Why Check a Definite-Integral Setup?
A correct integration method can still produce a wrong answer if the setup changes inconsistently along the way. Common sources of error include keeping the old bounds after changing variables, dropping a negative sign, or writing an expression in the new variable while leaving part of it in the old one.
These mistakes are often detectable before the integral is evaluated. A careful check asks whether the new limits correspond to the old endpoints, whether the transformed integrand and differential agree, and whether the sign and size of the answer make sense.
Here the main setting is substitution in a definite integral. The same habits are useful whenever an integral is rewritten: make each transformation explicit, then check what it implies about the value.
What Must Change Under Substitution?
Suppose \(u=g(x)\), where \(g\) is differentiable, and the integrand contains the matching factor \(g'(x)\). If \(f\) is continuous and \(g\) is continuously differentiable on \([a,b]\), then the substitution rule is
Three parts of the integral are linked: the expression inside \(f\) becomes \(u\), the factor \(g'(x)\,dx\) becomes \(du\), and each endpoint is changed by applying \(g\). A setup is incomplete if it changes only some of these parts.
The order of the transformed limits matters. If \(g\) decreases from \(a\) to \(b\), then \(g(a)>g(b)\), and the transformed integral has its limits in descending order. That order records the sign change associated with moving in the opposite direction; it is not an error to be silently ignored.
Why the Substitution Rule Works
The rule follows from the chain rule and the Fundamental Theorem of Calculus. Let \(F\) be an antiderivative of \(f\), so \(F'(u)=f(u)\). Then the chain rule gives
Thus \(F(g(x))\) is an antiderivative of the original integrand. Evaluating at the original endpoints gives
Because \(F'=f\), the right-hand side is also the value of \(\int_{g(a)}^{g(b)} f(u)\,du\). This argument keeps the endpoint order as it is, whether \(g\) is increasing or decreasing.
Worked Example: A Direct Substitution
Worked Example: A Direct Substitution
Evaluate \(\int_0^1 2x\cos(x^2)\,dx\), checking the transformed setup and the final value.
Set \(u=x^2\). Then \(du=2x\,dx\). Transform both endpoints: when \(x=0\), \(u=0^2=0\); when \(x=1\), \(u=1^2=1\). The setup becomes
The expression \(\cos(x^2)\) has become \(\cos(u)\), and the factor \(2x\,dx\) has become \(du\). No old \(x\) remains. Now integrate and evaluate:
As an antiderivative check in the original variable, \(\frac{d}{dx}\sin(x^2)=2x\cos(x^2)\). The endpoint evaluation gives \(\sin(1)-\sin(0)=\sin(1)\), agreeing with the substitution result. Also, the integrand is nonnegative on \([0,1]\), so a positive answer is expected.
Signs and the Direction of the Bounds
A substitution can make the new variable decrease as the original variable increases. In that case, the transformed limits appear reversed. There are two equivalent ways to handle this: keep the reversed limits and the sign in the differential, or reverse the limits and change the sign. Do not do both, and do not do neither.
Worked Example: A Decreasing Substitution
Worked Example: A Decreasing Substitution
Evaluate \(\int_0^\pi \sin(x)\,dx\) using \(u=\cos(x)\), and check the signs.
With \(u=\cos(x)\), \(du=-\sin(x)\,dx\), so \(\sin(x)\,dx=-du\). Transform the endpoints: when \(x=0\), \(u=\cos(0)=1\); when \(x=\pi\), \(u=\cos(\pi)=-1\). Therefore
Evaluate while keeping the limits in their transformed order:
Equivalently, reversing the limits gives \(-\int_1^{-1}1\,du=\int_{-1}^{1}1\,du=2\). As a check in the original variable, an antiderivative is \(-\cos(x)\), whose endpoint difference is \(-\cos(\pi)-(-\cos(0))=1-(-1)=2\). The integrand is positive on \((0,\pi)\), so the positive result also has the expected sign.
Notice that the transformed lower limit is \(1\), not \(0\), and the transformed upper limit is \(-1\), not \(\pi\). Both limits come from applying \(u=\cos(x)\) to the corresponding original endpoints.
