When an Antiderivative Does Not Give the Answer
A definite integral can be meaningful and useful even when its integrand has no elementary antiderivative. For example, the functions \(e^{-x^2}\) and \(\sin(x^2)\) have antiderivatives that cannot be expressed using elementary functions. That does not prevent us from learning about their definite integrals.
Sometimes symmetry gives an exact value or a useful identity. Sometimes interval properties let us split an integral into simpler pieces. In other cases, inequalities give rigorous bounds. These methods do not manufacture an elementary antiderivative; they use other information about the integrand and the interval.
Definite Integral Properties We Can Use
For a continuous function \(f\) on a closed interval, its definite integral exists. The Fundamental Theorem of Calculus evaluates the integral by an antiderivative when one is available in a usable form. When it is not, the integral still has its usual properties.
The comparison property lets us turn pointwise information about a function into information about its integral. The direction of the inequality does not change when integrating over an interval with its endpoints in increasing order.
Symmetry on a Symmetric Interval
An interval is symmetric about zero when it has the form \([-a,a]\), where \(a>0\). A function is even if \(f(-x)=f(x)\), and odd if \(f(-x)=-f(x)\), for every \(x\) in its domain. If the function is integrable on \([-a,a]\), these definitions give useful integral identities.
To see why, split the integral at zero. In the part from \(-a\) to \(0\), substitute \(u=-x\). The endpoints change from \(x=-a,0\) to \(u=a,0\), and \(dx=-du\). Thus \(\int_{-a}^0 f(x)\,dx=\int_0^a f(-u)\,du\). If \(f\) is even, this equals \(\int_0^a f(u)\,du\), so adding the part from \(0\) to \(a\) gives twice that integral. If \(f\) is odd, it equals \(-\int_0^a f(u)\,du\); the two halves cancel. This proves both identities.
Worked Example: Use Even Symmetry for a Fresnel-Type Integral
Consider \(\int_{-1}^{1}\sin(x^2)\,dx\). The integrand has no elementary antiderivative, but symmetry still gives a precise simplification.
First check its parity: $$ \sin((-x)^2)=\sin(x^2). $$ Therefore \(f(x)=\sin(x^2)\) is even. Apply the even-function identity with \(a=1\): $$ \int_{-1}^{1}\sin(x^2)\,dx =2\int_0^1\sin(x^2)\,dx. $$ This is an exact equality, not an approximation. It does not by itself provide an elementary value for the integral on \([0,1]\); it reduces the original integral to twice that one-sided value.
Check: Splitting at zero gives \(\int_{-1}^0\sin(x^2)\,dx+\int_0^1\sin(x^2)\,dx\). The substitution \(u=-x\) in the first term turns it into \(\int_0^1\sin(u^2)\,du\), confirming the factor of two.
Bounds from Monotonicity
When a function is monotone on an interval, rectangles give direct bounds for its integral. Divide \([a,b]\) into \(n\) equal pieces of width \(\Delta x=(b-a)/n\). If \(f\) is decreasing, its value at the right endpoint of each piece is no larger than its values within that piece, while its left-endpoint value is no smaller. Adding the corresponding rectangle areas gives $$ \Delta x\sum_{i=1}^{n} f(x_i) \leq \int_a^b f(x)\,dx \leq \Delta x\sum_{i=0}^{n-1} f(x_i), $$ where \(x_i=a+i\Delta x\). For an increasing function, left and right sums exchange their lower- and upper-bound roles.
These inequalities are useful even when evaluating an antiderivative is not possible. The bounds depend on the function values at chosen points; they do not assume that an antiderivative can be written down.
Worked Example: Bound a Gaussian Integral with Four Subintervals
Find lower and upper bounds for \(\int_0^2 e^{-x^2}\,dx\) using four equal subintervals. The width is \(\Delta x=(2-0)/4=1/2\), and the partition points are \(0,\frac12,1,\frac32,2\).
For \(x\geq0\), \(e^{-x^2}\) is decreasing: its derivative is \(-2xe^{-x^2}\leq0\). Therefore the right-endpoint sum is a lower bound, and the left-endpoint sum is an upper bound.
The lower sum is $$ \frac12\left(e^{-1/4}+e^{-1}+e^{-9/4}+e^{-4}\right) =\frac12(0.7788008+0.3678794+0.1053992+0.0183156) \approx 0.6351975. $$ The upper sum is $$ \frac12\left(1+e^{-1/4}+e^{-1}+e^{-9/4}\right) =\frac12(1+0.7788008+0.3678794+0.1053992) \approx 1.1260397. $$ Consequently, the unrounded bounds are the exact sums displayed above. To state a guaranteed decimal interval with outward rounding to four decimal places, round the lower bound down and the upper bound up: $$ 0.6351\leq\int_0^2 e^{-x^2}\,dx\leq1.1261. $$
Check: Each left-endpoint rectangle lies above the graph on its subinterval, and each right-endpoint rectangle lies below it because the integrand is decreasing. The lower bound is less than the upper bound, as required. The lower decimal was rounded downward, not to the nearest value, so the stated inequality remains guaranteed.
