Properties of Definite Integrals

Beyond direct evaluation via the Fundamental Theorem, definite integrals obey several

properties that often shortcut a computation -- particularly useful when an antiderivative is

hard to find directly but a symmetry in the integrand can be exploited instead. Two of the most

useful are proved here from the Fundamental Theorem (Section 7); the rest follow by very similar

reasoning and are stated for use.

P1: $\displaystyle\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$. Proof. Let $F$ be an

antiderivative of $f$. By the Second FTC, the left side is $F(b)-F(a)$, and the right side is

$-\big[F(a)-F(b)\big]=F(b)-F(a)$ -- identical. $\blacksquare$ In particular, taking $a=b$ gives

$\int_a^a f(x)\,dx=0$ (zero width, zero signed area).

P2 (additivity): $\displaystyle\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx$ for any

$c$. Proof. By the Second FTC, the right side is

$\big[F(c)-F(a)\big]+\big[F(b)-F(c)\big]=F(b)-F(a)$, exactly the left side. $\blacksquare$

P3: $\displaystyle\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$. Substitute $t=a+b-x$ (so

$dt=-dx$; when $x=a$, $t=b$, and when $x=b$, $t=a$): $\int_a^bf(a+b-x)\,dx =\int_b^af(t)(-dt)=\int_a^bf(t)\,dt$, using P1 to flip the limits back. The special case $a=0$

gives the very frequently used $\displaystyle\int_0^af(x)\,dx=\int_0^af(a-x)\,dx$ (P4), the tool

behind Exercise: Definite Integrals and Properties Q2.

P5 (even/odd functions): $\displaystyle\int_{-a}^{a}f(x)\,dx = 2\int_0^af(x)\,dx$ if $f$ is EVEN ($f(-x)=f(x)$), and $=0$ if $f$ is ODD ($f(-x)=-f(x)$). This follows by splitting the

integral at $0$ (P2) and applying P3-type reasoning to the $[-a,0]$ piece; the odd case underlies

Exercise Q3 ($\int_{-1}^1x^3\cos x\,dx=0$, since $x^3\cos x$ is odd, with no computation needed …