Integration as the Limit of a Sum

Understanding where integrals come from

SPEC 8.4 • INTEGRATION

The Area Problem

How do we find the exact area under a curve? The answer lies in approximation.

How do we find the shaded area under y = f(x) from x = a to x = b?

Key Idea

We can approximate the area by dividing it into rectangles, then take the limit as we use more and more rectangles.

Approximating with Rectangles

Better approximation

Key Insight

More rectangles = better approximation. As n → ∞, the sum of rectangle areas → exact area under the curve!

Setting Up the Sum

Each rectangle has width δx and height f(xᵢ)

Key Formulas

Strip width:

δ x = \frac{b - a}{n}

Area of one rectangle:

f (x_{i}) \times δ x

Total Area \approx \sum_{i = 1}^{n} f (x_{i}) δ x

The Limit Process

As we increase the number of rectangles:

n → ∞ (number of strips increases)

δx → 0 (strip width decreases)

The sum approaches the true area

Fundamental Result

Area = lim_{δ x \to 0} \sum f (x_{i}) δ x = \int_{a}^{b} f (x) d x

Key Point

The integral sign ∫ is an elongated "S" — it stands for "Sum"!

The notation dx represents an infinitesimally small δx.

Discrete SumContinuous Integral

Σ∫

δx (small width)dx (infinitesimal)

Standard Summation Results

To evaluate limits of sums, we need these standard results:

Key Formulas

\sum_{i = 1}^{n} 1 = n

Sum of n ones

\sum_{i = 1}^{n} i = \frac{n ( n + 1 )}{2}

Sum of first n integers

\sum_{i = 1}^{n} i^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

Sum of squares

\sum_{i = 1}^{n} i^{3} = [\frac{n ( n + 1 )}{2}]^{2}

Sum of cubes

Important

These formulas are given in the formula booklet — you don't need to memorise them, but you must know how to use them!

Worked Example01

Find an expression for the area under y = x² from x = 0 to x = 1 using n rectangles with right endpoints.

1

Width of each strip:

δ x = \frac{1 - 0}{n} = \frac{1}{n}

2

Right endpoint of strip i:

x_{i} = \frac{i}{n}

3

Height at right endpoint:

f (x_{i}) = (\frac{i}{n})^{2} = \frac{i ^{2}}{n ^{2}}

4

Area of strip i:

\frac{i ^{2}}{n ^{2}} \times \frac{1}{n} = \frac{i ^{2}}{n ^{3}}

5

Total area:

\sum_{i = 1}^{n} \frac{i ^{2}}{n ^{3}} = \frac{1}{n ^{3}} \sum_{i = 1}^{n} i^{2}

Worked Example02

Using the result from Example 01, evaluate the limit as n → ∞ to find the exact area.

1

From Example 01: Area =

\frac{1}{n ^{3}} \sum_{i = 1}^{n} i^{2}

2

Using

\sum i^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

:

3

Area =

\frac{1}{n ^{3}} \times \frac{n ( n + 1 ) ( 2 n + 1 )}{6} = \frac{( n + 1 ) ( 2 n + 1 )}{6 n ^{2}}

4

Expand:

= \frac{2 n ^{2} + 3 n + 1}{6 n ^{2}} = \frac{2}{6} + \frac{3}{6 n} + \frac{1}{6 n ^{2}}

5

As n → ∞:

\frac{3}{6 n} \to 0

and

\frac{1}{6 n ^{2}} \to 0

Area = $\frac{2}{6} = \frac{1}{3}$

Check: $\int_{0}^{1} x^{2} d x = [\frac{x ^{3}}{3}]_{0}^{1} = \frac{1}{3}$ ✓

Worked Example03

Use the limit of a sum to find $\int_{0}^{2} x d x$

1

Strip width:

δ x = \frac{2 - 0}{n} = \frac{2}{n}

2

Right endpoint:

x_{i} = \frac{2 i}{n}

3

Sum =

\sum_{i = 1}^{n} f (x_{i}) \cdot δ x = \sum_{i = 1}^{n} \frac{2 i}{n} \cdot \frac{2}{n} = \frac{4}{n ^{2}} \sum_{i = 1}^{n} i

4

Using

\sum i = \frac{n ( n + 1 )}{2}

: Sum =

\frac{4}{n ^{2}} \times \frac{n ( n + 1 )}{2} = \frac{2 ( n + 1 )}{n}

5

As n → ∞:

