Definite Integrals & Evaluation

MATDefinite Integrals

Definite-integral evaluation applies the Fundamental Theorem of Calculus: find an antiderivative $F$ of $f$ , then compute $F(b)-F(a)$ . The same toolbox — direct standard forms, substitution (with changed limits), integration by parts, and the symmetry properties — is used, but the final answer is a number, not a function, and no constant of integration is carried.

ISC tests this as the culmination of the chapter, often combining a technique (substitution or parts) with limit-evaluation in a single problem.

Fundamental Theorem of Calculus

\int_{a}^{b} f(x)\,dx = \big[F(x)\big]_{a}^{b} = F(b) - F(a)

F

is any antiderivative of

f

, i.e.

F'(x)=f(x)

, on

[a,b]

; the

+C

cancels in the subtraction.

Substitution with changed limits

\int_{a}^{b} f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(t)\,dt

Put

t=g(x)

; the new limits are

t=g(a)

and

t=g(b)

, so no back-substitution is needed.

Definite integration by parts

\int_{a}^{b} u\,dv = \big[uv\big]_{a}^{b} - \int_{a}^{b} v\,du

Evaluate the boundary term

[uv]_a^b

first, then subtract the remaining definite integral; e.g.

\int_0^1 x e^x\,dx = 1

Useful evaluated results

\int_{0}^{1} x e^{x}\,dx = 1, \qquad \int_{0}^{\pi/2} \sin^2 x\cos x\,dx = \frac{1}{3}

First by parts; second by

t=\sin x

. These illustrate combining a technique with limit substitution.

Find a correct antiderivative first; an error there propagates into the final number.
When substituting, change the limits to the new variable immediately — this is cleaner than substituting back to $x$ at the end.
Limits must match the variable: $x$ -limits with $dx$ , $t$ -limits with $dt$ ; never mix them.
For $\int_0^1 \dfrac{x}{x^2+1}\,dx$ use $t=x^2+1$ (limits $1\to 2$ ) to get $\dfrac{1}{2}\ln 2$ .
Rewrite trig products with identities before integrating: e.g. $\sin 2x\cos x = 2\sin x\cos^2 x$ , then substitute $t=\sin x$ or $t=\cos x$ .
Check whether a symmetry property (king-rule, even/odd) shortcuts the evaluation before grinding out an antiderivative.
Never write $+C$ in a definite integral; the result is a pure number and should be simplified (e.g. as a fraction or in terms of $\ln$ , $\pi$ ).
If the integrand is discontinuous inside $[a,b]$ , the FTC does not apply directly — ISC integrands are continuous on the given interval, so confirm that.

Where the marks go

Changing the variable by substitution but forgetting to change the limits (a very common ISC error).
Carrying a $+C$ into a definite integral or leaving the answer as a function instead of a number.
Evaluating $F(a)-F(b)$ instead of $F(b)-F(a)$ — wrong sign from reversing the order.
In parts, mishandling the boundary term $[uv]_a^b$ or forgetting to subtract the full remaining definite integral.

How the board asks it

Numericalthe fundamental theorem of calculus and standard forms
Evaluate $\displaystyle\int_0^{\pi/2} (2\cos x - 3\sin x)\,dx$ .
Numericalsubstitution with changed limits
Using the substitution $t = x^2 + 1$ , find the value of $\displaystyle\int_0^1 \frac{x}{x^2+1}\,dx$ .
Numericaldefinite integration by parts
Evaluate $\displaystyle\int_0^1 x\,e^{x}\,dx$ by the method of integration by parts.
Derive / provethe symmetry properties (king-rule)
Using properties of definite integrals, prove that $\displaystyle\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx = \frac{\pi}{4}$ .
Multiple choiceuseful evaluated results and even/odd symmetry
The value of $\displaystyle\int_{-\pi/2}^{\pi/2} \sin^3 x\,dx$ is $(a)\ \frac{2}{3}\quad(b)\ 0\quad(c)\ \frac{4}{3}\quad(d)\ \frac{\pi}{2}$ .

Practise this topic

PreviousIntegration by Parts NextProperties of Definite Integrals

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.