Why Change the Bounds?
A substitution can make an integrand simpler by replacing a complicated expression with a new variable. For an indefinite integral, one usually returns to the original variable before writing the answer. A definite integral offers another option: convert the endpoints to the new variable, then do the remaining integration and evaluation there.
This method keeps the variable and its bounds consistent from start to finish. If \(u=g(x)\), then the original endpoints \(x=a\) and \(x=b\) become \(u=g(a)\) and \(u=g(b)\). Once the substitution is made, the integral is evaluated using those new bounds; there is no need to express the antiderivative in terms of \(x\).
The Definite-Integral Substitution Rule
Suppose \(g\) is continuously differentiable on \([a,b]\), and \(f\) is continuous on an interval containing all the values \(g(x)\) for \(a\le x\le b\). Then the substitution rule is
The new lower bound is the value of \(u=g(x)\) at the original lower endpoint \(x=a\). The new upper bound is its value at \(x=b\). The order is determined by the original endpoints; if \(g(b)<g(a)\), the new bounds appear in decreasing order.
In an actual problem, the integrand may contain a constant multiple of \(g'(x)\), or a factor that must be rearranged before the substitution matches the rule. Keep that factor in the integral. For instance, if \(du=2x\,dx\), then \(x\,dx=\frac12\,du\), so a factor of \(\frac12\) remains.
Why the Rule Works
The rule follows from the chain rule and the Fundamental Theorem of Calculus. Let \(F\) be an antiderivative of \(f\), so \(F'(u)=f(u)\). By the chain rule,
Thus \(F(g(x))\) is an antiderivative of the original integrand. Applying the Fundamental Theorem of Calculus gives
On the other hand, \(F\) is an antiderivative of \(f\), so the integral in the new variable is
Both expressions equal the same difference, which proves the substitution rule. This argument also explains why the new bounds must be \(g(a)\) and \(g(b)\): those are exactly the endpoint values that appear when the chain-rule antiderivative is evaluated.
A Reliable Process
Use the following sequence to keep the variable, differential, and endpoints aligned.
Set \(u=g(x)\), selecting an inner expression whose derivative is present, or can be matched by rearranging the integrand.
Write \(du=g'(x)\,dx\), account for any constant factor, and calculate \(u=g(a)\) and \(u=g(b)\).
Express the integrand, differential, and both bounds using \(u\). Do not leave an \(x\) in the transformed integral.
Find an antiderivative in \(u\), then substitute the new upper and lower bounds into it.
Worked Examples
Worked Example: A Trigonometric Integrand
Evaluate \(\int_0^1 2x\cos(x^2)\,dx\) using substituted bounds.
Choose the inner expression \(u=x^2\). Then \(du=2x\,dx\). Convert the endpoints: when \(x=0\), \(u=0^2=0\); when \(x=1\), \(u=1^2=1\). Therefore,
The integral is now finished in \(u\); no change back to \(x\) is needed. To check, the corresponding antiderivative in the original variable is \(\sin(x^2)\), and
Evaluating this antiderivative at the original endpoints gives \(\sin(1^2)-\sin(0^2)=\sin(1)\), agreeing with the result.
Worked Example: A Rational Integrand and a Constant Factor
Evaluate \(\int_0^2 \frac{x}{x^2+1}\,dx\).
Let \(u=x^2+1\). Then \(du=2x\,dx\), so \(x\,dx=\frac12\,du\). The transformed bounds are \(u=0^2+1=1\) at \(x=0\), and \(u=2^2+1=5\) at \(x=2\). Substitute all of these into the integral:
Here \(u\) is positive throughout the new interval, so \(\ln u\) is defined. For a check, differentiate the corresponding antiderivative in \(x\):
Its endpoint difference is \(\frac12\ln(2^2+1)-\frac12\ln(0^2+1)=\frac12\ln5-\frac12\ln1=\frac12\ln5\), as obtained using the new bounds.
Worked Example: A Substitution That Reverses the Bounds
Evaluate \(\int_0^\pi \sin x\,e^{\cos x}\,dx\).
Set \(u=\cos x\), which gives \(du=-\sin x\,dx\), or \(\sin x\,dx=-du\). Convert the endpoints: at \(x=0\), \(u=\cos 0=1\); at \(x=\pi\), \(u=\cos\pi=-1\). The new bounds run from \(1\) down to \(-1\):
The change from \(-\int_1^{-1}\) to \(\int_{-1}^1\) uses the fact that reversing the bounds changes the sign of a definite integral. Alternatively, evaluate directly: \(-[e^u]_1^{-1}=-\bigl(e^{-1}-e\bigr)=e-e^{-1}\).
To check, differentiate \(-e^{\cos x}\):
Its endpoint difference is \(-e^{\cos\pi}-(-e^{\cos0})=-e^{-1}+e=e-e^{-1}\), matching the transformed evaluation.
Bounds Must Match the Variable
A definite integral pairs a variable with bounds expressed in that variable. Once the integral has been changed to \(u\), its bounds must also be values of \(u\). The original endpoints are values of \(x\), and using them unchanged with a \(u\)-integral generally changes the problem.
A second useful check is to inspect the transformed integral before evaluating it. It should contain only the new variable and its differential. If an \(x\) remains, or if the limits still represent \(x\)-values, the substitution has not been completed consistently.
Why This Method Matters
Changing the bounds can make a definite integral shorter and clearer because the antiderivative is evaluated directly in the variable that simplified the integrand. It also makes endpoint behavior visible: a substitution may send the original interval to a different interval, and that interval may run in the opposite direction.
The order of the new limits is not an arbitrary choice. The definite integral is oriented: switching its endpoints changes its sign. Thus, when \(u=g(x)\) decreases from \(x=a\) to \(x=b\), the transformed upper bound can be smaller than the lower bound. Preserve that order and the minus sign from the differential, or reverse the bounds and change the sign, but do not do both incorrectly.
Before finalizing an answer, check three things: the transformed differential includes every needed factor; each new bound is computed from its corresponding original endpoint; and the final evaluation uses the new variable consistently. If desired, differentiate an antiderivative in the original variable as an independent check, as in the examples.
Check Your Understanding
For each question, keep the new variable and its bounds consistent.
- For \(\int_1^3 2x(x^2+4)^3\,dx\), choose \(u=x^2+4\). What are the new bounds, and what is the transformed integral?
- Evaluate \(\int_0^1 6x e^{3x^2}\,dx\) using \(u=3x^2\). Show how the constant factor and bounds change.
- In \(\int_0^{\pi/2}\sin x\cos x\,dx\), use \(u=\cos x\). Why do the new bounds appear in decreasing order?
- What is inconsistent about changing the integrand to a function of \(u\) while leaving the original \(x\)-bounds unchanged?
- State the definite-integral substitution rule and identify its new lower and upper bounds.
