Evaluate in the New Variable
A substitution in a definite integral changes more than the expression being integrated. It also changes the variable and the endpoint values. Once the integrand and differential have both been rewritten in terms of a new variable \(u\), the integral can be completed entirely in \(u\): find an antiderivative with respect to \(u\), then evaluate it at the transformed endpoints.
This method does not require solving the substitution equation for \(x\). For example, if \(u=g(x)\), it may be inconvenient or impossible to write \(x\) explicitly as a function of \(u\). That causes no difficulty when the transformed integrand and the new endpoint values are available.
The Evaluation Rule
Suppose the integrand can be written as \(q(g(x))g'(x)\), where \(g\) is continuously differentiable on \([a,b]\), and \(q\) is continuous on an interval containing all values \(g(x)\) for \(x\in[a,b]\). Let \(Q\) be an antiderivative of \(q\), so \(Q'(u)=q(u)\). Then the definite integral is
The middle integral is written in the new variable \(u\). Its bounds are the values of \(u=g(x)\) at the original endpoints; the bounds are not found by inverting \(g\). In particular, if \(g\) is decreasing, then \(g(b)<g(a)\), and the transformed bounds appear in that order. The sign from the change in variable is accounted for by the oriented bounds.
Why No Inverse Formula Is Needed
The reason is the chain rule. The antiderivative in \(u\), when composed with \(g(x)\), becomes an antiderivative in \(x\). Therefore, the original integral can be evaluated from the endpoint values of that composition. Those values require \(g(a)\) and \(g(b)\), not a formula expressing \(x\) in terms of \(u\).
Thus \(Q(g(x))\) is an antiderivative of the original integrand. The Fundamental Theorem of Calculus gives
On the other hand, since \(Q\) is an antiderivative of \(q\), the integral in the new variable is
Both evaluations give the same endpoint difference. This proves the rule and shows exactly why inverting \(u=g(x)\) is unnecessary. The argument requires the stated continuity and differentiability conditions so that the chain rule and the Fundamental Theorem of Calculus apply.
A Reliable Procedure
For a definite integral, keep track of the variable and bounds together. The following steps make the evaluation explicit.
Write the substitution clearly as \(u=g(x)\), and compute \(du=g'(x)\,dx\).
Rewrite the integrand and differential so that no \(x\) remains.
Evaluate \(u=g(x)\) at the original lower and upper endpoints, in that order.
Find an antiderivative with respect to \(u\), then substitute the new bounds into it.
Differentiate the composed antiderivative to confirm it gives the original integrand, and check the endpoint order and sign.
Worked Examples
Worked Example: A Composite Radical
Evaluate
Set \(u=x^2+3\). Then \(du=2x\,dx\), so the numerator and differential become \(du\). Transform the endpoints: when \(x=0\), \(u=0^2+3=3\); when \(x=1\), \(u=1^2+3=4\). The integral is now entirely in \(u\):
No equation for \(x\) in terms of \(u\) is needed: the transformed integrand and both transformed endpoints suffice. To check, the derivative of \(2\sqrt{u}\) is \(u^{-1/2}\). Composing this antiderivative with \(u=x^2+3\) gives \(2\sqrt{x^2+3}\), whose derivative is \(2x/\sqrt{x^2+3}\), the original integrand.
Worked Example: An Exponential Substitution
Evaluate
Set \(u=1+e^x\), so \(du=e^x\,dx\). At \(x=0\), \(u=1+e^0=2\). At \(x=\ln 2\), \(u=1+e^{\ln 2}=1+2=3\). Thus
For a check, the derivative of \(-u^{-1}\) is \(u^{-2}\). After composing with \(u=1+e^x\), the derivative is \((1+e^x)^{-2}e^x\), exactly the original integrand. The endpoint calculation also gives a positive result, consistent with the positive integrand on the interval.
Worked Example: A Decreasing Substitution
Evaluate
Set \(u=1+\cos x\), so \(du=-\sin x\,dx\), or \(\sin x\,dx=-du\). Transform the endpoints in their original order: at \(x=0\), \(u=1+\cos 0=2\); at \(x=\pi/2\), \(u=1+\cos(\pi/2)=1\). Therefore,
The new bounds decrease from \(2\) to \(1\), and the differential contributes a minus sign. Both are retained in the first transformed integral. As a check, let \(H(u)=-\frac{2}{3}u^{3/2}\). Then \(H'(u)=-\sqrt{u}\), and \(u'(x)=-\sin x\). Consequently,
which is the original integrand.
A Common Bounds Error
Once the integrand has been rewritten in \(u\), the limits must also be in \(u\). Consider the exponential example. Its transformed integral is \(\int_2^3 u^{-2}\,du\). Leaving the old \(x\)-bounds in place would produce the invalid expression
The integrand uses \(u\), but \(0\) and \(\ln 2\) are the original \(x\)-values, not the corresponding \(u\)-values. This also leads to an undefined endpoint if one evaluates \([-u^{-1}]_0^{\ln 2}\). The correction is to transform the endpoints first: \(x=0\) gives \(u=2\), and \(x=\ln 2\) gives \(u=3\). The valid evaluation is \([-u^{-1}]_2^3=1/6\).
Why This Method Matters
Some substitutions are useful precisely because their inverse formulas are awkward. The new variable may simplify the integrand even when solving \(u=g(x)\) for \(x\) would add unnecessary algebra. For a definite integral, the endpoint values are usually obtained directly by applying \(g\) to the original endpoints.
The essential condition is that the substitution rewrite the whole integrand and differential in the new variable. A partial rewrite that leaves an \(x\)-dependent factor behind is not yet an integral entirely in \(u\). Once the rewrite is complete, use the transformed bounds, find an antiderivative in \(u\), and evaluate at those bounds. The chain-rule derivation explains why this procedure gives the original definite integral.
Check Your Understanding
Answer each question by keeping the transformed variable and its bounds together.
- If \(u=x^2+1\) and the original bounds are \(x=1\) and \(x=3\), what are the transformed bounds in the same order?
- For an integral transformed to \(\int_4^1 q(u)\,du\), should the bounds be reversed automatically? Explain what happens to the orientation.
- Why does the antiderivative \(Q(u)\) lead to the endpoint difference \(Q(g(b))-Q(g(a))\) in the original variable?
- In the substitution \(u=1+e^x\), why are \(0\) and \(\ln 2\) not valid bounds for an integrand written entirely in \(u\)?
- What must be true of the integrand and differential before an integral has been fully rewritten in \(u\)?
