It is the inverse operation to differentiation. If you differentiate a function and then integrate the result, you arrive back at the function you started with, apart from an unknown constant.

What is the difference between an indefinite and a definite integral?

The answer to an indefinite integral is another function , and it carries a constant of integration. The answer to a definite integral is a number , so no constant appears in it.

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Integration (Edexcel IGCSE Further Pure Maths): Flashcards

Exam code: 4PM1

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FrontIntroduction to Integration
What is integration?

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Cards in this collection (20)

What is integration?
It is the inverse operation to differentiation.
If you differentiate a function and then integrate the result, you arrive back at the function you started with, apart from an unknown constant.
What does the notation $\int \dots d x$ mean?
It means the integral, with respect to $x$ , of whatever sits between the integral sign and the $d x$ .
The $d x$ is not decoration: it names the variable you are integrating with respect to.
What is the difference between an indefinite and a definite integral?
The answer to an indefinite integral is another function, and it carries a constant of integration.
The answer to a definite integral is a number, so no constant appears in it.
Given that the derivative of $x^{3} + \frac{1}{x}$ is $3 x^{2} - \frac{1}{x^{2}}$ , write down $\int (3 x^{2} - \frac{1}{x^{2}}) d x$ .
It is $x^{3} + \frac{1}{x} + c$ .
No integrating is needed: the two operations are inverses, so the answer can be read straight off the derivative given.
The constant of integration still has to be written in.
True or False?
Differentiating the answer to $\int f' (x) d x$ gives $f' (x)$ .
True.
The integral comes to $f (x) + c$ , and differentiating that returns $f' (x)$ once more.
This is what makes it possible to check any integration by differentiating the answer.
Complete the rule for integrating a power of $x$ :
$\int x^{n} d x = \frac{x^{______}}{______} + c$
The completed rule is:
$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c$
Raise the power by one and divide by the new power, which is exactly the reverse of differentiating.
For which value of $n$ does the rule for integrating $x^{n}$ fail?
At $n = - 1$ , because raising the power by one gives $0$ and the rule would divide by zero.
So $\int \frac{1}{x} d x$ cannot be found this way, and it is not required on this course.
What is $\int a d x$ ?
It is $a x + c$ .
A constant is $a x^{0}$ , so the same rule applies: raising the power by one turns it into $a x^{1}$ .
How do you integrate $\sqrt{x}$ and $\frac{1}{x^{2}}$ ?
Rewrite each as a power of $x$ first, as $x^{\frac{1}{2}}$ and $x^{- 2}$ .
The rule then gives $\frac{2}{3} x^{\frac{3}{2}} + c$ and $- x^{- 1} + c$ , the second of which is $- \frac{1}{x} + c$ .
Why can $8 x^{2} (2 x - 3)$ not be integrated as it stands?
Because there is no rule for integrating a product, and multiplying the two separate integrals together does not work.
Expand it to $16 x^{3} - 24 x^{2}$ first, and then integrate term by term.
Complete the two trigonometric integrals:
$\int \sin a x d x =______\cos a x + c and \int \cos a x d x =______\sin a x + c$
The completed integrals are:
$\int \sin a x d x = - \frac{1}{a} \cos a x + c and \int \cos a x d x = \frac{1}{a} \sin a x + c$
Both pick up a factor of $\frac{1}{a}$ , and this time the minus sign belongs to $\sin$ , the other way round from differentiating.
Given $f' (x) = 4 \cos 3 x - \frac{1}{2} \sin 2 x$ , find $f (x)$ .
Integrating term by term gives $4 (\frac{1}{3} \sin 3 x) - \frac{1}{2} (- \frac{1}{2} \cos 2 x) + c$ .
That simplifies to $f (x) = \frac{4}{3} \sin 3 x + \frac{1}{4} \cos 2 x + c$ , where the two minus signs have cancelled.
True or False?
Integrating $\sin 2 x$ gives $- \cos 2 x + c$ .
False.
The factor of $\frac{1}{a}$ has been left out, so the correct integral is $- \frac{1}{2} \cos 2 x + c$ .
Differentiating $- \cos 2 x$ gives $2 \sin 2 x$ rather than $\sin 2 x$ , which shows the answer is wrong.
Complete the integral of the exponential function:
$\int e^{a x} d x =______e^{______} + c$
The completed integral is:
$\int e^{a x} d x = \frac{1}{a} e^{a x} + c$
The exponential is unchanged and picks up a factor of $\frac{1}{a}$ , where differentiating would have multiplied by $a$ instead.
Given $f' (x) = \frac{e^{2 x} - e^{- 3 x}}{2}$ , find $f (x)$ .
Split the fraction first, so the integrand is $\frac{1}{2} e^{2 x} - \frac{1}{2} e^{- 3 x}$ .
Integrating gives $\frac{1}{2} (\frac{1}{2} e^{2 x}) - \frac{1}{2} (- \frac{1}{3} e^{- 3 x}) + c = \frac{1}{4} e^{2 x} + \frac{1}{6} e^{- 3 x} + c$ .
Why does an indefinite integral need a constant of integration?
Because the derivative of any constant is zero, so many different functions share one derivative.
Each of $x^{3} - 5 x$ , $x^{3} - 5 x + 7$ and $x^{3} - 5 x - 498$ differentiates to $3 x^{2} - 5$ , and integrating cannot tell you which one you started from.
What do different values of $c$ look like on a graph?
They are vertical translations of the same curve.
The answer to an indefinite integral is therefore a whole family of identically shaped curves, stacked above and below one another.
What extra information do you need to find the value of $c$ ?
The value of the function at one value of $x$ .
That is usually given as a point the graph passes through, but it may equally be given in words, and either way you substitute and solve for $c$ .
The graph of $y = f (x)$ passes through $(3, - 4)$ and $f' (x) = 3 x^{2} - 4 x - 4$ ; find $f (x)$ .
Integrating gives $f (x) = x^{3} - 2 x^{2} - 4 x + c$ .
Substituting the point, $27 - 18 - 12 + c = - 4$ , so $c = - 1$ .
Therefore $f (x) = x^{3} - 2 x^{2} - 4 x - 1$ .
True or False?
You need two points on the curve to find the constant of integration.
False.
One is enough, because integrating has already fixed everything except $c$ , leaving a single unknown.
That is unlike finding the equation of a straight line, where two points are needed to get both the gradient and the intercept.

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