Comparison and Interval Splitting
Monotonicity is not the only source of bounds. If simple functions lie below and above an integrand, the comparison property transfers those inequalities to their integrals. Interval splitting can also combine bounds obtained on separate pieces. One must keep the endpoints in increasing order when applying comparison; reversing an integral reverses its sign.
Worked Example: Bound a Fresnel-Type Integral
We will bound \(\int_0^1\sin(x^2)\,dx\) using four equal subintervals. On \([0,1]\), \(x^2\) increases from \(0\) to \(1\), and \(\sin t\) is increasing on \([0,1]\), since its derivative \(\cos t\) is positive there. Thus \(\sin(x^2)\) is increasing on \([0,1]\).
The partition points are \(0,\frac14,\frac12,\frac34,1\), with width \(\frac14\). The left-endpoint sum is a lower bound: $$ L=\frac14\left(\sin(0)+\sin\left(\frac1{16}\right)+\sin\left(\frac14\right)+\sin\left(\frac9{16}\right)\right). $$ Using \(\sin(0)=0\), \(\sin(1/16)\approx0.0624593\), \(\sin(1/4)\approx0.2474040\), and \(\sin(9/16)\approx0.5226872\), we get $$ L\approx\frac14(0+0.0624593+0.2474040+0.5226872) =\frac14(0.8325505)\approx0.2081376. $$ The right-endpoint sum is an upper bound: $$ U=\frac14\left(\sin\left(\frac1{16}\right)+\sin\left(\frac14\right)+\sin\left(\frac9{16}\right)+\sin(1)\right). $$ Since \(\sin(1)\approx0.8414710\), $$ U\approx\frac14(0.0624593+0.2474040+0.5226872+0.8414710) =\frac14(1.6740215)\approx0.4185054. $$ Thus the exact sums \(L\) and \(U\) give the rigorous bounds \(L\leq\int_0^1\sin(x^2)\,dx\leq U\). Rounded outward to four decimal places, they imply $$ 0.2081\leq\int_0^1\sin(x^2)\,dx\leq0.4186. $$
Check: The integrand is increasing, so the left sum must be below the integral and the right sum above it. The computed lower estimate is less than the upper estimate. Outward rounding preserves both inequalities.
What These Methods Do—and Do Not—Establish
A symmetry identity may produce an exact value, as happens when an odd integrand is integrated over a symmetric interval. It may instead produce an exact relation between integrals, as in the example with \(\sin(x^2)\). A bound gives an interval known to contain the integral, not necessarily its exact value. Increasing the number of subintervals generally sharpens monotone rectangle bounds, but a finite set of bounds should not be mistaken for an exact evaluation.
It is also important to distinguish “no elementary antiderivative” from “no exact answer.” A special function, a symmetry identity, or a specified definite integral may provide an exact description without an elementary formula. Conversely, a decimal approximation alone does not establish an exact value. State clearly whether a result is an identity, a rigorous bound, or an approximation.
A Common Mistake: Treating Rounded Bounds as Guaranteed Bounds
Suppose a lower rectangle sum is approximately \(0.6351975\). Rounding this to the nearest four decimal places gives \(0.6352\), which is slightly larger than the actual lower sum. It is therefore not valid to write \(0.6352\leq\int_0^2 e^{-x^2}\,dx\) on the strength of that lower sum: the sum only guarantees that the integral is at least \(0.6351975\).
For a guaranteed decimal interval, round the lower endpoint down and the upper endpoint up. Here, \(0.6351\) is below the lower sum and \(1.1261\) is above the upper sum, so \(0.6351\leq\int_0^2 e^{-x^2}\,dx\leq1.1261\) is valid. If the goal is simply to report nearest-decimal approximations to the sums, write that the sums are approximately \(0.6352\) and \(1.1260\), without presenting those rounded values as rigorous integral bounds.
Choosing a Method
Before trying to find an antiderivative, inspect the integrand and the interval. A short sequence of questions can identify what a definite integral property can accomplish.
If it is symmetric about zero, test whether the integrand is even or odd.
Apply symmetry or split the interval where a useful property changes.
Use comparison or endpoint sums to obtain justified bounds.
Distinguish an exact value or identity from a bound and from a numerical approximation.
These strategies are particularly valuable for definite integrals whose antiderivatives are non-elementary. They let us extract information from structure, even when the usual endpoint-substitution route is unavailable.
Check Your Understanding
Answer each question by identifying the property or reasoning that applies.
- If \(f\) is odd and integrable on \([-a,a]\), what is \(\int_{-a}^{a}f(x)\,dx\), and why?
- For a decreasing function on \([a,b]\), which endpoint sums give lower and upper bounds?
- Why does a non-elementary antiderivative not prevent a definite integral from having an exact value?
- When rounding a computed lower bound to obtain a guaranteed decimal interval, which direction should you round?
- What does a pair of valid upper and lower sums establish if they are not equal?
