Substitution and definite integrals
A substitution can make an integral easier to evaluate by expressing its integrand in terms of a new variable. For a definite integral, however, the endpoints also belong to the variable of integration. If the integrand is rewritten using \(u\), its bounds must be rewritten in \(u\) as well—or the antiderivative must be changed back to \(x\) before using the original bounds.
The Fundamental Theorem of Calculus connects an antiderivative to the value of a definite integral: evaluate the antiderivative at the upper endpoint and subtract its value at the lower endpoint. Combining that theorem with substitution gives a reliable way to evaluate integrals whose integrands contain a composite function.
The substitution formula for definite integrals
Suppose \(g\) is continuously differentiable on \([a,b]\), and \(F\) is continuous on an interval containing all values of \(g(x)\) for \(a\leq x\leq b\). If \(A\) is an antiderivative of \(F\), so \(A'(u)=F(u)\), then the chain rule and the Fundamental Theorem give
The bounds \(g(a)\) and \(g(b)\) are the images of the original endpoints under the substitution. They are not the original endpoint numbers unless the substitution happens to leave them unchanged.
In practice, one often writes \(u=g(x)\) and \(du=g'(x)\,dx\). The differential notation records how the integrand changes; the endpoint values record how the interval changes. Together, they express the same equality above.
Why the formula works
The reason substitution works for a definite integral is the chain rule. If \(A'(u)=F(u)\), then differentiating the composite function \(A(g(x))\) gives
Thus \(A(g(x))\) is an antiderivative, in the variable \(x\), of the original integrand. The Fundamental Theorem then gives the integral as the change in that antiderivative between \(a\) and \(b\).
Applying the Fundamental Theorem directly in the \(u\)-variable gives \(\int_{g(a)}^{g(b)}F(u)\,du=A(g(b))-A(g(a))\). The two evaluations therefore agree. This argument also explains why the transformed endpoints remain in their actual order: if \(g(b)<g(a)\), the resulting integral has reversed bounds and carries the corresponding sign.
Worked Example: Transforming the bounds
Worked Example: Transforming the bounds
Evaluate \(\int_0^1 2x\cos(x^2)\,dx\) by substitution.
The inner function is \(x^2\), and its derivative is \(2x\). Set \(u=x^2\), so \(du=2x\,dx\). Transform the endpoints: when \(x=0\), \(u=0^2=0\); when \(x=1\), \(u=1^2=1\). Therefore,
An antiderivative of \(\cos(u)\) is \(\sin(u)\). By the Fundamental Theorem,
To check, differentiate the corresponding antiderivative in the original variable: \(\frac{d}{dx}\sin(x^2)=2x\cos(x^2)\), which is the original integrand. Evaluating it at the original bounds gives \(\sin(1^2)-\sin(0^2)=\sin(1)-\sin(0)=\sin(1)\), as required.
Two consistent ways to finish
After the substitution, there are two valid ways to complete a definite-integral calculation. The first is to transform the bounds and evaluate the new-variable antiderivative there. The second is to find the antiderivative in the new variable, replace \(u\) by \(g(x)\), and evaluate at the original \(x\)-bounds. The choice is a matter of convenience; consistency is essential.
For example, after setting \(u=g(x)\), suppose an antiderivative in \(u\) is \(A(u)\). The transformed-bound method calculates \(A(g(b))-A(g(a))\). The substitute-back method calculates the same quantity by writing \(A(g(x))\) and evaluating it at \(b\) and \(a\). These are not different answers or different theorems; they are two forms of the same endpoint calculation.
Write \(u=g(x)\) and identify \(du=g'(x)\,dx\), including any constant factor needed to match the integrand.
If you will evaluate in \(u\), calculate \(u=g(a)\) and \(u=g(b)\). Otherwise, keep the original bounds and return to \(x\) before evaluating.
Find an antiderivative in the variable you are using and apply the Fundamental Theorem with the matching bounds.
Worked Example: Accounting for a constant factor
Worked Example: Accounting for a constant factor
Evaluate \(\int_1^2 \frac{x}{(x^2+1)^2}\,dx\).
