Why Definite Substitution Works
A substitution in a definite integral is more than a change of notation. It follows from a relationship between derivatives and accumulated change: if an antiderivative is composed with an inner function, the chain rule describes the derivative of that composition. The Fundamental Theorem of Calculus then turns that derivative relationship into an equality of definite integrals.
The key formula is familiar in form, but its hypotheses matter. We will derive it carefully, then use it to evaluate several integrals. The aim is to understand why the endpoints in the result are the values of the inner function at the original endpoints.
The Formula and Its Conditions
Let \(a<b\), let \(g\) be continuously differentiable on \([a,b]\), and let \(f\) be continuous on an interval containing every value \(g(x)\) for \(x\in[a,b]\). Then \(f(g(x))g'(x)\) is continuous on \([a,b]\), so its definite integral exists. The definite substitution formula is:
Notice the two essential features of the left-hand integrand. It contains the composite \(f(g(x))\), and it contains the factor \(g'(x)\). The formula also specifies the new endpoints: they are \(g(a)\) and \(g(b)\), in that order.
In this theorem, “continuously differentiable” means that \(g'\) exists and is continuous on the interval. This condition ensures that the left-hand integrand is continuous, allowing us to use the Fundamental Theorem of Calculus in its standard continuous-integrand form. Differentiability of \(g\) alone does not guarantee continuity of \(f(g(x))g'(x)\), so it does not by itself justify that application.
Derivation from the Fundamental Theorem of Calculus
Since \(f\) is continuous, it has an antiderivative \(F\) on the interval of interest. Thus \(F'(u)=f(u)\). Compose \(F\) with \(g\). By the chain rule, the derivative of this composite is exactly the integrand on the left side of the theorem.
Because \(f\) is continuous, take \(F\) such that \(F'(u)=f(u)\) throughout the interval containing the values of \(g\).
The chain rule gives \(\frac{d}{dx}F(g(x))=F'(g(x))g'(x)=f(g(x))g'(x)\).
The derivative just found is continuous, so its integral from \(a\) to \(b\) is the change in \(F(g(x))\) between those endpoints.
Writing out the last step gives the complete derivation:
The first equality uses the chain rule and the fact that \(F'=f\). The second equality is the Fundamental Theorem of Calculus applied on \([a,b]\). The last equality is the same theorem applied to \(F\) between \(g(a)\) and \(g(b)\). This proves the formula.
The proof also explains why substitution is valid: both integrals describe the same change in the same antiderivative. It is not necessary to guess that the formula works from a collection of examples; the chain rule and the Fundamental Theorem force it to hold under the stated conditions.
Worked Examples
Worked Example: A Trigonometric Composite
Evaluate \(\int_0^1 2x\cos(x^2)\,dx\) using the definite substitution formula.
Set \(g(x)=x^2\), so \(g'(x)=2x\), and let \(f(u)=\cos u\). The integrand has the required form because \(f(g(x))g'(x)=\cos(x^2)(2x)\). The endpoints for the integral on the right are \(g(0)=0^2=0\) and \(g(1)=1^2=1\). Therefore:
To check, an antiderivative of the original integrand is \(\sin(x^2)\), since \(\frac{d}{dx}\sin(x^2)=\cos(x^2)(2x)=2x\cos(x^2)\). Evaluating it at the original endpoints gives \(\sin(1)-\sin(0)=\sin(1)\), as required.
Worked Example: A Rational Composite
Evaluate \(\int_0^1 \frac{4x^3}{1+x^4}\,dx\).
Choose \(g(x)=1+x^4\), so \(g'(x)=4x^3\), and set \(f(u)=1/u\). The inner values lie between \(g(0)=1\) and \(g(1)=2\), so \(f\) is continuous throughout the relevant interval. The integrand matches \(f(g(x))g'(x)\):
The theorem now gives:
For a check, \(\ln(1+x^4)\) is an antiderivative of the original integrand because its derivative is \(\frac{1}{1+x^4}(4x^3)\). Its endpoint difference is \(\ln(2)-\ln(1)=\ln(2)\), agreeing with the substitution result.
Worked Example: A Linear Inner Function
Evaluate \(\int_0^1 6(3x+1)^2\,dx\) using substitution.
Let \(g(x)=3x+1\), so \(g'(x)=3\). To match the factor in the theorem, choose \(f(u)=2u^2\). Then:
The new endpoint values are \(g(0)=3(0)+1=1\) and \(g(1)=3(1)+1=4\). Thus:
To check, differentiate \(\frac{2}{3}(3x+1)^3\): its derivative is \(\frac{2}{3}\cdot3(3x+1)^2\cdot3=6(3x+1)^2\). At the original endpoints, the antiderivative difference is \(\frac{2}{3}(4^3)-\frac{2}{3}(1^3)=42\), confirming the result.
What the Hypotheses Protect
The continuity assumptions are not decorative. The proof uses the Fundamental Theorem of Calculus on the composite \(F(g(x))\). The chain rule identifies its derivative as \(f(g(x))g'(x)\); continuity of \(f\) and \(g'\) ensures that this derivative is continuous, so the standard evaluation form of the theorem applies.
There are broader versions of substitution under weaker assumptions, but they require additional care about integrability. For the result derived here, it is enough to check that \(f\) is continuous on the relevant range and that \(g\) is continuously differentiable on the original interval. In particular, merely knowing that \(g\) is differentiable does not automatically establish all the continuity needed for this proof.
A Common Mistake: Omitting the Derivative Factor
A frequent error is to replace a composite \(f(g(x))\) with \(f(u)\) while forgetting the factor \(g'(x)\). For instance, consider \(\int_0^1 \cos(x^2)\,dx\). A mistaken calculation might set \(u=x^2\) and write \(\int_0^1\cos u\,du=\sin(1)\). That step is not justified: the original integrand is \(\cos(g(x))\), not \(\cos(g(x))g'(x)\).
Here \(g'(x)=2x\), so the integrand needed for the theorem would be \(2x\cos(x^2)\). The formula therefore evaluates \(\int_0^1 2x\cos(x^2)\,dx\), as in the first example; it does not say that this equals \(\int_0^1\cos(x^2)\,dx\). The correction is to check the full product \(f(g(x))g'(x)\) before applying the theorem. If the derivative factor is absent, the theorem in this form does not apply.
Why This Result Matters
The definite substitution theorem connects two central ideas of calculus. The chain rule explains how a rate changes when its input is itself changing; the Fundamental Theorem of Calculus relates that rate to total accumulated change. Substitution combines them, allowing an integral involving a composite expression to be evaluated through an antiderivative of the simpler outer function.
The theorem also gives a reliable way to check a calculation. Identify \(f\) and \(g\), verify the derivative factor, compute the inner-function values at the endpoints, and compare the result with an antiderivative when one is available. Each check corresponds to a specific part of the proof.
Check Your Understanding
Use the theorem and its proof to answer these questions.
- State the continuity conditions used in the definite substitution theorem presented here.
- In the proof, what derivative does the chain rule produce when applied to \(F(g(x))\)?
- For \(\int_0^2 2x e^{x^2}\,dx\), identify \(f\), \(g\), and the two endpoint values used on the right.
- Why is \(\int_0^1 \cos(x^2)\,dx=\int_0^1\cos u\,du\) not justified by taking \(u=x^2\)?
- Which step of the derivation uses the Fundamental Theorem of Calculus on the original interval \([a,b]\)?
