Definite Integrals (including Improper Integrals)

Introduction

A definite integral produces a numerical value rather than a function. It represents the signed area between a curve and the $x$ -axis over a specified interval. In the 9709 exam, definite integrals appear throughout — in area and volume calculations, in kinematics, and as standalone evaluation questions. Mastering precise limit substitution and handling simple improper integrals (where one limit is $\infty$ ) is directly examined and earns method marks at every stage.

Core Concept

From Indefinite to Definite

Recall that the indefinite integral gives a family of antiderivatives:

\int f(x)\,\mathrm{d}x = F(x) + c

A definite integral fixes two limits of integration, $a$ (lower) and $b$ (upper), and evaluates the antiderivative at each:

\int_a^b f(x)\,\mathrm{d}x = \Big[F(x)\Big]_a^b = F(b) - F(a)

The constant of integration $c$ cancels in every definite integral, so it is never written.

The Square-Bracket Notation

The notation $\Big[F(x)\Big]_a^b$ means: substitute $x = b$ , then subtract the result of substituting $x = a$ .

Simple Improper Integrals

An improper integral arises when one (or both) limits of integration is $\pm\infty$ . For the 9709 syllabus, only simple cases are required — typically an upper limit of $+\infty$ .

The technique is to replace the infinite limit with a finite parameter $t$ , evaluate the integral, then take the limit as $t \to \infty$ :

\int_a^{\infty} f(x)\,\mathrm{d}x = \lim_{t \to \infty}\int_a^{t} f(x)\,\mathrm{d}x = \lim_{t \to \infty}\Big[F(x)\Big]_a^t

If this limit exists and is finite, the improper integral converges to that value. If the expression grows without bound, it diverges (no finite answer exists).

Key Formulae & Definitions

Definite integral — Fundamental Theorem of Calculus:

\int_a^b f(x)\,\mathrm{d}x = F(b) - F(a), \quad \text{where } F'(x) = f(x)

Standard power rule (for definite integrals, $n \neq -1$ ):

\int_a^b x^n\,\mathrm{d}x = \left[\frac{x^{n+1}}{n+1}\right]_a^b = \frac{b^{n+1}}{n+1} - \frac{a^{n+1}}{n+1}

Improper integral definition:

\int_a^{\infty} f(x)\,\mathrm{d}x = \lim_{t \to \infty}\Big[F(x)\Big]_a^t = \lim_{t \to \infty} F(t) - F(a)

Key limit facts needed:

Expression	Limit as $t \to \infty$
$t^n$ for $n > 0$	$\infty$ (diverges)
$t^{-n}$ for $n > 0$	$0$
$\dfrac{1}{t^n}$ for $n > 0$	$0$

Worked Examples

Example 1 — Standard Definite Integral

Evaluate $\displaystyle\int_1^4 \left(3x^2 - 4x + 1\right)\mathrm{d}x$ .

Step 1 — Find the antiderivative (omit $c$ ):

F(x) = x^3 - 2x^2 + x

Step 2 — Apply the limits using square-bracket notation:

\Big[x^3 - 2x^2 + x\Big]_1^4

Step 3 — Substitute $x = 4$ :

F(4) = 64 - 32 + 4 = 36

Step 4 — Substitute $x = 1$ :

F(1) = 1 - 2 + 1 = 0

Step 5 — Subtract:

\int_1^4 \left(3x^2 - 4x + 1\right)\mathrm{d}x = 36 - 0 = \boxed{36}

Example 2 — Integral Involving Negative and Fractional Powers

Evaluate $\displaystyle\int_1^9 \left(\sqrt{x} + \frac{2}{x^2}\right)\mathrm{d}x$ .

Step 1 — Rewrite in index form:

\int_1^9 \left(x^{1/2} + 2x^{-2}\right)\mathrm{d}x

Step 2 — Find the antiderivative:

F(x) = \frac{x^{3/2}}{\tfrac{3}{2}} + \frac{2x^{-1}}{-1} = \frac{2}{3}x^{3/2} - 2x^{-1}

Step 3 — Apply the limits:

\left[\frac{2}{3}x^{3/2} - \frac{2}{x}\right]_1^9

Step 4 — Substitute $x = 9$ :

\frac{2}{3}(27) - \frac{2}{9} = 18 - \frac{2}{9} = \frac{162 - 2}{9} = \frac{160}{9}

Step 5 — Substitute $x = 1$ :

\frac{2}{3}(1) - 2 = \frac{2}{3} - 2 = -\frac{4}{3}

Step 6 — Subtract:

\frac{160}{9} - \left(-\frac{4}{3}\right) = \frac{160}{9} + \frac{12}{9} = \frac{172}{9} \approx 19.1

\boxed{\dfrac{172}{9}}

Example 3 — Simple Improper Integral

Evaluate $\displaystyle\int_2^{\infty} \frac{3}{x^4}\,\mathrm{d}x$ , or show it diverges.

Step 1 — Replace $\infty$ with parameter $t$ :

\int_2^{t} 3x^{-4}\,\mathrm{d}x

Step 2 — Find the antiderivative:

\left[\frac{3x^{-3}}{-3}\right]_2^t = \left[-x^{-3}\right]_2^t

Step 3 — Apply the limits:

= -t^{-3} - \left(-2^{-3}\right) = -\frac{1}{t^3} + \frac{1}{8}

Step 4 — Take the limit as $t \to \infty$ :

\lim_{t \to \infty}\left(-\frac{1}{t^3} + \frac{1}{8}\right) = 0 + \frac{1}{8} = \boxed{\dfrac{1}{8}}

The integral converges to $\dfrac{1}{8}$ .