Set \(u=x^2+1\). Then \(du=2x\,dx\), so \(x\,dx=\frac12\,du\). Transform the bounds: at \(x=1\), \(u=1^2+1=2\); at \(x=2\), \(u=2^2+1=5\). The integral becomes
An antiderivative of \(u^{-2}\) is \(-u^{-1}\). Applying the bounds,
Check in the original variable. Substituting \(u=x^2+1\) into the antiderivative and including the factor \(\frac12\) gives \(-\frac{1}{2(x^2+1)}\). Its derivative is
Evaluating this antiderivative at \(x=2\) and \(x=1\) gives \(-\frac{1}{10}-(-\frac14)=\frac{3}{20}\). Both the derivative and the endpoint calculation confirm the result.
When the substitution reverses the bounds
A substitution need not send the lower endpoint to a smaller number than the upper endpoint. If \(g\) decreases over the interval, it is possible that \(g(a)>g(b)\). The transformed integral then has its bounds in descending order. Keep that order and use the usual rule “upper endpoint value minus lower endpoint value”; the resulting sign is part of the calculation.
Worked Example: A decreasing substitution
Worked Example: A decreasing substitution
Evaluate \(\int_0^\pi \sin(x)e^{\cos(x)}\,dx\).
Set \(u=\cos(x)\), so \(du=-\sin(x)\,dx\). At the original lower endpoint \(x=0\), \(u=\cos(0)=1\). At the original upper endpoint \(x=\pi\), \(u=\cos(\pi)=-1\). The substitution therefore gives
An antiderivative of \(e^u\) is \(e^u\), so
For a check in \(x\), an antiderivative is \(-e^{\cos(x)}\), because
Evaluating at the original bounds gives \(-e^{\cos(\pi)}-(-e^{\cos(0)})=-e^{-1}+e=e-e^{-1}\), agreeing with the transformed calculation.
A common bounds mistake
In the last example, a tempting but incorrect step would be to keep the bounds \(0\) and \(\pi\) after changing the variable to \(u\), writing \(-\int_0^\pi e^u\,du\). This mixes \(x\)-values with a \(u\)-integral. It produces \(-(e^\pi-1)\), which is not the value of the original integral. The correct \(u\)-bounds are \(1\) and \(-1\), obtained by applying \(u=\cos(x)\) to the original endpoints.
Another possible error is to reverse the bounds to \(-1\) and \(1\) without changing the sign. Reversing bounds changes the sign of an integral: \(\int_1^{-1}e^u\,du=-\int_{-1}^1e^u\,du\). Since the substitution already contributes a negative sign through \(du=-\sin(x)\,dx\), both signs must be accounted for. Keeping the original transformed order avoids losing track of them.
Why the combined method matters
Substitution can expose the basic antiderivative hidden inside a composite integrand, while the Fundamental Theorem turns that antiderivative into a definite value. The method is useful only when the transformation is carried through consistently: the differential, any constant factor, and the endpoints all contribute to the result.
The chain-rule derivation also gives a useful check. Once an antiderivative has been proposed, differentiating its composition with \(g(x)\) should recover the original integrand. This checks the antiderivative itself, while evaluating it at the proper endpoints checks the definite-integral calculation. The checks address different possible errors, so both are valuable.
Check Your Understanding
For each question, keep track of the variable used for the antiderivative and the bounds used for evaluation.
- For \(\int_0^2 2x(x^2+1)^3\,dx\), what substitution and transformed bounds make the integral an integral of a power of \(u\)?
- State the substitution formula for \(\int_a^b F(g(x))g'(x)\,dx\), including the required relationship between \(A\) and \(F\).
- If \(u=g(x)\) and \(g(a)>g(b)\), what order should the transformed bounds have?
- For \(\int_0^1 2x\cos(x^2)\,dx\), how can you check the answer by differentiating an antiderivative in \(x\)?
- Explain why retaining the original \(x\)-bounds while evaluating an antiderivative written in \(u\) is inconsistent.
