Integrals · Lesson 8 of 9

Properties and Symmetry of Definite Integrals

“Use interval structure, reflection, and parity to simplify definite integrals before calculating.”

Learning Objectives

• Justify the basic limit-reversal, zero-length, and interval-splitting properties. • Derive reflection identities by changing variables. • Use midpoint symmetry and even or odd functions to simplify an integral. • Split absolute-value integrals at the correct sign changes. • Combine symmetry with substitution and standard antiderivatives.

The variable name and the interval play different roles

Once the endpoints are fixed, the integration variable is internal to the calculation. Renaming it does not change the value. The endpoints, however, control the interval and its direction. Understanding these roles lets us reorganise an integral before choosing an antiderivative.

Basic interval propertiesLaTeX

\int_{a}^{b} f (x) d x \int_{a}^{b} f (x) d x \int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t, = - \int_{b}^{a} f (x) d x, \int_{a}^{a} f (x) d x = 0, = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x .

Use these identities when the integrals involved exist. For the usual geometric split take

a \leq c \leq b

; with oriented limits the algebraic identity is more general.

Renaming the internal variable changes no heights or widths. For the other identities let $F$ be an antiderivative. Reversing limits replaces $F (b) - F (a)$ by $F (a) - F (b)$ , giving the negative. Equal limits give zero. Splitting follows because $[F (c) - F (a)] + [F (b) - F (c)] = F (b) - F (a)$ : the middle endpoint cancels.

Absolute values require sign information

The rule $∣ q (x) ∣ = q (x)$ applies where $q (x) \geq 0$ and $∣ q (x) ∣ = - q (x)$ where it is negative. Locate the zeros and determine the sign between them before integrating. Integrating first and taking an absolute value at the end generally measures something different.

Example — Three sign intervals

Problem
Evaluate $\int_{- 1}^{2} ∣ x^{3} - x ∣ d x$ .

1.Factor $x^{3} - x = x (x - 1) (x + 1)$ . Its zeros in the interval are $- 1, 0, 1$ ; the sign is positive on $(- 1, 0)$ , negative on $(0, 1)$ , and positive on $(1, 2)$ .
2.Split the integral as $\int_{- 1}^{0} (x^{3} - x) d x - \int_{0}^{1} (x^{3} - x) d x + \int_{1}^{2} (x^{3} - x) d x$ .
3.Use $F (x) = x^{4} /4 - x^{2} /2$ . The three contributions are $1/4$ , $1/4$ , and $9/4$ .
4.The total is $11/4$ . Every contribution is nonnegative because the absolute-value integrand is nonnegative.

Sign changes and absolute value— The negative middle segment is reflected upward. Split at zero and one before integrating.

Reflecting an interval

A reflection exchanges points at the same distance from the midpoint. For the interval $[a, b]$ , the reflected point corresponding to $x$ is $a + b - x$ . This change of variable often exchanges sine and cosine or turns an awkward numerator into a useful partner. Adding the original and reflected versions can be much easier than integrating either one directly.

Reflection identitiesLaTeX

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x, \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x

The second is the first with lower endpoint zero. These identities require the reflected expression to be considered on the same interval.

To derive the first, put $t = a + b - x$ , so $d t = - d x$ . When $x = a$ , $t = b$ ; when $x = b$ , $t = a$ . The substitution gives $- \int_{b}^{a} f (a + b - t) d t$ , and reversing its limits removes the minus sign. Rename $t$ as $x$ to obtain the displayed identity. Setting the lower endpoint to zero gives the second version, after renaming the upper endpoint.

Example — Sine and cosine become partners

Problem
Evaluate $I = \int_{0}^{π /2} sin^{4} x / (sin^{4} x + cos^{4} x) d x$ .

1.Reflect with $x \mapsto π /2 - x$ . This exchanges sine and cosine, so $I = \int_{0}^{π /2} cos^{4} x / (sin^{4} x + cos^{4} x) d x$ .
2.Add the two expressions: $2 I = \int_{0}^{π /2} 1 d x = π /2$ .
3.Therefore $I = π /4$ . The denominator is positive throughout the interval, so the pairing is legitimate.

