Define the trapezium rule .

A numerical method for estimating the value of a definite integral . The region is divided into strips of equal width, and each strip is treated as a trapezium rather than as its exact shape.

True or False? If the graph is a straight line, the trapezium rule gives the exact value of the integral, however few strips are used.

True. A trapezium's sloping top joins two points on the graph, so for a straight line it lies along the graph itself. The trapezium and the region are then identical, and even a single strip gives the exact answer.

How does increasing the number of strips change a trapezium rule estimate?

It makes the estimate more accurate . Narrower strips mean each sloping top stays closer to the curve, so less area is wrongly included or left out.

ASMathsCambridge (CIE)Pure 2FlashcardsIntegrationFurther Integration

Further Integration (Cambridge (CIE) AS Maths: Pure 2): Flashcards

Exam code: 9709

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FrontIntegrating Other Functions
Complete these three standard integrals:
$\int \sin x d x =______+ c$
$\int \sec^{2} x d x =______+ c$
$\int e^{x} d x =______+ c$

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Cards in this collection (14)

Complete these three standard integrals:
$\int \sin x d x =______+ c$
$\int \sec^{2} x d x =______+ c$
$\int e^{x} d x =______+ c$
The completed integrals are:
$\int \sin x d x = - \cos x + c$
$\int \sec^{2} x d x = \tan x + c$
$\int e^{x} d x = e^{x} + c$
Each one is a standard derivative read backwards: $\tan x$ appears, for instance, because its derivative is $\sec^{2} x$ .
What is $\int \frac{1}{x} d x$ , and why does the answer need a modulus sign?
The integral is:
$\int \frac{1}{x} d x = \ln |x| + c$
$\ln x$ is only defined for $x > 0$ , but $\frac{1}{x}$ exists for negative values of $x$ as well, so the modulus is what lets the result cover both.
True or False?
Differentiating $\cos x$ introduces a minus sign, but integrating $\cos x$ does not.
True.
$\frac{d}{d x} (\cos x) = - \sin x$ , whereas $\int \cos x d x = \sin x + c$ .
The minus sign belongs to differentiating $\cos x$ and to integrating $\sin x$ , not to $\cos x$ in both directions.
Find $\int e^{5 x} d x$ .
The integral is:
$\int e^{5 x} d x = \frac{1}{5} e^{5 x} + c$
Differentiating $e^{5 x}$ would multiply it by 5, so integrating has to divide by 5 to undo that.
True or False?
$\int \frac{4}{x^{2}} d x = 4 \ln |x^{2}| + c$
False.
$\frac{4}{x^{2}}$ is $4 x^{- 2}$ , an ordinary power of $x$ , and integrating it gives $- \frac{4}{x} + c$ .
A logarithm appears only when the power of $x$ in the denominator is exactly 1.
Find $\int \frac{1}{2 x + 3} d x$ .
The integral is:
$\int \frac{1}{2 x + 3} d x = \frac{1}{2} \ln |2 x + 3| + c$
The whole linear bracket goes inside the logarithm unchanged, and the $\frac{1}{2}$ is there because $x$ has a coefficient of 2.
Why does reversing a standard derivative work for $e^{5 x}$ but not for $e^{x^{2}}$ ?
Because differentiating $e^{5 x}$ brings out a constant factor, whereas differentiating $e^{x^{2}}$ brings out $2 x$ .
A constant can be compensated for, but a factor containing $x$ cannot.
This is why the standard results are only quoted for linear brackets of the form $a x + b$ .
Define the trapezium rule.
A numerical method for estimating the value of a definite integral.
The region is divided into strips of equal width, and each strip is treated as a trapezium rather than as its exact shape.
Complete the trapezium rule, filling in the two missing groups of $y$ -values:
$\int_{a}^{b} y d x \approx \frac{1}{2} h [(______) + 2 (______)]$
The completed rule is:
$\int_{a}^{b} y d x \approx \frac{1}{2} h [(y_{0} + y_{n}) + 2 (y_{1} + y_{2} + \dots + y_{n - 1})]$
The first and last $y$ -values are counted once, and every one in between is counted twice.
A definite integral from $x = 1$ to $x = 9$ is to be estimated using 4 strips. What is the strip width, and how many $y$ -values are needed?
The strip width is $h = \frac{9 - 1}{4} = 2$ , from $h = \frac{b - a}{n}$ .
Five $y$ -values are needed, $y_{0}$ to $y_{4}$ , because $n$ strips have $n + 1$ edges.
Questions often call the strips intervals instead.
The curve $y = f (x)$ bends upwards throughout the interval, like $y = x^{2}$ . Does the trapezium rule over-estimate or under-estimate $\int_{a}^{b} f (x) d x$ , and why?
It over-estimates it.
The top of each strip is a straight chord joining two points on the curve, and where the curve bends upwards it sags below that chord, so every trapezium carries a sliver of extra area.
A curve bending the other way, such as $y = \ln x$ , gives an under-estimate.
True or False?
If the graph is a straight line, the trapezium rule gives the exact value of the integral, however few strips are used.
True.
A trapezium's sloping top joins two points on the graph, so for a straight line it lies along the graph itself.
The trapezium and the region are then identical, and even a single strip gives the exact answer.
Why is the trapezium rule needed to evaluate $\int_{0}^{2} \sqrt{1 + x^{3}} d x$ ?
Because $\sqrt{1 + x^{3}}$ cannot be matched to any of the standard integrals, so there is no exact answer to be found by reversing a derivative.
The integral still has a definite value, and the trapezium rule estimates it from a small number of $y$ -values.
How does increasing the number of strips change a trapezium rule estimate?
It makes the estimate more accurate.
Narrower strips mean each sloping top stays closer to the curve, so less area is wrongly included or left out.

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