How can rectangles be used to estimate the area under a curve?

Divide the interval into strips and draw a rectangle on each, taking its height from the curve. Adding up the areas of those rectangles gives an estimate of the area under the curve.

Why does using more rectangles improve the estimate?

Each rectangle overshoots or undershoots the curve only across its own width, so narrower rectangles leave less unaccounted for. As the number of rectangles rises, the total error shrinks towards zero.

True or False? With enough rectangles, the sum of their areas is exactly the area under the curve.

False. Any finite number of rectangles gives only an estimate, however many there are. It is the limit as the number tends to infinity that equals the area exactly, which is why integration is defined as a limit rather than as a sum.

What does the reverse chain rule undo?

A differentiation that used the chain rule , so the integrand is a composite function multiplied by the derivative of its inside. Recognising that shape is what lets you integrate by inspection, without setting up a formal substitution.

What are the steps of the reverse chain rule?

Spot the main function, write down what would differentiate to give it, then adjust and compensate for any constant that the chain rule would have produced. Simplify at the end.

How can you check an integration answer?

Differentiate it. You should get back exactly what you set out to integrate. This is worth doing whenever the integral was reached by inspection rather than by a formal method, because inspection is where a constant is most easily dropped.

What is different about a harder substitution question?

The substitution is given to you , because it is not one you would be expected to spot. The method that follows is exactly the same as before; it is the algebra in between that gets heavier.

True or False? Being given the substitution makes the question easier than having to find it yourself.

False. The substitution is given precisely because it is not one you would be expected to find, and the algebra that follows it is heavier than in a standard substitution question. What you are handed removes one difficulty and signals another.

Why do you sometimes need a trigonometric identity before integrating?

Because the expression as written is not one of the standard integrals, but an identity can turn it into one that is. Most often it is a squared trigonometric term that has to be rewritten.

Which differentiation rule does integration by parts reverse?

The product rule , which is why it is the method for integrating a product of two functions. That makes it the counterpart of the reverse chain rule, which undoes the chain rule instead.

True or False? Integration by parts can be applied more than once in the same question.

True. If the second integral is still a product, apply the formula to that as well. It is rare to need it more than twice, so a third application that still does not finish usually means something went wrong earlier.

When should you integrate using partial fractions?

When the integrand is a fraction whose denominator is degree 2 or more and factorises into linear factors. Splitting it turns one integral you cannot do into two or three that you can.

What changes when both boundaries of a region are curves?

Nothing about the method: it is still the integral of upper minus lower , taken between the intersections. What is lost is the shortcut of using a triangle or trapezium formula for one boundary, since neither of them is a straight line any more.

Two curves meet at three points. How many integrals does the enclosed area need?

Two , one for each enclosed region: from the first intersection to the second, and from the second to the third. Each region is bounded separately, so each needs its own integral with its own pair of limits.

Why must you check which curve is on top for each region separately?

Because the two curves swap over at every point where they cross. A curve that was above before an intersection is below after it, so "upper minus lower" means a different subtraction in each region.

Why is a sketch essential here?

Because it is the only reliable way to see how many separate regions there are, and which curve is on top in each of them. The algebra gives you the intersections; only the picture tells you what to do with them.

True or False? The total area between two curves is the integral of their difference across the whole interval.

False. Wherever the curves swap over, the difference changes sign, so one region subtracts from another and the total comes out too small. Integrate each region separately, take each as a positive area, and then add them.

What is the first thing to check when deciding how to integrate?

Whether it is already a standard integral , or can be turned into one just by rewriting. Expanding brackets, splitting a fraction or simplifying a quotient often removes the need for any technique at all.

Why look again for the reverse chain rule after using an identity?

Because rewriting an expression changes its shape, and a reverse chain rule that was not available before may be available now. This is the easiest thing to miss, since applying the identity feels like the answer rather than a step towards it.

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Further Integration (Edexcel A Level Maths: Pure): Flashcards

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How can rectangles be used to estimate the area under a curve?

