“Teach A Level Maths” Vol. 2: A2 Core Modules 22: Integrating the Simple Functions © Christine Crisp

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Before we look again at integration we need to remind ourselves how to differentiate the simple functions. What goes here?

We also need to know that multiplying constants just “tag along” e.g. and that terms like the above can be differentiated independently when they appear in sums and differences. e.g.

Indefinite integration is just the reverse of differentiation, so, reading the differentiation table from right to left, we get: We don’t want to remember the formula with , so we use

Indefinite integration is just the reverse of differentiation, so, reading the differentiation table from right to left, we get: is only defined for x > 0, so we write which means negative signs are ignored. We don’t want the minus sign

SUMMARY Which function is “missing” from the l.h.s. and why?

SUMMARY We can’t yet integrate since we haven’t found a function that differentiates to give .

ANS: We can’t divide by zero. Reminder: To find we write If, by mistake, we do a similar thing with ( forgetting that it gives ), we get . Then, using the 1st rule Why is this impossible? ANS: We can’t divide by zero. We will next practise using the integrals of the simple functions by evaluating some definite integrals and finding some areas.

e.g. 1. Evaluate the following integrals: (b) (c) Solutions: (a) Be careful here . . . Substituting x = 0 does not give 0.

e.g. 1. Evaluate the following integrals: (b) (c) Solutions: (a) The integral gives the shaded area.

In part (a) we needed to remember that What other function are you likely to meet that doesn’t give 0 when x = 0? ANS: since

(b) Radians! The definite integral can give an area, so this result may seem surprising. However, the graph shows us why it is correct. The areas above and below the axis are equal, . . . This part gives a positive integral but the integral for the area below is negative. This part gives a negative integral

(b) Radians! The definite integral can give an area, so this result may seem surprising. However, the graph shows us why it is correct. This part gives a positive integral How would you find the area? Ans: Find the integral from 0 to and double it. This part gives a negative integral

(c) Since the limits are positive, the mod sign makes no difference so we can now omit it.

Exercises Evaluate the following integrals: 1. 2. In each case sketch a graph and briefly explain how your answer relates to area.

Solutions: 1. The areas above and below the axis are equal, but the integral for the area below is negative.

2. N.B. The area is above the axis, so the integral gives the entire area.

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

SUMMARY We can’t yet integrate since we haven’t found a function that differentiates to give .

e.g. 1. Evaluate the following integrals: Solutions: (a) (b) (c) The integral gives the shaded area.

The graph shows us why it is correct. This part gives a negative integral This part gives a positive integral The areas above and below the axis are equal, but the integral for the area below is negative. To find the area, find the integral from 0 to and double it. (b) Radians!

(c) Since the limits are positive, the mod sign makes no difference so we can now omit it. N.B. When working out definite integrals we need to remember that some functions don’t give 0 when x = 0. In particular,