Example — Reflection on an interval not starting at zero

Problem
Evaluate $\int_{π /6}^{π /3} d x / (1 + tan x)$ .

1.Rewrite the integrand as $cos x / (cos x + sin x)$ . The interval midpoint is $π /4$ , so reflection is $x \mapsto π /2 - x$ .
2.The reflected integrand is $sin x / (sin x + cos x)$ . Adding the two integrals gives $2 I = \int_{π /6}^{π /3} 1 d x = π /6$ .
3.Thus $I = π /12$ . Here the symmetry reduces the answer to half the interval length.

Symmetry around a midpoint

Split an interval $[0, 2 a]$ at its midpoint $a$ , then reflect its second half into the first. Matching values add, while equal and opposite values cancel. These observations lead to two useful formulas and explain the conditions that must accompany them.

Folding the second half onto the firstLaTeX

\int_{0}^{2 a} f (x) d x = \int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x

Split at

a

, then use

t = 2 a - x

in the second integral. The result compares the values at reflected points.

Symmetric and antisymmetric casesLaTeX

f (2 a - x) = f (x) f (2 a - x) = - f (x) ⟹ \int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x, ⟹ \int_{0}^{2 a} f (x) d x = 0.

The first condition means equal heights on either side of the midpoint; the second means equal and opposite signed contributions.

Example — Removing a factor of the integration variable

Problem
Evaluate $I = \int_{0}^{π} x sin x / (1 + cos^{2} x) d x$ .

1.The factor $g (x) = sin x / (1 + cos^{2} x)$ satisfies $g (π - x) = g (x)$ . Reflecting gives $I = \int_{0}^{π} (π - x) g (x) d x$ .
2.Add the two versions: $2 I = π \int_{0}^{π} g (x) d x$ .
3.Put $t = cos x$ in the remaining integral. The limits change from $1$ to $- 1$ and the negative differential reverses them, giving $\int_{- 1}^{1} d t / (1 + t^{2}) = π /2$ .
4.Therefore $I = (π /2) (π /2) = π^{2} /4$ . Symmetry removed the factor $x$ before substitution was attempted.

Even and odd functions on symmetric intervals

Reflection about zero is especially easy to recognise. An even function has the same value at $x$ and $- x$ ; its graph is symmetric about the vertical axis. An odd function has opposite values at those points, so its graph is symmetric about the origin. These statements concern the complete integrand, including any products or quotients.

Parity and integrationLaTeX

f (- x) = f (x) f (- x) = - f (x) ⟹ \int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x, ⟹ \int_{- a}^{a} f (x) d x = 0.

The first condition defines evenness and the second oddness. Assume the integral exists over the symmetric interval; a singularity at zero cannot be ignored.

To derive these rules, split at zero. In the integral from $- a$ to $0$ , put $t = - x$ . It becomes $\int_{0}^{a} f (- t) d t$ . Adding the positive-half integral gives $\int_{0}^{a} [f (- t) + f (t)] d t$ . For an even function this is twice the positive-half integral; for an odd function it is zero.

Example — An even squared sine

Problem
Evaluate $\int_{- π /4}^{π /4} sin^{2} x d x$ .

1.Since $sin^{2} (- x) = sin^{2} x$ , double the integral from $0$ to $π /4$ .
2.Use $2 sin^{2} x = 1 - cos 2 x$ , giving $[x - sin 2 x /2]_{0}^{π /4}$ .
3.The value is $π /4 - 1/2$ .

Example — An odd product

Problem
Evaluate $\int_{- 1}^{1} sin^{5} x cos^{4} x d x$ .

1.Sine to an odd power is odd, while cosine to any integer power is even.
2.Thus $f (- x) = (- sin x)^{5} (cos x)^{4} = - f (x)$ . The integrand is continuous on the interval.
3.The two halves cancel and the integral is $0$ . A zero signed integral does not mean the integrand is zero or that its total geometric area is zero.

Even and odd symmetry— Equal reflected heights add for an even function. Opposite heights cancel for an odd function.

A logarithmic integral from reflection and rescaling

Some useful integrals are not evaluated by finding an elementary antiderivative. Reflection can relate one unknown definite integral to another copy of itself. A rescaling and a midpoint symmetry then produce an equation for its value, extending the same pairing idea used earlier.

