Evaluation of Definite Integrals

Evaluating a definite integral $\int_a^b f(x)\,dx$ combines everything built so far: an

indefinite-integration technique (Sections 2-6) to find an antiderivative, the Fundamental

Theorem (Section 7) to convert it into a number, and -- often -- a property (Section 8) to

shortcut the work entirely.

The direct route. (i) Find any antiderivative $F(x)$ of $f(x)$, using substitution, partial

fractions, by parts, or a standard form, exactly as for an indefinite integral; (ii) evaluate

$F(b)-F(a)$. For $\int_1^3(2x+1)\,dx$: $F(x)=x^2+x$, so the value is

$F(3)-F(1)=(9+3)-(1+1)=12-2=10$.

Definite integration BY SUBSTITUTION -- changing the limits. When a substitution $u=g(x)$ is

used inside a definite integral, there are two equally valid routes: either substitute back to

$x$ at the end (as for an indefinite integral) and then apply the original limits $a,b$; OR --

usually faster -- change the limits to $u$-values immediately, using $u=g(a)$ and $u=g(b)$, and

never convert back to $x$ at all, evaluating the antiderivative in $u$ directly at these new

limits.

When to reach for a property instead. Direct evaluation is always available, but three

signals suggest checking Section 8's properties first: (a) the integrand contains $f(x)$ AND

$f(a-x)$-type structure over $[0,a]$ (try P4, Exercise Q2); (b) the integrand is visibly an ODD

or EVEN function over a symmetric interval $[-a,a]$ (try P5, Exercise Q3-Q4) -- often collapsing

the integral to $0$ with no antiderivative needed at all, or halving the work; (c) a

trigonometric identity simplifies the integrand into directly integrable pieces first, as in

Miscellaneous Q3, where $\tan^2x=\sec^2x-1$ turns $\int_0^{\pi/4}\tan^2x\,dx$ into

$\big[\tan x-x\big]_0^{\pi/4}=1-\pi/4$.

A caution on the constant of integration. For a DEFINITE integral, the arbitrary constant $C$ …