NDA Maths · Definite Integration

Fundamental Theorem, Periodicity and the Leibniz Rule

A definite integral is the change in an antiderivative across the limits; periodic integrands repeat over each period, and the Leibniz rule differentiates an integral with a variable limit.

All Definite Integration notes NDA Maths strategy

Why this matters

Start here — these are the foundations the rest of the chapter builds on. 11 PYQs, mostly EASY/MODERATE. The three ideas: the Fundamental Theorem (evaluate by antiderivative at the limits), periodicity (an integral over many periods is a multiple of one period), and the Leibniz rule (differentiate an integral whose limit is a variable). All quick marks once recognised.

Concept 1 of 3

The Fundamental Theorem of Calculus

Intuition

A definite integral measures the net change of an antiderivative between two points. So to evaluate one, find any antiderivative and subtract its values at the limits. Two corollaries the NDA loves: integrating a derivative back gives the original change, and a function that returns to its start over the interval integrates its derivative to zero.

Definition

The theorem and its corollaries:

FTC: if $F'=f$ , then $\displaystyle\int_a^b f(x)\,dx = F(b)-F(a)$ .
$\displaystyle\int_a^b f'(x)\,dx = f(b)-f(a)$ . So if $f(a)=f(b)$ , the integral is 0.
$\displaystyle\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)|$ — spot a derivative over its function.
Simplify the integrand first: $e^{\ln(\cos x)} = \cos x$ ; $\frac{d}{dx}\tan^{-1}\frac1x = \frac{-1}{1+x^2}$ .

Fundamental Theorem of Calculus and its corollaries

\int_a^b f(x)\,dx = F(b)-F(a)\;\;(F'=f) \qquad \int_a^b f'(x)\,dx = f(b)-f(a) \qquad \int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)|

Worked example

Evaluate

\displaystyle\int_1^3 (2x-1)\,dx

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 1Definite IntegrationMODERATE

\int_a^b x^3\,dx=0

and

\int_a^b x^2\,dx=2/3

, then a and b are respectively?

[Q93 · Apr · 2018]

Integrating a derivative is not always trivial

\int_{-1}^{1}\frac{d}{dx}\big(\tan^{-1}\frac1x\big)\,dx

is NOT

[\tan^{-1}\frac1x]_{-1}^1

naively, because the function jumps at

x=0

. Compute the derivative

\frac{-1}{1+x^2}

first, then integrate to get

-\frac{\pi}{2}

Drill 5 more on the fundamental theorem of calculus

Concept 2 of 3

Integrals of periodic functions

Intuition

If a function repeats every period T, its integral over a whole number of periods is just that many copies of the integral over one period. So an integral over a huge interval collapses to one period times a count.

Definition

The periodicity rule:

If $f$ has period $T$ , then $\displaystyle\int_0^{nT} f(x)\,dx = n\int_0^{T} f(x)\,dx$ .
Find the period first: $\sin^4x+\cos^4x$ has period $\tfrac{\pi}{2}$ ; $|\sin x|$ has period $\pi$ .
Two integrals can be equal without being computed — a substitution like $x=e^t$ can turn one into the other.

Integral of a periodic function

\int_0^{nT} f(x)\,dx = n\int_0^{T} f(x)\,dx \qquad \int_{a}^{a+nT} f(x)\,dx = n\int_0^{T} f(x)\,dx

Worked example

Given

|\sin x|

has period

\pi

and

\int_0^{\pi}|\sin x|\,dx = 2

, find

\int_0^{8\pi}|\sin x|\,dx

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 2Definite IntegrationMODERATE

\int_0^{\pi/2}(\sin^4x+\cos^4x)\,dx = k

, then what is the value of

\int_0^{20\pi}(\sin^4x+\cos^4x)\,dx

[Q95 · Sep · 2022]

Use the function's period, not the trig argument's

\sin^4x+\cos^4x

does NOT have period

2\pi

— squaring and adding shrinks the period to

\tfrac{\pi}{2}

. Always determine the actual period of the whole integrand before counting how many fit.

Drill 2 more on integrals of periodic functions

Concept 3 of 3

The Leibniz rule — differentiating an integral

Intuition

When an integral has a variable in its limit, it defines a function of that variable — and you can differentiate it without computing the integral. The derivative is just the integrand evaluated at the moving limit, times the limit's own derivative.

Definition

The Leibniz (variable-limit) rule:

$\displaystyle \frac{d}{dx}\int_{a}^{g(x)} f(t)\,dt = f(g(x))\cdot g'(x)$ .
More generally with both limits varying: $\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)\,dt = f(v)\,v' - f(u)\,u'$ .
The lower constant limit contributes nothing to the derivative.

Leibniz rule (variable upper limit)

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

Worked example

\phi(x)=\displaystyle\int_0^{x^2} \sin t\,dt

, find

\phi'(x)

Practice this conceptself-check · 2 quick reps

From the bank · past-year question

Example 3Definite IntegrationEASY

Let

\phi(a)=\displaystyle\int_a^{a+100\pi}|\sin x|\,dx

What is

\phi'(a)

equal to?

[Q90 · Apr · 2024]

Don't forget the chain-rule factor

\frac{d}{dx}\int_a^{g(x)} f

f(g(x))\cdot g'(x)

, NOT just

f(g(x))

. The

g'(x)

factor is the most commonly dropped term.

Drill 1 more on the leibniz rule — differentiating an integral

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The Fundamental Theorem of Calculus
Fundamental Theorem of Calculus and its corollaries
$\int_a^b f(x)\,dx = F(b)-F(a)\;\;(F'=f) \qquad \int_a^b f'(x)\,dx = f(b)-f(a) \qquad \int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)|$
Integrals of periodic functions
Integral of a periodic function
$\int_0^{nT} f(x)\,dx = n\int_0^{T} f(x)\,dx \qquad \int_{a}^{a+nT} f(x)\,dx = n\int_0^{T} f(x)\,dx$
The Leibniz rule — differentiating an integral
Leibniz rule (variable upper limit)
$\frac{d}{dx}\int_{a}^{g(x)} f(t)\,dt = f(g(x))\,g'(x)$

Watch out for (3)

Integrating a derivative is not always trivial
→ The Fundamental Theorem of Calculus
Use the function's period, not the trig argument's
→ Integrals of periodic functions
Don't forget the chain-rule factor
→ The Leibniz rule — differentiating an integral

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