10.1.5 Trapezium Rule (Numerical Integration)

Numerical Integration

Upper and Lower Bounds using rectangle

Upper Bound

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Concept:
The area under a curve $y = x^2$ from $x = 0$ to $x = 5$ is approximated using rectangles.
The tops of these rectangles lie above the curve, leading to an overestimate.
Calculation:
The sum of the areas of these rectangles is calculated:

A \approx 1 + 4 + 9 + 16 + 25 = 55

This gives an approximate area of 55.
Conclusion:
The integral $\int_0^5 x^2 \, dx$ is less than or equal to $55$ .
This is an upper bound.

Lower Bound

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Concept:
Here, the left-hand side (LHS) of the rectangles touches the curve, leading to an underestimate.
The tops of these rectangles lie beneath the curve.
Calculation:
The sum of the areas of these rectangles is:

A \approx 0 + 1 + 4 + 9 + 16 = 30

\therefore 30 \lt \int_0^5 x^2 \, dx \lt 55

Conclusion:
This result provides a lower bound for the integral $\int_0^5 x^2 \, dx$ . If we use an infinite number of infinitely thin retanlges

LB = \int_0^5 x^2 \, dx = UB

Note:

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This is theory that underpins integrations,

Find an Upper and Lower Bound using Rectangles
Make the Rectangles Thinner
$\int_a^b f(x)\ dx$ is the limit as $h \rightarrow 0$ where h is the rectangle width.

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The diagram shows the curve with the equation $y = e^{x^2}$ , for $0 \leq x \leq 1$ . The region under the curve between these limits is divided into four strips of equal width. The area of this region under the curve is A.

(i) Upper Bound Calculation:

By considering the set of rectangles indicated in the diagram, show that an upper bound for A is 1.71. Solution:

\text{Upper Bound} = \frac{1}{4} \left( e^{0.25^2} + e^{0.5^2} + e^{0.75^2} + e^{1^2} \right) \approx 1.71

(ii) Lower Bound Calculation:

By considering an appropriate set of four rectangles, find a lower bound for A. Solution:

\text{Lower Bound} = \frac{1}{4} \left( e^{0^2} + e^{0.25^2} + e^{0.5^2} + e^{0.75^2} \right) \approx 1.28

Numerical Integration: The Trapezium Rule

These trapeziums will give an overestimate for the area $\int_{0}^{5} x^2 \, dx$ because the tops of the trapeziums lie above the curve. This is true for any positive convex curve.

$x$	0	1	2	3	4	5
$y$	$0^2$	$1^2$	$2^2$	$3^2$	$4^2$	$5^2$

In this calculation, we are using 5 strips/trapezia. 5 strips require 6 ordinates ( $x$ -values).

\frac{1}{2} \times (0^2 + 1^2) \times 1 + \frac{1}{2} \times (1^2 + 2^2) \times 1 + \frac{1}{2} \times (2^2 + 3^2) \times 1 + \frac{1}{2} \times (3^2 + 4^2) \times 1 + \frac{1}{2} \times (4^2 + 5^2) \times 1

\frac{1}{2} \times (0^2 + 1^2 + 1^2+ 2^2 + 2^2 + 3^2 + 3^2 + 4^2 + 4^2 + 5^2) \times 1

\frac{1}{2} \times h \times \left(\text{First } y + \text{Last } y + 2 \times (\text{Sum of other } y \text{'s})\right)

Exam Answer

Given:

$h = 1$ | $x$ | 0 | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---|---| | $y$ | $0^2$ | $1^2$ | $2^2$ | $3^2$ | $4^2$ | $5^2$ |

Using the Trapezium Rule:

\frac{1}{2} \times 1 \times \left( 0^2 + 5^2 + 2 \times (1^2 + 2^2 + 3^2 + 4^2) \right) = \frac{85}{2}

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Note: To improve the estimate given by the trapezium rule, use more/thinner strips.

Trapezium Rule (Numerical Integration) Simplified Revision Notes for A-Level OCR Maths Pure

10.1.5 Trapezium Rule (Numerical Integration)

Numerical Integration

Upper and Lower Bounds using rectangle

Upper Bound

Lower Bound

Note:

Numerical Integration: The Trapezium Rule

Exam Answer

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