Why the Bounds Must Change
A definite integral has both an integrand and limits that describe the variable over which the accumulation is measured. If a substitution replaces \(x\) by a new variable \(u=g(x)\), the limits must describe the corresponding values of \(u\). Keeping the original numerical limits while changing the integration variable generally changes the interval and therefore the integral.
The rule is simple: evaluate the substitution at the original lower and upper endpoints. If \(x=a\) corresponds to \(u=g(a)\), and \(x=b\) corresponds to \(u=g(b)\), then the integral in \(u\) has endpoints \(g(a)\) and \(g(b)\), in that order. The differential must also be expressed in terms of \(u\), as in the earlier material on substitution.
The Definite Substitution Rule
Here is a precise version of the endpoint rule. Let \(a<b\), let \(g\) be continuously differentiable on \([a,b]\), and suppose its values lie in an interval \(I\). Let \(h\) be continuous on \(I\). Then the integrand \(h(g(x))g'(x)\) is continuous, and
The formula does not require \(g\) to be increasing. The endpoints on the right are always written as \(g(a)\) followed by \(g(b)\), so their order is determined by the images of the original endpoints. In the examples below, the substitutions are increasing over the intervals used.
In practice, substitution is useful when the integrand can be written as a function of \(g(x)\) multiplied by \(g'(x)\), possibly with a constant factor that can be handled separately. Once \(u=g(x)\) and \(du=g'(x)\,dx\) are identified, the endpoint calculations are direct substitutions into \(g\).
Why the Rule Works
The proof connects the substitution rule to the Fundamental Theorem of Calculus and the chain rule. Because \(h\) is continuous, it has an antiderivative \(H\) on \(I\), with \(H'(u)=h(u)\). The chain rule gives the derivative of the composite function \(H(g(x))\):
Thus \(H(g(x))\) is an antiderivative of the original integrand. Applying the Fundamental Theorem of Calculus and then using \(H'=h\) gives
This proves the rule, including why the transformed endpoints are the values of \(g\) at the original endpoints. The same endpoint values appear because the antiderivative is evaluated at \(x=b\) and \(x=a\), producing \(H(g(b))\) and \(H(g(a))\).
Worked Examples
Worked Example: A Polynomial Substitution
Evaluate \(\displaystyle \int_0^2 2x(x^2+1)^3\,dx\) by setting \(u=x^2+1\).
First, differentiate the substitution: \(du=2x\,dx\). The original endpoints are \(x=0\) and \(x=2\), so transform each one:
The factor \(2x\,dx\) is \(du\), and \((x^2+1)^3=u^3\). Therefore
Check: an antiderivative in \(x\) is \(\frac{(x^2+1)^4}{4}\), since
Its endpoint difference is \(\frac{5^4}{4}-\frac{1^4}{4}=156\), agreeing with the transformed integral.
Worked Example: An Affine Substitution
Evaluate \(\displaystyle \int_0^2 3(3x-1)^4\,dx\) using \(u=3x-1\).
Differentiate to get \(du=3\,dx\). Transform the endpoints by evaluating the affine expression at each original endpoint:
Since \(3\,dx=du\) and \((3x-1)^4=u^4\), the integral becomes
Check: differentiating \(\frac{(3x-1)^5}{5}\) gives \(\frac{1}{5}\cdot5(3x-1)^4\cdot3=3(3x-1)^4\). Evaluating this antiderivative at the original endpoints gives \(\frac{5^5-(-1)^5}{5}=\frac{3126}{5}\), as required.
Worked Example: A Trigonometric Substitution
Evaluate \(\displaystyle \int_0^{\pi/3}\frac{\cos x}{(2+\sin x)^2}\,dx\) using \(u=2+\sin x\).
The derivative is \(du=\cos x\,dx\). Evaluate \(u\) at both original endpoints. At \(x=0\), \(\sin 0=0\), so \(u=2\). At \(x=\pi/3\), \(\sin(\pi/3)=\sqrt{3}/2\), so \(u=2+\sqrt{3}/2\). Thus
Check: an antiderivative in \(x\) is \(-1/(2+\sin x)\), because
Evaluating it at \(x=\pi/3\) and \(x=0\) gives \(-1/(2+\sqrt{3}/2)+1/2\), the same value as the transformed integral.
A Reliable Endpoint Procedure
Writing the endpoint transformation explicitly is a useful way to prevent a variable mismatch. A complete substitution can be organized as follows:
Write \(u=g(x)\) and identify how the remaining factors in the integrand relate to \(g'(x)\,dx\).
Calculate \(g(a)\) and \(g(b)\) from the original limits \(x=a\) and \(x=b\).
Express the integrand, differential, and limits in terms of \(u\), so no original \(x\)-limits remain.
Find the transformed integral and, when useful, verify the answer using an antiderivative in the original variable.
Common Mistakes
A frequent error is to change the integrand and differential but keep the old numerical limits. For example, in the first worked example, the substitution is \(u=x^2+1\), and the original \(x\)-limits are \(0\) and \(2\). The incorrect calculation is
This calculation integrates \(u^3\) over \(0\leq u\leq2\), but the substitution maps the original endpoints to \(u=1\) and \(u=5\). It therefore uses the wrong interval. The corrected calculation is
Another mistake is to transform the limits but leave part of the integrand in \(x\). For instance, after setting \(u=3x-1\), the expression \((3x-1)^4\) becomes \(u^4\), and \(3\,dx\) becomes \(du\). The transformed integral must use \(u\) throughout; writing \(u^4\,dx\) would still mix variables and would not be the result of the substitution.
When the Endpoint Rule Applies
The theorem applies when the integrand has the form \(h(g(x))g'(x)\), with \(g\) continuously differentiable on the original closed interval and \(h\) continuous on an interval containing the values of \(g\). In practical problems, the derivative factor may appear exactly as \(g'(x)\,dx\), as in the examples, or be identified after algebraic rearrangement. If it appears multiplied by a constant, that factor must be accounted for rather than silently discarded.
These conditions ensure that the integrand is continuous and that the Fundamental Theorem of Calculus and chain rule can be used as in the proof. The rule is not a license to replace an arbitrary expression by a new variable without checking the derivative factor and the limits. A correct substitution accounts for all three: the inner expression, its differential, and the images of the endpoints.
Check Your Understanding
For each question, keep track of which variable is being integrated and calculate endpoint images before evaluating.
- Under \(u=x^2+3\), what are the transformed endpoints when the original interval is \(0\leq x\leq2\)?
- For \(\displaystyle\int_1^4 2(2x+1)^3\,dx\), set \(u=2x+1\). What are the new limits and the resulting integral in \(u\)?
- State the continuity and differentiability conditions used in the definite substitution rule.
- Explain why \(\displaystyle\int_0^2 u^3\,du\) does not represent the first worked example after substituting \(u=x^2+1\).
- For \(u=1+\sin x\) on \(0\leq x\leq\pi/6\), calculate the transformed lower and upper endpoints.
- In one sentence, explain why \(g(a)\) and \(g(b)\) are the limits after the substitution \(u=g(x)\).
