Why These Definite Integrals Have a Direct Structure
When an integrand contains an exponential or logarithm applied to an inner function, the derivative of that inner function often appears elsewhere in the integrand. This pairing makes substitution especially effective: it turns the composite expression into a simpler function of a new variable, and the original endpoints determine the new limits of integration.
For example, the derivative of \(e^{g(x)}\) is \(e^{g(x)}g'(x)\). The derivative of \(\ln(g(x))\), when \(g(x)>0\), is \(g'(x)/g(x)\). These derivative patterns lead to direct formulas for definite integrals. Their hypotheses matter: the inner function must have enough regularity for the ordinary definite integral and the Fundamental Theorem of Calculus to apply, and a logarithm requires a positive argument throughout the interval.
The Definite Substitution Rule for These Patterns
Suppose \(g\in C^1([a,b])\). For an exponential composite, the chain rule gives \(\frac{d}{dx}e^{g(x)}=e^{g(x)}g'(x)\). Because \(g\) and \(g'\) are continuous on the closed interval, this integrand is continuous, so the Fundamental Theorem of Calculus applies:
For the logarithmic pattern, suppose in addition that \(g(x)>0\) for every \(x\in[a,b]\). Then \(\ln(g(x))\) is defined throughout the interval, and the chain rule gives \(\frac{d}{dx}\ln(g(x))=g'(x)/g(x)\). Since \(g\) is continuous and positive on a closed interval, it has a positive minimum there; consequently \(g'/g\) is continuous. The Fundamental Theorem of Calculus therefore gives:
These are not merely patterns to memorize. They are instances of substitution with the new variable \(u=g(x)\). In each case the endpoints in the \(u\)-integral are \(g(a)\) and \(g(b)\), not the original \(x\)-endpoints.
Why the Substitution Formula Works
A short argument explains both the formula and its regularity requirements. Let \(F\) be continuous on an interval containing the range of \(g\), and choose a fixed number \(u_0\) in that interval. Define
By the Fundamental Theorem of Calculus, \(H'(u)=F(u)\). Since \(g\in C^1([a,b])\), the chain rule applies to \(H(g(x))\) and gives
The integrand on the right is continuous. Applying the Fundamental Theorem of Calculus on \([a,b]\) yields the definite substitution formula:
This proof does not require \(g\) to be increasing. If \(g(b)<g(a)\), the transformed integral has its upper limit below its lower limit, and the usual signed orientation of a definite integral handles that case.
Taking \(F(u)=e^u\) gives the exponential formula. Taking \(F(u)=1/u\), on an interval of positive \(u\)-values, gives the logarithmic formula. The continuity and domain conditions in those applications are precisely what allow the argument to use the Fundamental Theorem of Calculus.
Worked Examples
Worked Example: An Exponential of a Polynomial
Evaluate \(\int_0^1 2x e^{x^2+1}\,dx\).
The inner function is \(g(x)=x^2+1\), and \(g'(x)=2x\), which matches the factor multiplying the exponential. The function \(g\) is in \(C^1([0,1])\). Set \(u=x^2+1\), so \(du=2x\,dx\). Transform the bounds: when \(x=0\), \(u=1\); when \(x=1\), \(u=2\). Thus
Check by differentiating the corresponding antiderivative \(e^{x^2+1}\): its derivative is \(e^{x^2+1}(2x)=2x e^{x^2+1}\). Evaluating it at \(1\) and \(0\) gives \(e^2-e^1=e^2-e\), as obtained by substitution.
Worked Example: A Power of a Logarithmic Composite
Evaluate \(\int_1^e \frac{(\ln x)^3}{x}\,dx\).
The interval \([1,e]\) lies in the domain \(x>0\) of the logarithm. Choose \(u=\ln x\), so \(du=dx/x\). The new limits are \(u=0\) at \(x=1\) and \(u=1\) at \(x=e\). Therefore
To check, differentiate \(\frac{(\ln x)^4}{4}\). The chain rule gives \(\frac14\cdot4(\ln x)^3\cdot(1/x)=(\ln x)^3/x\). Its endpoint difference is \(1/4-0=1/4\).
Worked Example: A Logarithm of a Positive Composite
Evaluate \(\int_0^1 2x\ln(1+x^2)\,dx\).
