Why Check a Definite Integral a Second Way?
A definite integral can be evaluated correctly by several different routes. One route may use an antiderivative; another may use symmetry, a substitution, or a split of the interval. When the routes produce the same exact value, that agreement is useful evidence that the calculation is sound.
The aim is not to generate a second answer by repeating the first set of steps. It is to find a representation whose important steps differ. A useful check tracks the changes to the integrand, the interval, and any signs or factors. It then compares exact expressions, not merely rounded decimal approximations.
For the results below, assume the integrands are continuous on the intervals of integration. This condition ensures that the definite integrals exist and that the Fundamental Theorem of Calculus and the substitution and additivity formulas apply in the stated forms.
What Makes an Alternate Check Useful?
Each representation has its own details that can be checked. Symmetry depends on using the correct interval and on how the integrand behaves under reflection. Substitution changes both the variable and the limits. Interval additivity requires the pieces to meet at the chosen split point, with no gap or overlap.
A second representation is most informative when it makes those details visible. For example, a substitution may expose a missing minus sign, while a symmetry argument may reveal that an odd term contributes zero over a symmetric interval. If two routes use exactly the same antiderivative evaluation and the same endpoint arithmetic, they are less likely to uncover an error in those shared steps.
Symmetry as an Alternate Representation
For a continuous function \(f\) on \([-a,a]\), where \(a>0\), splitting at zero and reflecting the left half gives the identity
This formula expresses the whole integral using values on just the right half of the interval. It also gives two familiar consequences. If \(f\) is even, so \(f(-x)=f(x)\), the integral is \(2\int_0^a f(x)\,dx\). If \(f\) is odd, so \(f(-x)=-f(x)\), the integral is zero.
Worked Example: Check a Polynomial Integral by Symmetry
Evaluate \(\int_{-1}^{1}(x^3+2x^2+1)\,dx\) by an antiderivative, then check the result using symmetry.
First, an antiderivative is \(F(x)=\frac{x^4}{4}+\frac{2x^3}{3}+x\). Evaluate the endpoints:
Thus the antiderivative evaluation gives \(\frac{23}{12}-(-\frac{17}{12})=\frac{40}{12}=\frac{10}{3}\).
For the alternate check, \(x^3\) is odd, so its integral from \(-1\) to \(1\) is zero. The remaining function \(2x^2+1\) is even, so its integral is twice its integral from \(0\) to \(1\):
The two representations agree exactly. As a further check of the endpoint arithmetic, \(\frac{23}{12}-(-\frac{17}{12})=\frac{40}{12}=\frac{10}{3}\), matching the symmetry result.
Substitution: Track the Bounds and the Sign
A change of variable provides another representation. In the standard definite-integral substitution formula, if \(g\) is continuously differentiable on \([a,b]\) and \(h\) is continuous on the range of \(g\), then
The bounds on the right are the images of the original endpoints. In particular, if \(g\) decreases from \(a\) to \(b\), then \(g(a)>g(b)\); the order of the new bounds is not automatically reversed. Keeping the order and the transformed differential together is essential to a reliable check.
Worked Example: Check a Logarithmic Integral by Substitution
Evaluate \(\int_0^\pi \frac{\sin x}{2+\cos x}\,dx\) and check the result using a substitution.
For the first evaluation, an antiderivative is \(-\ln(2+\cos x)\), since \(\frac{d}{dx}[-\ln(2+\cos x)]=\frac{\sin x}{2+\cos x}\). The logarithm is defined throughout the interval because \(2+\cos x\geq 1\). At the endpoints,
Therefore the integral is \(0-(-\ln3)=\ln3\).
For an independent representation, set \(u=2+\cos x\). Then \(du=-\sin x\,dx\). The transformed endpoints are \(u=3\) at \(x=0\) and \(u=1\) at \(x=\pi\). Consequently,
Both routes give the exact value \(\ln3\). The substitution check accounts for the negative sign in \(du\) and keeps the transformed bounds in their correct order.
Interval Additivity: Check by Splitting
If \(f\) is continuous on \([a,b]\) and \(c\) lies between \(a\) and \(b\), interval additivity states
This also follows from the Fundamental Theorem of Calculus: if \(F'=f\), the right side is \(F(c)-F(a)+F(b)-F(c)=F(b)-F(a)\), which is the left side. A split can check the full evaluation by making the endpoint calculations on smaller intervals and then adding the exact results.
Worked Example: Check a Polynomial Integral by Splitting the Interval
Evaluate \(\int_{-2}^{3}(x^2-x)\,dx\), then check it by splitting at zero.
An antiderivative is \(F(x)=\frac{x^3}{3}-\frac{x^2}{2}\). Its endpoint values are
Thus the full-interval evaluation is
Now evaluate the pieces. On \([-2,0]\),
On \([0,3]\),
Adding the pieces gives \(\frac{14}{3}+\frac92=\frac{28}{6}+\frac{27}{6}=\frac{55}{6}\), exactly the full-interval result. The split verifies both the additivity and the arithmetic of the endpoint evaluation.
A Common Mistake: Losing the Substitution Sign
In the logarithmic example, a tempting but incorrect substitution calculation is
The bounds \(3\) and \(1\) are correct, but this calculation omitted \(du=-\sin x\,dx\). The transformed integral must include the minus sign:
This error is exactly what an alternate check can expose: the antiderivative evaluation gives \(\ln3\), while the incomplete substitution gives \(-\ln3\). Rather than choosing one result based on appearance, retrace the transformation and find the missing differential factor.
A Practical Checking Routine
When you have an exact value for a definite integral, choose an alternate representation that changes the work in a meaningful way. Use symmetry when reflection simplifies the integrand, substitution when the integrand contains a useful composite form, and interval additivity when a split makes endpoint calculations easier to inspect.
Keep its integrand and endpoints visible so the alternate calculation can be compared with the same quantity.
Check the symmetry condition, the substitution formula's hypotheses, or that the subintervals join at the selected point.
For substitution, show the differential and transformed endpoints. For splitting, show each endpoint difference. For symmetry, identify the even and odd parts.
Simplify fractions, logarithms, or other exact expressions and check that the two routes produce the same value.
A mismatch is a reason to inspect the calculations, not to average the answers or select the more convenient one.
Why Exact Agreement Matters
A decimal approximation can hide a small arithmetic or sign error, especially if it is rounded early. Exact forms preserve the structure of the calculation: \(\frac{55}{6}\) can be checked by common denominators, and \(\ln3\) can be traced to endpoint values of a logarithm. Keep exact expressions until the comparison is complete.
Agreement between two valid representations is a strong diagnostic, but it is not a substitute for checking the reasoning. If both routes share the same incorrect endpoint value, they may agree for the wrong reason. The most effective check uses a genuinely different route and verifies its assumptions, transformations, and arithmetic explicitly.
Check Your Understanding
For each question, identify the transformation or condition that must be checked, not only the final value.
- What condition on \(f\) makes \(\int_{-a}^{a}f(x)\,dx=2\int_0^a f(x)\,dx\) valid?
- In the substitution \(u=g(x)\), what happens to the definite-integral bounds?
- How can interval additivity check a value computed over a whole interval?
- For \(\int_0^\pi \frac{\sin x}{2+\cos x}\,dx\), what sign appears when \(u=2+\cos x\), and why?
- Why is it useful for two checks to use different representations rather than repeat the same endpoint calculation?