\frac{2 ( n + 1 )}{n} = 2 + \frac{2}{n} \to 2

Area = 2

Worked Example04

Use the limit of a sum to show that $\int_{0}^{1} x^{3} d x = \frac{1}{4}$

1

Strip width:

δ x = \frac{1}{n}

, right endpoint:

x_{i} = \frac{i}{n}

2

Sum =

\sum_{i = 1}^{n} (\frac{i}{n})^{3} \cdot \frac{1}{n} = \frac{1}{n ^{4}} \sum_{i = 1}^{n} i^{3}

3

Using

\sum i^{3} = [\frac{n ( n + 1 )}{2}]^{2}

:

4

Sum =

\frac{1}{n ^{4}} \times \frac{n ^{2} ( n + 1 ) ^{2}}{4} = \frac{( n + 1 ) ^{2}}{4 n ^{2}}

5

As n → ∞:

\frac{( n + 1 ) ^{2}}{4 n ^{2}} = \frac{n ^{2} + 2 n + 1}{4 n ^{2}} \to \frac{1}{4}

Area = $\frac{1}{4}$ as required ✓

Worked Example05

Show that $\int_{0}^{a} x^{2} d x = \frac{a ^{3}}{3}$ using limits of sums.

1

Strip width:

δ x = \frac{a}{n}

, right endpoint:

x_{i} = \frac{ai}{n}

2

Sum =

\sum_{i = 1}^{n} (\frac{ai}{n})^{2} \cdot \frac{a}{n} = \frac{a ^{3}}{n ^{3}} \sum_{i = 1}^{n} i^{2}

3

=

\frac{a ^{3}}{n ^{3}} \times \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

4

=

\frac{a ^{3} ( n + 1 ) ( 2 n + 1 )}{6 n ^{2}} = \frac{a ^{3} ( 2 n ^{2} + 3 n + 1 )}{6 n ^{2}}

5

As n → ∞:

\to \frac{2 a ^{3}}{6} = \frac{a ^{3}}{3}

$\int_{0}^{a} x^{2} d x = \frac{a ^{3}}{3}$ as required ✓

Practice

Question 1: Set up a limit of sum expression for

\int_{0}^{1} 2 x d x

and evaluate it.

1

δx = 1/n, xᵢ = i/n

2

Sum =

\sum_{i = 1}^{n} 2 \cdot \frac{i}{n} \cdot \frac{1}{n} = \frac{2}{n ^{2}} \sum_{i = 1}^{n} i

3

=

\frac{2}{n ^{2}} \times \frac{n ( n + 1 )}{2} = \frac{n + 1}{n}

4

As n → ∞:

\frac{n + 1}{n} \to 1

Answer: 1

Practice

Question 2: Use the limit of a sum to verify that

\int_{0}^{2} x^{2} d x = \frac{8}{3}

1

δx = 2/n, xᵢ = 2i/n

2

Sum =

\sum_{i = 1}^{n} (\frac{2 i}{n})^{2} \cdot \frac{2}{n} = \frac{8}{n ^{3}} \sum_{i = 1}^{n} i^{2}

3

=

\frac{8}{n ^{3}} \times \frac{n ( n + 1 ) ( 2 n + 1 )}{6} = \frac{4 ( n + 1 ) ( 2 n + 1 )}{3 n ^{2}}

4

=

\frac{4 ( 2 n ^{2} + 3 n + 1 )}{3 n ^{2}} \to \frac{8}{3}

as n → ∞

Answer: $\frac{8}{3}$ ✓

Practice

Question 3: The expression

\frac{1}{n} \sum_{i = 1}^{n} (\frac{i}{n})^{4}

represents an area. Write the corresponding definite integral.

1

Rewrite as:

\sum_{i = 1}^{n} (\frac{i}{n})^{4} \cdot \frac{1}{n}

2

Identify: δx = 1/n, so interval width = 1

3

The term (i/n)⁴ means f(x) = x⁴ with x = i/n going from 0 to 1

4

Interval is [0, 1]

Answer: $\int_{0}^{1} x^{4} d x$

Summary

Key Concepts

ConceptFormula

Strip width

δ x = \frac{b - a}{n}

Sum of areas

\sum_{i = 1}^{n} f (x_{i}) δ x

Definite integral

lim_{δ x \to 0} \sum f (x_{i}) δ x = \int_{a}^{b} f (x) d x

Key Exam Skills

Set up limit of sum expressions correctly

Use standard summation formulas from the formula booklet

Simplify algebraically before taking limits

Interpret sums as definite integrals and vice versa