Interpreting the endpoints

The function $ln (sin x)$ tends to negative infinity as $x$ approaches zero from the right. The following source example uses the finite limit of integrals with the endpoint approached from within the interval. It is not an application of the continuous-function endpoint theorem at a point where the logarithm is undefined.

Example — A logarithmic symmetry calculation

Problem
Evaluate $I = \int_{0}^{π /2} ln (sin x) d x$ , with endpoint values interpreted by limits.

1.Reflection gives $I = \int_{0}^{π /2} ln (cos x) d x$ . Add to obtain $2 I = \int_{0}^{π /2} ln (sin x cos x) d x$ .
2.Use $sin x cos x = sin 2 x /2$ , giving $2 I = \int_{0}^{π /2} ln (sin 2 x) d x - (π /2) ln 2$ .
3.With $t = 2 x$ , the first integral is $\frac{1}{2} \int_{0}^{π} ln (sin t) d t$ . Since $sin (π - t) = sin t$ , midpoint symmetry makes this equal to $I$ .
4.Therefore $2 I = I - (π /2) ln 2$ , and $I = - (π /2) ln 2$ . The logarithmic endpoint singularities are integrable: near zero, $ln (sin x) - ln x$ has a finite limit, and $x ln x - x$ has a finite limit as $x$ approaches zero from above.

Common mistake

Symmetry is useful only after checking the correct condition on the complete integrand and the existence of the integral. In particular, $1/ x$ is odd but undefined at zero, so the continuous odd-function rule does not justify an integral across zero. For an absolute value, split where the inside expression changes sign before using a polynomial antiderivative.

Quick check

If $f$ is continuous and odd, what is $\int_{- 3}^{3} f (x) d x$ ?

Quick check

If $f$ is continuous and even, what is $\int_{- a}^{a} f (x) d x$ for $a > 0$ ?

Quick check

What is the reflection of $x$ in the midpoint of $[2, 8]$ ?

Quick check

What is $\int_{0}^{π /2} sin^{2} x / (sin^{2} x + cos^{2} x) d x$ ?

Quick check

Where should $\int_{0}^{2} ∣ x^{2} - x ∣ d x$ be split inside its interval?

Practice Problems

Evaluate $\int_{- 5}^{5} ∣ x + 2∣ d x$ .
Evaluate $\int_{2}^{8} ∣ x - 5∣ d x$ .
Evaluate $\int_{0}^{2} ∣ x^{2} - x ∣ d x$ .
Evaluate $\int_{0}^{π /2} cos^{5} x / (sin^{5} x + cos^{5} x) d x$ using reflection.
Evaluate $\int_{- π /2}^{π /2} sin^{7} x d x$ without finding an antiderivative.
Evaluate $\int_{0}^{1} x (1 - x)^{n} d x$ for a positive integer $n$ . Use a substitution or expand only if it helps.
If $g (a - x) = g (x)$ , prove $\int_{0}^{a} xg (x) d x = (a /2) \int_{0}^{a} g (x) d x$ .
For continuous $f, g$ on $[0, a]$ with $f (a - x) = f (x)$ and $g (x) + g (a - x) = 4$ , prove $\int_{0}^{a} f (x) g (x) d x = 2 \int_{0}^{a} f (x) d x$ .
Use the logarithmic symmetry result to evaluate $\int_{0}^{π} ln (1 + cos x) d x$ , interpreting the final endpoint by a limit.

Key Takeaways

The integration variable can be renamed, but changing the direction of limits changes the sign. Split intervals to handle sign changes and absolute values. Reflection pairs points around a midpoint and can turn two difficult integrands into a simple sum. Even functions double and odd functions cancel on symmetric intervals when the integral exists. Symmetry, substitution, and algebra often work best together.

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Definite Integrals and the Fundamental Theorem

Next · Lesson 9

Chapter Summary and Practice

The variable name and the interval play different roles

Absolute values require sign information

Reflecting an interval

Symmetry around a midpoint

Even and odd functions on symmetric intervals

A logarithmic integral from reflection and rescaling

Quiz

Practice Problems

Key Takeaways