Use Bounds to Test the Size of an Answer
A sign check is useful, but a result can have the right sign and still be far too large or too small. For a continuous function \(f\) on \([a,b]\), if \(m\leq f(x)\leq M\) throughout the interval and \(a<b\), then
This follows by integrating the inequalities \(m\leq f(x)\leq M\) over an interval of length \(b-a\). In particular, if \(0\leq f(x)\leq M\), then the integral must lie between \(0\) and \(M(b-a)\). These estimates do not usually give the exact value; they give a range that an exact or approximate answer must satisfy.
Worked Example: Checking a Logarithmic Result
Worked Example: Checking a Logarithmic Result
Evaluate \(\int_0^2 \frac{x}{x^2+1}\,dx\) by substitution and use an estimate to assess its size.
Set \(u=x^2+1\). Then \(du=2x\,dx\), so \(x\,dx=\frac12du\). Transform the endpoints: \(x=0\) gives \(u=0^2+1=1\), and \(x=2\) gives \(u=2^2+1=5\). Thus
Since the limits are positive, an antiderivative of \(1/u\) is \(\ln(u)\). Evaluation gives
For a size check, the original integrand is nonnegative on \([0,2]\). Also, \(\frac{x}{x^2+1}\leq\frac12\), because this inequality is equivalent to \(2x\leq x^2+1\), or \((x-1)^2\geq0\). The interval has length \(2\), so
The exact value is \(\frac12\ln(5)\approx0.8047\), rounded to four decimal places, which lies in the required range. Differentiating \(\frac12\ln(x^2+1)\) gives \(\frac{x}{x^2+1}\), providing an additional check of the evaluation.
A Common Setup Error: Mixing the Old and New Bounds
In the preceding example, \(u=x^2+1\) sends the original interval \(0\leq x\leq2\) to \(1\leq u\leq5\). Keeping the original bounds after changing the variable is inconsistent. For example, the incorrect line
uses \(u\) in the integrand but leaves the limits \(0\) and \(2\) from the \(x\)-interval. Worse, this integral diverges at \(u=0\), while the original integrand is continuous on \([0,2]\) and has a finite integral. That contradiction is a clear warning that the setup is wrong.
The correction is to transform the endpoints using the same substitution: \(u(0)=1\) and \(u(2)=5\). The consistent expression is \(\frac12\int_1^5 \frac{1}{u}\,du\). Checking what happens at the endpoints can reveal an error before any further evaluation is attempted.
A Short Audit Before Evaluation
Before evaluating a transformed definite integral, pause to match each part of the setup. This need not be a long calculation; it is a way to catch structural errors early.
Write the new variable explicitly, such as \(u=g(x)\), and compute its differential.
Calculate \(u=g(a)\) and \(u=g(b)\) separately. Keep their order unless you deliberately reverse the limits and change the sign.
Check that the expression and differential use the new variable consistently, with no leftover \(x\) unless it has been rewritten in terms of \(u\).
Use the sign of the integrand and any simple upper or lower bounds to constrain the value.
When practical, differentiate the proposed antiderivative and check its endpoint difference against the transformed calculation.
Why These Checks Matter
Definite integrals encode both an integrand and an interval. Changing variables changes how that interval is described, but it does not give permission to change the quantity being calculated. Correct bounds preserve the endpoints; a consistent differential preserves the integrand’s contribution; and careful sign handling preserves orientation.
Estimates provide a separate form of evidence. If an integrand is nonnegative, a negative answer is impossible. If its values never exceed \(M\) on an interval of length \(L\), an answer greater than \(ML\) is impossible. A discrepancy does not by itself identify which line is wrong, but it tells you to inspect the setup rather than accept an implausible result.
Check Your Understanding
For each question, focus on the consistency of the setup before trying to evaluate.
- For \(\int_1^3 2x(x^2+1)^4\,dx\), use \(u=x^2+1\). What are the transformed limits, and what is the complete transformed integral?
- For \(\int_0^{\pi/2}\cos(x)\,dx\), use \(u=\sin(x)\). What are the transformed limits, and does the substitution produce a negative differential factor?
- If \(0\leq f(x)\leq 4\) on \([1,3]\), what lower and upper bounds follow for \(\int_1^3 f(x)\,dx\)?
- In a substitution calculation, the new integrand is written entirely in \(u\), but one endpoint is still an \(x\)-value. What should be checked?
- Why is it useful to compare a computed answer with the sign and size bounds implied by the original integrand?
