Properties of Definite Integrals

MATDefinite Integrals

The properties of definite integrals let you evaluate integrals that have no elementary antiderivative, or that simplify dramatically, by exploiting symmetry and reflection rather than brute-force integration. The workhorse is the king-rule $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$ , which (added to the original) collapses ratios like $\dfrac{\sin x}{\sin x+\cos x}$ .

ISC examines these heavily because they reward method over computation, and the same definite integral also models accumulated quantities such as distance from velocity or total cost from marginal cost.

King property (reflection)

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

Valid for any integrable

f

[a,b]

; the special case

\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx

is the most used.

Even/odd symmetry on

[-a,a]

\int_{-a}^{a} f(x)\,dx = \begin{cases} 2\int_{0}^{a} f(x)\,dx & f(-x)=f(x) \\ 0 & f(-x)=-f(x) \end{cases}

f

even means

f(-x)=f(x)

;

f

odd means

f(-x)=-f(x)

. E.g.

\int_{-2}^{2} x^2\,dx = 2\int_0^2 x^2\,dx

Property over

[0,2a]

\int_{0}^{2a} f(x)\,dx = \begin{cases} 2\int_{0}^{a} f(x)\,dx & f(2a-x)=f(x) \\ 0 & f(2a-x)=-f(x) \end{cases}

Used to fold integrals over

[0,2a]

onto

[0,a]

; e.g. powers of

\sin x

\cos x

over

[0,\pi]

Interchange of limits and zero-width

\int_{a}^{b} f(x)\,dx = -\int_{b}^{a} f(x)\,dx, \qquad \int_{a}^{a} f(x)\,dx = 0

Swapping the limits changes the sign; equal limits give

0

Accumulation (net change)

\int_{a}^{b} F'(x)\,dx = F(b) - F(a)

Distance

=\int v\,dt

, total cost change

=\int C'(x)\,dx

; the definite integral accumulates a rate over an interval.

King-rule technique: write $I=\int_a^b f(x)\,dx$ , form $I=\int_a^b f(a+b-x)\,dx$ , add the two so the integrand simplifies (often to a constant), then divide by $2$ .
For $\int_0^{\pi/2}\dfrac{\sin x}{\sin x+\cos x}\,dx$ , applying $x\to\dfrac{\pi}{2}-x$ and adding gives $2I=\int_0^{\pi/2}1\,dx$ , so $I=\dfrac{\pi}{4}$ .
Before using the even/odd rule, always check symmetry by computing $f(-x)$ ; only an even function doubles and only an odd function vanishes.
$\int_0^{\pi} x\,f(\sin x)\,dx = \dfrac{\pi}{2}\int_0^{\pi} f(\sin x)\,dx$ because $\sin(\pi-x)=\sin x$ — this kills the stray $x$ factor.
The classic result $\int_0^{\pi/2}\log(\sin x)\,dx = -\dfrac{\pi}{2}\log 2$ is proved by $x\to\dfrac{\pi}{2}-x$ symmetry plus the identity $\sin 2x = 2\sin x\cos x$ .
In application problems the integrand is a rate: distance travelled $=\int_{t_1}^{t_2} v(t)\,dt$ , total cost increase $=\int_{x_1}^{x_2} C'(x)\,dx$ .
Definite integrals need no constant of integration; the $+C$ cancels in $F(b)-F(a)$ .
A definite integral is a number depending only on $a$ , $b$ and $f$ — the variable of integration is a dummy: $\int_a^b f(x)\,dx = \int_a^b f(t)\,dt$ .

Where the marks go

Applying the even/odd shortcut without verifying symmetry — e.g. assuming $\int_{-a}^a f\,dx = 0$ when $f$ is not actually odd.
When velocity changes sign on $[t_1,t_2]$ , distance is $\int |v|\,dt$ (split at the zeros), not $\int v\,dt$ which gives only displacement.
Reflecting with the wrong substitution: over $[0,a]$ use $x\to a-x$ , but over $[a,b]$ you must use $x\to a+b-x$ .
Forgetting to divide by $2$ after adding $I$ to its reflected copy, or carrying a $+C$ into a definite integral.

How the board asks it

Numericalthe king-rule (reflection property)
Evaluate $\displaystyle\int_0^{\pi/2}\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$ .
Numericalking-rule with a follow-up substitution on $[0,\pi]$
Evaluate $\displaystyle\int_0^{\pi}\dfrac{x\,\sin x}{1+\cos^2 x}\,dx$ .
Derive / provethe $\int_0^{\pi} x\,f(\sin x)\,dx$ identity
Prove that $\displaystyle\int_0^{\pi}\dfrac{x}{1+\sin x}\,dx=\pi$ using a suitable property of definite integrals.
Numericaleven/odd symmetry on $[-a,a]$
Evaluate $\displaystyle\int_{-\pi/4}^{\pi/4}\left(x^3+x\cos x+\tan^5 x\right)dx$ , justifying your use of symmetry.
Give reasonschecking symmetry before the odd-function shortcut
State, with reasons, whether $\displaystyle\int_{-2}^{2}\dfrac{x^2}{1+5^{x}}\,dx=0$ , and hence evaluate the integral.
Applicationaccumulation of a rate (net change)
A particle moves with velocity $v(t)=t^2-4t+3$ (in $\text{m/s}$ ) for $0\le t\le 3$ s. Find the total distance travelled, taking care where $v(t)$ changes sign.

Practise this topic

PreviousDefinite Integrals & Evaluation

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.