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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.

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.

Test yourself — a quick 18-question Definite Integration recall quiz, instant score.

Concept 1 of 3: The Fundamental Theorem of Calculus

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′=fF'=f, then ∫abf(x) dx=F(b)−F(a)\displaystyle\int_a^b f(x)\,dx = F(b)-F(a).
  • ∫abf′(x) dx=f(b)−f(a)\displaystyle\int_a^b f'(x)\,dx = f(b)-f(a). So if f(a)=f(b)f(a)=f(b), the integral is 0.
  • ∫f′(x)f(x) dx=ln⁡∣f(x)∣\displaystyle\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| — spot a derivative over its function.
  • Simplify the integrand first: eln⁡(cos⁡x)=cos⁡xe^{\ln(\cos x)} = \cos x; ddxtan⁡−11x=−11+x2\frac{d}{dx}\tan^{-1}\frac1x = \frac{-1}{1+x^2}.

Fundamental Theorem of Calculus and its corollaries

∫abf(x) dx=F(b)−F(a)    (F′=f)∫abf′(x) dx=f(b)−f(a)∫f′(x)f(x) dx=ln⁡∣f(x)∣\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)|
ab∫ f dxy = f(x)

Worked example

Evaluate ∫13(2x−1) dx\displaystyle\int_1^3 (2x-1)\,dx.
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

NDA · Apr 2018 · Q93Moderate

Example 1 · Definite Integration · Fundamental Theorem, Periodic Integrals, and Leibniz Rule

If ∫abx3 dx=0\int_a^b x^3\,dx=0 and ∫abx2 dx=2/3\int_a^b x^2\,dx=2/3, then a and b are respectively?

Integrating a derivative is not always trivial

∫−11ddx(tan⁡−11x) dx\int_{-1}^{1}\frac{d}{dx}\big(\tan^{-1}\frac1x\big)\,dx is NOT [tan⁡−11x]−11[\tan^{-1}\frac1x]_{-1}^1 naively, because the function jumps at x=0x=0. Compute the derivative −11+x2\frac{-1}{1+x^2} first, then integrate to get −π2-\frac{\pi}{2}.

Concept 2 of 3: Integrals of periodic functions

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 ff has period TT, then ∫0nTf(x) dx=n∫0Tf(x) dx\displaystyle\int_0^{nT} f(x)\,dx = n\int_0^{T} f(x)\,dx.
  • Find the period first: sin⁡4x+cos⁡4x\sin^4x+\cos^4x has period π2\tfrac{\pi}{2}; ∣sin⁡x∣|\sin x| has period π\pi.
  • Two integrals can be equal without being computed — a substitution like x=etx=e^t can turn one into the other.

Integral of a periodic function

∫0nTf(x) dx=n∫0Tf(x) dx∫aa+nTf(x) dx=n∫0Tf(x) dx\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∣|\sin x| has period π\pi and ∫0π∣sin⁡x∣ dx=2\int_0^{\pi}|\sin x|\,dx = 2, find ∫08π∣sin⁡x∣ dx\int_0^{8\pi}|\sin x|\,dx.
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

NDA · Sep 2022 · Q95Moderate

Example 2 · Definite Integration · Fundamental Theorem, Periodic Integrals, and Leibniz Rule

If ∫0π/2(sin⁡4x+cos⁡4x) dx=k\int_0^{\pi/2}(\sin^4x+\cos^4x)\,dx = k, then what is the value of ∫020π(sin⁡4x+cos⁡4x) dx\int_0^{20\pi}(\sin^4x+\cos^4x)\,dx?

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

sin⁡4x+cos⁡4x\sin^4x+\cos^4x does NOT have period 2π2\pi — squaring and adding shrinks the period to π2\tfrac{\pi}{2}. Always determine the actual period of the whole integrand before counting how many fit.

Concept 3 of 3: The Leibniz rule — differentiating an integral

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:

  • ddx∫ag(x)f(t) dt=f(g(x))⋅g′(x)\displaystyle \frac{d}{dx}\int_{a}^{g(x)} f(t)\,dt = f(g(x))\cdot g'(x).
  • More generally with both limits varying: ddx∫u(x)v(x)f(t) dt=f(v) v′−f(u) u′\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)

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

Worked example

If ϕ(x)=∫0x2sin⁡t dt\phi(x)=\displaystyle\int_0^{x^2} \sin t\,dt, find ϕ′(x)\phi'(x).
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

NDA · Apr 2024 · Q90Easy

Example 3 · Definite Integration · Fundamental Theorem, Periodic Integrals, and Leibniz Rule

Let ϕ(a)=∫aa+100π∣sin⁡x∣ dx\phi(a)=\displaystyle\int_a^{a+100\pi}|\sin x|\,dx.
What is ϕ′(a)\phi'(a) equal to?

Don't forget the chain-rule factor

ddx∫ag(x)f\frac{d}{dx}\int_a^{g(x)} f is f(g(x))⋅g′(x)f(g(x))\cdot g'(x), NOT just f(g(x))f(g(x)). The g′(x)g'(x) factor is the most commonly dropped term.

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

    ∫abf(x) dx=F(b)−F(a)    (F′=f)∫abf′(x) dx=f(b)−f(a)∫f′(x)f(x) dx=ln⁡∣f(x)∣\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

    ∫0nTf(x) dx=n∫0Tf(x) dx∫aa+nTf(x) dx=n∫0Tf(x) dx\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)

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

Watch out for (3)

Test yourself on Definite Integration

15 past NDA questions from this chapter, timed at 19 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.