Common Mistakes & Examiner Pitfalls

Forgetting to subtract $F(a)$ , or computing $F(a) - F(b)$ in the wrong order. The upper limit is always substituted first: $F(b) - F(a)$ .
Including $+\,c$ in a definite integral. The constant of integration is never written — it cancels automatically.
Errors with negative index arithmetic. When integrating $x^{-2}$ , the result is $\dfrac{x^{-1}}{-1} = -x^{-1}$ , not $+x^{-1}$ . Always check the sign carefully.
Not checking convergence in improper integrals. If $\lim_{t\to\infty}F(t)$ grows to $\pm\infty$ (e.g., integrating $x^2$ to $\infty$ ), the integral diverges — you must state this clearly rather than writing a numerical answer.
Assuming all improper integrals converge. For example, $\int_1^\infty x^{-1}\,\mathrm{d}x = \lim_{t\to\infty}[\ln t - \ln 1]$ diverges because $\ln t \to \infty$ . The power must be strictly less than $-1$ (i.e., $n < -1$ ) for $\int_a^\infty x^n\,\mathrm{d}x$ to converge.
Mishandling fractional powers at the lower limit. Check that the integrand is defined at every point in $[a, b]$ — for instance, $x^{-1/2}$ is undefined at $x = 0$ .

Practice Questions

Q1. Evaluate $\displaystyle\int_0^3 (2x^3 - 5x)\,\mathrm{d}x$ .

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Antiderivative: $F(x) = \dfrac{x^4}{2} - \dfrac{5x^2}{2}$

Upper limit $x=3$ : $\dfrac{81}{2} - \dfrac{45}{2} = \dfrac{36}{2} = 18$

Lower limit $x=0$ : $0 - 0 = 0$

\int_0^3(2x^3-5x)\,\mathrm{d}x = 18 - 0 = \boxed{18}

Q2. Evaluate $\displaystyle\int_4^{16} \frac{1}{\sqrt{x}}\,\mathrm{d}x$ .

Show answer

Rewrite: $x^{-1/2}$

Antiderivative: $\dfrac{x^{1/2}}{1/2} = 2x^{1/2} = 2\sqrt{x}$

Apply limits:

\Big[2\sqrt{x}\Big]_4^{16} = 2(4) - 2(2) = 8 - 4 = \boxed{4}

Q3. Evaluate $\displaystyle\int_1^{\infty} \frac{5}{x^3}\,\mathrm{d}x$ , or show it diverges.

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Replace $\infty$ with $t$ : $\displaystyle\int_1^t 5x^{-3}\,\mathrm{d}x$

Antiderivative: $\left[\dfrac{5x^{-2}}{-2}\right]_1^t = \left[-\dfrac{5}{2x^2}\right]_1^t$

Apply limits:

= -\frac{5}{2t^2} + \frac{5}{2}

Take $t \to \infty$ :

\lim_{t\to\infty}\left(-\frac{5}{2t^2} + \frac{5}{2}\right) = 0 + \frac{5}{2} = \boxed{\dfrac{5}{2}}

The integral converges.

Q4. Determine whether $\displaystyle\int_1^{\infty} \frac{1}{\sqrt{x}}\,\mathrm{d}x$ converges or diverges. Justify your answer.

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Replace $\infty$ with $t$ :

\int_1^t x^{-1/2}\,\mathrm{d}x = \Big[2x^{1/2}\Big]_1^t = 2\sqrt{t} - 2

Take $t \to \infty$ :

\lim_{t\to\infty}(2\sqrt{t} - 2) = \infty

The limit does not exist (finite), so the integral diverges. No numerical answer can be given.

Q5. Given that $\displaystyle\int_1^k (6x^2 + 2)\,\mathrm{d}x = 26$ , find the value of $k$ (where $k > 1$ ).

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Antiderivative: $F(x) = 2x^3 + 2x$

Apply limits:

\Big[2x^3 + 2x\Big]_1^k = (2k^3 + 2k) - (2 + 2) = 2k^3 + 2k - 4

Set equal to 26:

2k^3 + 2k - 4 = 26

2k^3 + 2k = 30

k^3 + k = 15

k^3 + k - 15 = 0

Testing $k = 2$ : $8 + 2 - 15 = -5$ (not zero). Testing $k = 2.3$ : $12.167 + 2.3 - 15 = -0.533$ (close). Testing $k = 2.34$ : $12.81 + 2.34 - 15 \approx 0.15$ .

By inspection or a sign-change argument, the solution is $k \approx 2.33$ .

For exact form: $k^3 + k - 15 = 0$ . Since this cubic has one real root, by the intermediate value theorem $k \approx 2.33$ (to 3 s.f.).

\boxed{k \approx 2.33}

Connections

Prerequisites — build directly on these:

Integration as the Reverse of Differentiation: the antiderivative $F(x)$ found by reversing the power rule is the essential building block for every definite integral.

Leads directly into:

Area Under a Curve: definite integrals are used to calculate the area between a curve and the $x$ -axis, requiring careful attention to sign when the curve dips below the axis.
Area Between Two Curves: extends the definite integral to compute the area enclosed between $y = f(x)$ and $y = g(x)$ over an interval.
Volumes of Revolution: the formula $V = \pi\int_a^b y^2\,\mathrm{d}x$ depends entirely on evaluating definite integrals of squared functions.
Kinematics using Calculus: displacement is recovered from velocity by evaluating a definite integral between two time values.

Introduction

Core Concept

From Indefinite to Definite

The Square-Bracket Notation

Simple Improper Integrals

Key Formulae & Definitions

Worked Examples

Example 1 — Standard Definite Integral

Example 2 — Integral Involving Negative and Fractional Powers

Example 3 — Simple Improper Integral

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

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