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How can rectangles be used to estimate the area under a curve?
Divide the interval into strips and draw a rectangle on each, taking its height from the curve.
Adding up the areas of those rectangles gives an estimate of the area under the curve.
Why does using more rectangles improve the estimate?
Each rectangle overshoots or undershoots the curve only across its own width, so narrower rectangles leave less unaccounted for.
As the number of rectangles rises, the total error shrinks towards zero.
As the number of rectangles increases, complete what happens to each quantity:
$δ x \to______$ and $n \to______$
The completed statement is:
$δ x \to 0$ and $n \to \infty$
Here $δ x$ is the width of one rectangle and $n$ is how many there are, so making each one thinner means needing more of them.
How does a sum of rectangle areas become a definite integral?
The sum $\sum f (x) δ x$ approaches a limit as $δ x \to 0$ , and that limit is written $\int_{a}^{b} f (x) d x$ .
The summation sign becomes the integral sign and the $δ x$ becomes the $d x$ , which is where the notation comes from.
True or False?
With enough rectangles, the sum of their areas is exactly the area under the curve.
False.
Any finite number of rectangles gives only an estimate, however many there are.
It is the limit as the number tends to infinity that equals the area exactly, which is why integration is defined as a limit rather than as a sum.
Complete the two exponential integrals:
$\int e^{x} d x =______$
$\int e^{k x} d x =______$
The completed integrals are:
$\int e^{x} d x = e^{x} + c$
$\int e^{k x} d x = \frac{1}{k} e^{k x} + c$
Differentiating $e^{k x}$ multiplies by $k$ , so integrating it divides by $k$ instead.
What is $\int \frac{1}{x} d x$ , and why is it a special case?
$\ln |x| + c$ .
It is the one power the power rule cannot handle, because raising $- 1$ by one gives zero and the rule would then divide by it.
The modulus is needed because $\ln$ is undefined for negative numbers, whereas $\frac{1}{x}$ is perfectly well defined there.
Complete the two trigonometric integrals:
$\int \sin x d x =______$
$\int \cos x d x =______$
The completed integrals are:
$\int \sin x d x = - \cos x + c$
$\int \cos x d x = \sin x + c$
The minus sits on the sine integral, which is the opposite of where it sits when you differentiate.
What is $\int \sec^{2} x d x$ ?
$\tan x + c$ .
Every standard integral of this kind is a derivative read backwards, and $\tan x$ is what differentiates to give $\sec^{2} x$ .
True or False?
$\int \tan x d x = \ln |\sec x| + c$
True.
It is a standard result, and it is given in the formulae booklet.
Unlike $\sin$ and $\cos$ , the integral of $\tan$ is not another trigonometric function at all, which is why it has to be looked up rather than guessed at.
What does the reverse chain rule undo?
A differentiation that used the chain rule, so the integrand is a composite function multiplied by the derivative of its inside.
Recognising that shape is what lets you integrate by inspection, without setting up a formal substitution.
What are the steps of the reverse chain rule?
Spot the main function, write down what would differentiate to give it, then adjust and compensate for any constant that the chain rule would have produced.
Simplify at the end.
What does it mean to "adjust and compensate" when integrating?
Put in the constant your answer needs, then multiply by its reciprocal outside, so that nothing has actually been changed.
Integrating $e^{5 x}$ needs a $5$ from the chain rule, so you write $\frac{1}{5} e^{5 x}$ : the $\frac{1}{5}$ compensates for the $5$ that differentiating would bring out.
How can you check an integration answer?
Differentiate it. You should get back exactly what you set out to integrate.
This is worth doing whenever the integral was reached by inspection rather than by a formal method, because inspection is where a constant is most easily dropped.
True or False?
The reverse chain rule works whenever the integrand is a composite function.
False.
The derivative of the inside function has to be present as well, at least up to a constant multiple.
$\int 2 x e^{x^{2}} d x$ works because the $2 x$ is the derivative of $x^{2}$ , whereas $\int e^{x^{2}} d x$ cannot be done at all by elementary means.
Complete the standard result:
$\int \frac{f' (x)}{f (x)} d x =______$
The completed result is:
$\int \frac{f' (x)}{f (x)} d x = \ln |f (x)| + c$
The modulus is there for the same reason as in $\int \frac{1}{x} d x$ : the logarithm needs a positive argument.
How do you test whether a fraction is of the form $\frac{f' (x)}{f (x)}$ ?
Differentiate the denominator and compare the result with the numerator.
Ignore any coefficients while comparing: if the two match apart from a constant multiple, the form applies.
The numerator is a constant multiple of the derivative of the denominator, but not equal to it. What do you do?
Adjust for that constant, exactly as in the reverse chain rule.
In $\int \frac{x}{x^{2} + 1} d x$ the denominator differentiates to $2 x$ , so write it as $\frac{1}{2} \int \frac{2 x}{x^{2} + 1} d x$ , giving $\frac{1}{2} \ln |x^{2} + 1| + c$ .
True or False?
$\int \frac{2 x + 1}{x^{2} + x} d x = \ln |x^{2} + x| + c$
True.
Differentiating $x^{2} + x$ gives exactly $2 x + 1$ , which is the numerator, so there is no constant to adjust for.
