Some Properties of Definite Integrals

7.10 Some Properties of Definite Integrals

Definite integrals can often be evaluated more easily using properties that relate integrals over different intervals or with transformed integrands. These properties are not just shortcuts — they reveal deep symmetries in integration. Understanding why each one works matters more than memorising the formula.

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Property P₀: Change of Variable Name

The definite integral depends only on the function $f$ and the limits $a$, $b$ — not on the letter used for the variable of integration. The variable $x$ is a dummy variable.

Proof: Substitute $x = t$; then $dx = dt$ and the limits remain unchanged.

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Property P₁: Reversing Limits

Proof: With $F$ an antiderivative of $f$ (by the Second Fundamental Theorem):

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Property P₂: Splitting the Interval

›Proof

With $F$ an antiderivative of $f$:

$\int_a^c f(x)\,dx + \int_c^b f(x)\,dx = [F(c) - F(a)] + [F(b) - F(c)] = F(b) - F(a) = \int_a^b f(x)\,dx$

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Property P₃: Reflection about the Midpoint

This reflects the integrand about the midpoint $\frac{a+b}{2}$ of the interval.

›Proof

Let $t = a + b - x$, so $dt = -dx$; when $x = a$, $t = b$ and when $x = b$, $t = a$.

$\int_a^b f(x)\,dx = \int_b^a f(a+b-t)\,(-dt) = \int_a^b f(a+b-t)\,dt = \int_a^b f(a+b-x)\,dx$

(using P₁ and P₀).

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Property P₄: Reflection on $[0,a]$ (Special Case of P₃)

This is P₃ with $a=0$, $b=a$.

›Proof

Put $t = a - x$, so $dt = -dx$; when $x=0$, $t=a$ and when $x=a$, $t=0$.

$\int_0^a f(x)\,dx = \int_a^0 f(a-t)\,(-dt) = \int_0^a f(a-t)\,dt = \int_0^a f(a-x)\,dx$

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Property P₅: Splitting $[0,2a]$

›Proof

By P₂, $\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_a^{2a} f(x)\,dx$. In the second integral let $t = 2a - x$, so $dt = -dx$; when $x = a$, $t = a$ and when $x = 2a$, $t = 0$:

$\int_a^{2a} f(x)\,dx = \int_a^0 f(2a-t)\,(-dt) = \int_0^a f(2a-x)\,dx$

Therefore $\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a-x)\,dx$.

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Property P₆: Symmetry about $x = a$

Proof: By P₅, $\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a-x)\,dx$.

• If $f(2a-x) = f(x)$: the sum is $2\int_0^a f(x)\,dx$.
• If $f(2a-x) = -f(x)$: the two integrals cancel, giving $0$.

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Property P₇: Even and Odd Functions

›Proof

By P₂, $\int_{-a}^a f(x)\,dx = \int_{-a}^0 f(x)\,dx + \int_0^a f(x)\,dx$. In the first integral let $t = -x$, so $dt = -dx$; when $x = -a$, $t = a$ and when $x = 0$, $t = 0$: …