Set \(u=1+x^2\), so \(du=2x\,dx\). On the entire interval, \(1+x^2\geq1>0\), so the logarithm is defined. The new limits are \(u=1\) at \(x=0\) and \(u=2\) at \(x=1\). Then
For a check in the original variable, an antiderivative is \((1+x^2)\ln(1+x^2)-(1+x^2)\). Differentiating gives \(2x\ln(1+x^2)+2x-2x=2x\ln(1+x^2)\). At \(x=1\) its value is \(2\ln2-2\); at \(x=0\) its value is \(-1\). Their difference is \(2\ln2-1\).
Worked Example: An Exponential with a Base Other Than e
Evaluate \(\int_0^1 6x\,2^{3x^2}\,dx\).
For a positive base \(c\), \(c^u=e^{u\ln c}\), so \(\frac{d}{du}c^u=(\ln c)c^u\). In particular, an antiderivative of \(2^u\) is \(2^u/\ln2\). Set \(u=3x^2\), giving \(du=6x\,dx\). The bounds become \(0\) and \(3\), so
Check in \(x\): differentiating \(2^{3x^2}/\ln2\) gives \((2^{3x^2}\ln2)(6x)/\ln2=6x\,2^{3x^2}\). Its endpoint difference is \((8-1)/\ln2=7/\ln2\).
Bounds and Domain Checks Are Part of the Method
When the variable changes, the bounds must change with it. If \(u=g(x)\), the transformed integral uses \(u=g(a)\) and \(u=g(b)\). Keeping the original bounds after replacing \(x\) by \(u\) generally describes a different integral. Separately, every logarithm must have a positive argument throughout the original interval, not merely at the endpoints.
Look for \(g(x)\) in an exponential or logarithm and check whether the remaining factor includes \(g'(x)\), possibly with a constant multiple.
For these formulas, require \(g\in C^1([a,b])\). For a logarithm of \(g(x)\), also verify \(g(x)>0\) on all of \([a,b]\).
Compute the new lower and upper bounds by evaluating \(g\) at \(a\) and \(b\), respectively.
Integrate in the new variable, evaluate at the transformed endpoints, and check the result using the chain rule.
Common Mistake: Keeping the Original Bounds
Consider \(\int_1^2 2x e^{x^2}\,dx\). With \(u=x^2\), the differential is \(du=2x\,dx\), and the transformed bounds are \(u=1\) and \(u=4\). A wrong calculation keeps the bounds \(1\) and \(2\) after changing the variable:
The error is that \(u=2\) does not correspond to the original upper endpoint \(x=2\); rather, \(u=2^2=4\). The correct calculation is
Check by differentiating \(e^{x^2}\), which gives \(2x e^{x^2}\). Its endpoint difference is \(e^{2^2}-e^{1^2}=e^4-e\). The incorrect result came solely from using the wrong transformed upper limit.
Why the Conditions Matter
The regularity condition is not decorative. Differentiability of \(g\) alone does not guarantee that \(e^{g(x)}g'(x)\) is continuous or that its ordinary Riemann integral exists. Requiring \(g\in C^1([a,b])\) ensures that the integrand is continuous, so the proof using the Fundamental Theorem of Calculus is justified. Likewise, positivity is necessary for the real logarithm \(\ln(g(x))\) to be defined throughout the interval.
These formulas are useful because they convert a composite integrand into an endpoint calculation. The inner function determines the new bounds, while its derivative supplies the differential factor. Keeping those roles distinct helps prevent both domain errors and incorrect evaluations.
Check Your Understanding
For each question, identify the substitution and state any domain condition needed before evaluating.
- Evaluate \(\int_0^1 4x e^{2x^2}\,dx\), transforming the bounds explicitly.
- Evaluate \(\int_1^{e^2} \frac{(\ln x)^2}{x}\,dx\).
- For which real \(x\) is \(\ln(3-x^2)\) defined? In particular, is it defined throughout \([-1,1]\)?
- State a sufficient regularity condition on \(g\) for \(\int_a^b e^{g(x)}g'(x)\,dx=e^{g(b)}-e^{g(a)}\) to follow from the Fundamental Theorem of Calculus.
- What are the transformed bounds when \(u=1-x^2\), \(x\) runs from \(0\) to \(1\), and why is the order of those bounds relevant?