This is the cleanest form the pattern takes.
Why does this pattern integrate to a logarithm?
Because differentiating $\ln (f (x))$ gives $\frac{f' (x)}{f (x)}$ , by the chain rule.
The integral is that result read backwards, which is why no separate rule has to be learned for it.
Define integration by substitution.
Replacing part of an integrand with a single new variable $u$ , so that the integral becomes one you can do.
It is the formal counterpart of the reverse chain rule: the same underlying process, written out in full rather than spotted.
How do you choose a substitution when none is given?
Look for the "second" function rather than the main one, and let $u$ be that.
In $\int 2 x {(x^{2} + 1)}^{5} d x$ the main function is the fifth power, so the substitution is $u = x^{2} + 1$ , the expression inside it.
Why can $\frac{d u}{d x}$ be treated like a fraction here?
It is shorthand for swapping $d x$ for $d u$ , and it is a licensed step in a substitution even though $\frac{d y}{d x}$ must never be split when rearranging an implicit derivative.
From $u = x^{2} + 1$ , $\frac{d u}{d x} = 2 x$ is used in the form $d u = 2 x d x$ .
What has to be replaced when you substitute?
Everything. Every $x$ term must become a $u$ term, including the $d x$ .
An integral containing both letters cannot be integrated, so anything left in $x$ has to be dealt with before you go on.
For a definite integral, what happens to the limits under a substitution?
They must be converted from $x$ values into $u$ values, using the substitution itself.
Doing so lets you evaluate straight away in $u$ ; leaving them means substituting $x$ back in first.
True or False?
After integrating in $u$ , you must always substitute $x$ back in.
False.
For an indefinite integral you must, because the answer has to be a function of $x$ .
For a definite integral whose limits have already been converted to $u$ , you can evaluate in $u$ and never return to $x$ at all.
What is different about a harder substitution question?
The substitution is given to you, because it is not one you would be expected to spot.
The method that follows is exactly the same as before; it is the algebra in between that gets heavier.
You are given the substitution $u = \sqrt{x + 1}$ . How do you get $d x$ in terms of $d u$ ?
Rearrange the substitution first, then differentiate: squaring gives $x = u^{2} - 1$ , so $\frac{d x}{d u} = 2 u$ and $d x = 2 u d u$ .
Rearranging before differentiating is usually easier than differentiating a root as it stands.
Why is it useful to rearrange a given substitution to make $x$ the subject?
Because the integrand usually contains $x$ terms that are not part of the obvious swap, and those have to be converted too.
With $u = \sqrt{x + 1}$ , rearranging to $x = u^{2} - 1$ gives a ready replacement for every one of them.
True or False?
Being given the substitution makes the question easier than having to find it yourself.
False.
The substitution is given precisely because it is not one you would be expected to find, and the algebra that follows it is heavier than in a standard substitution question.
What you are handed removes one difficulty and signals another.
How do you know a harder substitution has been carried out correctly?
The integral should contain only $u$ and $d u$ , with no $x$ left anywhere, and it should be something you can actually integrate.
If it is no simpler than what you started with, the substitution has been applied wrongly rather than chosen wrongly, since it was chosen for you.
Why do you sometimes need a trigonometric identity before integrating?
Because the expression as written is not one of the standard integrals, but an identity can turn it into one that is.
Most often it is a squared trigonometric term that has to be rewritten.
How do you integrate $\sin^{2} x$ or $\cos^{2} x$ ?
Rewrite them with the double angle identity for $\cos 2 A$ , which turns a square into a linear expression in $\cos 2 x$ .
From $\cos 2 x = 1 - 2 \sin^{2} x$ you get $\sin^{2} x = \frac{1}{2} (1 - \cos 2 x)$ , which integrates term by term.
How do you integrate $\sin 3 x \cos 3 x$ ?
Use $\sin 2 A = 2 \sin A \cos A$ backwards, which gives $\sin 3 x \cos 3 x = \frac{1}{2} \sin 6 x$ .
That is a standard integral, so the answer is $- \frac{1}{12} \cos 6 x + c$ .
Complete the identities used to integrate these squared functions:
$\tan^{2} x =______- 1$
$\cot^{2} x =______- 1$
The completed identities are:
$\tan^{2} x = \sec^{2} x - 1$
$\cot^{2} x = {cosec}^{2} x - 1$
Each turns a square you cannot integrate directly into one you can, since $\sec^{2} x$ and ${cosec}^{2} x$ are both standard integrals.
True or False?
$\int \sin^{4} x \cos x d x$ needs a trigonometric identity.
False.
It looks as though it should, but $\cos x$ is the derivative of $\sin x$ , so this is a reverse chain rule integral and the answer is $\frac{1}{5} \sin^{5} x + c$ .
Anything of the form $\sin^{n} k x \cos k x$ behaves the same way and needs no identity at all.
Complete the integration by parts formula:
$\int u \frac{d v}{d x} d x = u v -______$
The completed formula is:
$\int u \frac{d v}{d x} d x = u v - \int v \frac{d u}{d x} d x$
Note that the product being integrated is made from $u$ and $\frac{d v}{d x}$ , not from $u$ and $v$ .
Which differentiation rule does integration by parts reverse?
The product rule, which is why it is the method for integrating a product of two functions.
