively, use \(u\) to find an antiderivative, then replace \(u\) by \(g(x)\) in that antiderivative. Once the expression is back in \(x\), keep the original \(x\)-bounds. This route is often convenient when the antiderivative is easy to recognize after substitution, but evaluating it in terms of the original endpoints is clearer.
Worked Example: Convert the Antiderivative Back to \(x\)
Evaluate \(\int_0^1 \frac{2x}{1+x^2}\,dx\), returning to \(x\) before applying the bounds.
Set \(u=1+x^2\), so \(du=2x\,dx\). The integral becomes \(\int \frac{1}{u}\,du=\ln|u|+C\). Returning to \(x\) gives \(\ln|1+x^2|+C\). On \([0,1]\), \(1+x^2\) is positive, so this is \(\ln(1+x^2)+C\). Now use the original bounds:
Check by differentiating the proposed antiderivative:
It matches the integrand, and the endpoint difference is \(\ln(2)\). For comparison, the all-\(u\) route gives bounds \(u=1\) and \(u=2\), and \(\int_1^2 \frac{1}{u}\,du=\ln(2)-\ln(1)=\ln(2)\) as well.
Choosing Between the Routes
Neither route is inherently more correct. Choose the one that makes the variables and endpoint evaluations easiest to manage, while keeping them consistent. Staying in \(u\) is natural when the transformed integrand and bounds are simple. Returning to \(x\) is natural when the antiderivative is easy to express in \(x\), or when it makes the original endpoints especially clear.
| Route | Expression at evaluation | Bounds to use |
|---|---|---|
| Stay in \(u\) | An antiderivative written in \(u\) | \(g(a)\) and \(g(b)\) |
| Return to \(x\) | An antiderivative written in \(x\) | \(a\) and \(b\) |
In either route, compute the new endpoint values directly from the substitution. Do not assume the new bounds are the minimum and maximum values attained by \(g\); for the substitution formula, they are \(g(a)\) and \(g(b)\), in that order.
Endpoint Bounds and a Nonmonotone Substitution
A useful detail is that the endpoint bounds need not describe the full range of \(g\) in increasing order. When \(g\) is not monotone, the transformed integral still records the signed change along the original interval. The matched substitution formula remains valid under the stated conditions, but replacing its endpoint bounds with the minimum and maximum of the range can change the answer.
Worked Example: The Endpoints Can Have the Same Image
Evaluate \(\int_0^2 2(x-1)e^{(x-1)^2}\,dx\) using \(u=(x-1)^2\).
Set \(u=(x-1)^2\). Then \(du=2(x-1)\,dx\). At both original endpoints, \(u=1\): at \(x=0\), \(u=(-1)^2=1\); at \(x=2\), \(u=1^2=1\). Thus the transformed integral has equal bounds:
Check by returning to \(x\): an antiderivative is \(e^{(x-1)^2}\), because its derivative is \(e^{(x-1)^2}\cdot 2(x-1)\). Its endpoint difference is \(e^{(2-1)^2}-e^{(0-1)^2}=e-e=0\). Although \(u\) takes values from \(0\) to \(1\) on the interval, using \(0\) and \(1\) as the transformed bounds would give \(\int_0^1e^u\,du=e-1\), which is not the value of the original integral.
A Common Mistake: Changing the Variable but Not the Bounds
The bounds are part of the variable change. If an integral is written in \(u\), its bounds must be \(u\)-values. Keeping the original \(x\)-bounds generally evaluates the wrong integral, even if the antiderivative itself is correct.
Worked Example: Correcting Unchanged Bounds
Consider again \(\int_1^2 2x\cos(x^2)\,dx\). A mistaken calculation sets \(u=x^2\), \(du=2x\,dx\), but keeps the bounds \(1\) and \(2\):
The error is that the original \(x\)-endpoint \(x=2\) corresponds to \(u=4\), not \(u=2\). The correct calculation is:
The check in \(x\) gives \(\sin(x^2)\big|_1^2=\sin(4)-\sin(1)\), confirming the corrected result. The incorrect calculation paired a \(u\)-antiderivative with bounds that still referred to \(x\).
A Reliable Consistency Check
Before evaluating a substituted definite integral, pause and identify the variable in every part of the expression. If the differential is \(du\), the integrand and bounds should be expressed in \(u\). If the antiderivative has been rewritten in \(x\), use the original \(x\)-bounds. A quick check is to substitute each original endpoint into \(u=g(x)\) and verify that the resulting endpoint values appear in the transformed integral in the correct order.
It is also useful to differentiate the final antiderivative when using the return-to-\(x\) route. This checks that the chain-rule factor was retained. If the transformed integral has reversed or equal bounds, preserve them as written: reversing bounds changes the sign, and equal bounds give zero.
Check Your Understanding
For each question, keep track of which variable the bounds refer to.
- For \(\int_0^2 3x^2\sin(x^3)\,dx\), what are the transformed bounds if \(u=x^3\)?
- State the two consistent evaluation routes for a definite integral after setting \(u=g(x)\).
- Why does the standard substitution formula used here assume \(g\in C^1([a,b])\) and continuity of \(h\) on the range of \(g\)?
- If \(g(a)=g(b)\), what does the transformed integral \(\int_{g(a)}^{g(b)}h(u)\,du\) equal?
- What is wrong with keeping the original \(x\)-bounds after rewriting the integral in \(u\)?