That makes it the counterpart of the reverse chain rule, which undoes the chain rule instead.
How do you choose $u$ and $\frac{d v}{d x}$ ?
Take $u$ to be the part that becomes simpler when differentiated, and $\frac{d v}{d x}$ to be a part you can integrate easily.
No rule always works, so if the second integral comes out harder than the first, swap the two choices over and start again.
Why are $e^{x}$ and $\sin x$ awkward choices for $u$ ?
Because they cycle: differentiating them repeatedly never makes them any simpler.
$e^{x}$ returns to itself every time, and $\sin x$ runs through $\cos x$ , $- \sin x$ and $- \cos x$ before coming back.
How do you integrate $\ln x$ , which is not a product at all?
Write it as $1 \times \ln x$ , then take $u = \ln x$ and $\frac{d v}{d x} = 1$ .
That gives $x \ln x - \int x \times \frac{1}{x} d x = x \ln x - x + c$ .
True or False?
Integration by parts can be applied more than once in the same question.
True.
If the second integral is still a product, apply the formula to that as well.
It is rare to need it more than twice, so a third application that still does not finish usually means something went wrong earlier.
When should you integrate using partial fractions?
When the integrand is a fraction whose denominator is degree 2 or more and factorises into linear factors.
Splitting it turns one integral you cannot do into two or three that you can.
Why does integrating partial fractions usually give logarithms?
Because each piece has a linear denominator, which puts it in the $\frac{f' (x)}{f (x)}$ form up to a constant.
Each one therefore integrates to $\ln$ of its own denominator.
Integrate $\frac{5}{(x - 3) (x + 2)}$ .
Split it into $\frac{1}{x - 3} - \frac{1}{x + 2}$ , then integrate each piece separately.
That gives $\ln |x - 3| - \ln |x + 2| + c$ , which tidies to $\ln |\frac{x - 3}{x + 2}| + c$ .
How do you integrate a partial fraction such as $\frac{3}{2 x + 1}$ ?
Adjust for the coefficient of $x$ : the denominator differentiates to $2$ , so the answer is $\frac{3}{2} \ln |2 x + 1| + c$ .
Dropping that factor is the commonest slip once the splitting has been done correctly.
True or False?
Any fraction with a quadratic denominator can be integrated using partial fractions.
False.
The denominator has to factorise into linear factors first.
$\frac{1}{x^{2} + 1}$ does not split at all, and it integrates to an inverse trigonometric function instead, but that is a method beyond this course.
What changes when both boundaries of a region are curves?
Nothing about the method: it is still the integral of upper minus lower, taken between the intersections.
What is lost is the shortcut of using a triangle or trapezium formula for one boundary, since neither of them is a straight line any more.
Two curves meet at three points. How many integrals does the enclosed area need?
Two, one for each enclosed region: from the first intersection to the second, and from the second to the third.
Each region is bounded separately, so each needs its own integral with its own pair of limits.
Why must you check which curve is on top for each region separately?
Because the two curves swap over at every point where they cross.
A curve that was above before an intersection is below after it, so "upper minus lower" means a different subtraction in each region.
Why is a sketch essential here?
Because it is the only reliable way to see how many separate regions there are, and which curve is on top in each of them.
The algebra gives you the intersections; only the picture tells you what to do with them.
True or False?
The total area between two curves is the integral of their difference across the whole interval.
False.
Wherever the curves swap over, the difference changes sign, so one region subtracts from another and the total comes out too small.
Integrate each region separately, take each as a positive area, and then add them.
What is the first thing to check when deciding how to integrate?
Whether it is already a standard integral, or can be turned into one just by rewriting.
Expanding brackets, splitting a fraction or simplifying a quotient often removes the need for any technique at all.
The integrand is a product of two functions. Which methods should you consider?
Reverse chain rule first, if one factor is the derivative of something sitting inside the other.
If it is not, then integration by parts, or a substitution where one factor suggests an obvious $u$ .
The integrand is a fraction. What does its denominator tell you?
A linear denominator points towards a logarithm, and one that factorises points towards partial fractions.
If the numerator is close to the derivative of the denominator, it is the $\frac{f' (x)}{f (x)}$ form.
Why look again for the reverse chain rule after using an identity?
Because rewriting an expression changes its shape, and a reverse chain rule that was not available before may be available now.
This is the easiest thing to miss, since applying the identity feels like the answer rather than a step towards it.
True or False?
If a substitution does not work, the integral cannot be done by substitution at all.
False.
A substitution failing almost always means the wrong $u$ was chosen, not that the method is unavailable.
The usual fix is to substitute a different part of the integrand and try again.

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