Please agree to accept our cookies. If you continue to use the site, we'll assume you're happy to accept them.

# Secondary Magazine - Issue 122: Sixth Sense

Created on 29 May 2015 by ncetm_administrator
Updated on 23 June 2015 by ncetm_administrator

# Sixth Sense

In my last article, I touched on the strategy of guess and check to integrate functions:

Question: What did I differentiate to get $\dpi{80} \fn_jvn \small 16 \times \left (x ^{2}-5 \right )^{3}$?

Guess: $\dpi{80} \fn_jvn \small \left (x ^{2}-5 \right )^{4}$

Check: The derivative of this is $\dpi{80} \fn_jvn \small 4 \left (x ^{2}-5 \right )^{3}\times 2x$

Thought: So I need to multiply this guess by 2 to get the answer I’m looking for

Therefore: $\dpi{80} \fn_jvn \small \int 16x \left (x ^{2}-5 \right )^{3}dx=2\left (x ^{2}-5 \right )^{4}+c$

Developing this thought process as early as possible in your students is well worth doing, because they can return to it time and again during their A level studies. One can start immediately after teaching differentiation:

Question 1: What did I differentiate to get $\dpi{80} \fn_jvn \small x^{3}$

Guess: The power of 3 makes me think of $\dpi{80} \fn_jvn \small x^{4}$

Check: When I differentiate $\dpi{80} \fn_jvn \small x^{4}$, I get $\dpi{80} \fn_jvn \small 4x^{3}$

Thought: This is 4 multiplied by the answer I was hoping for

New guess: I’ll divide my first guess by 4 , to get $\dpi{80} \fn_jvn \small \frac{x^4{}}{4}$

Check: The derivative of this is $\dpi{80} \fn_jvn \small \frac{1}{4}\times 4x^{3}$

Therefore: The answer to the question is $\dpi{80} \fn_jvn \small \frac{x^4{}}{4}$

[with constants still to be discussed!]

Question 2: What did I differentiate to get $\dpi{80} \fn_jvn \small 5x^{3}$?

Guess: Given what I’ve just done, I’ll try $\dpi{80} \fn_jvn \small \frac{5x^{4}}{4}$

Check: If I differentiate, I get $\dpi{80} \fn_jvn \small \frac{5}{4}\times 4x^{3}$

Therefore: Right first time!

Having practised lots of AS-level problems of the form $\dpi{80} \fn_jvn \small \int ax^{b}dx$, your students are now ready for the types they’ll meet at A2, such as the one at the top of the article. In time, most of the method should take place in the students’ heads – they need to develop procedural fluency, and not have to write a short script every time they integrate - but they must always be able to vocalise the process when asked, so that you can check that they are developing conceptual understanding.

Question: $\dpi{80} \fn_jvn \small \int \left ( x-1 \right )\left ( x^{2}-2x+7 \right )^{4}dx=$?

Guess: I see a power of 4, so I’ll try $\dpi{80} \fn_jvn \small \left ( x^{2}-2x+7 \right )^{5}$

Check: This differentiates to $\dpi{80} \fn_jvn \small 5\left ( x^{2}-2x+7 \right )^{4}\times \left ( 2x-2 \right )$

Thought: Ah-ha: if I factorise the second bracket, this is 10 times what I want, so I need to divide my guess by 10.

Therefore: $\dpi{80} \fn_jvn \small \int \left ( x-1 \right )\left ( x^{2}-2x+7 \right )^{4}dx = \frac{\left ( x^{2}-2x+7 \right )^{5}}{10}+c$

Now, once the existence and significance of the function $\dpi{80} \fn_jvn \small e^{x}$ have been discussed, and your students are happy and confident with the result (as explored last time) that $\dpi{80} \fn_jvn \small \frac{d}{dx}\left ( e^{function} \right )=e^{function}\times functiondifferentiated$, then they can approach integrals such as $\dpi{80} \fn_jvn \small \int 2e^{7x}dx$ and $\dpi{80} \fn_jvn \small \int 5xe^{4x^{2}-3}dx$ using the guess and check structure :

Guess: $\dpi{80} \fn_jvn \small e^{7x}$

Check: This differentiates to $\dpi{80} \fn_jvn \small e^{7x}\times 7$, so I need to divide by 7 and multiply by 2.

Therefore: $\dpi{80} \fn_jvn \small \int 2e^7x{}dx=\frac{2e^{7x}}{7}+c$

Guess: $\dpi{80} \fn_jvn \small e^{4x^{2}-3}$

Check: This differentiates to $\dpi{80} \fn_jvn \small e^{4x^{2}-3}\times 8x$, so the guess is 8 times too big.

Therefore: $\dpi{80} \fn_jvn \small \int 5xe^{4x^{2}-3}dx=\frac{5e^{4x^{2}-3}}{8}+c$

It is, of course, important to see that this method has its limitations – in the same way that the “rule” for differentiating $\dpi{80} \fn_jvn \small \left ( function \right )^{6}$ does not apply to $\dpi{80} \fn_jvn \small x^{2}\left ( 4x-1 \right )^{6}$, the guess and check procedure will not work easily for certain integrals:

Question: $\dpi{80} \fn_jvn \small \int x^{2}\left (x ^{2}-8 \right )^{7}dx=$?

Guess: I see a power of 7, so I’ll try $\dpi{80} \fn_jvn \small \left (x ^{2}-8 \right )^{8}$

Check: The derivative of this is $\dpi{80} \fn_jvn \small 8\left (x ^{2}-8 \right )^{7}\times \left ( 2x \right )$

Thought: I need to divide by 16 and multiply by $\dpi{80} \fn_jvn \small x$

New guess: $\dpi{80} \fn_jvn \small \frac{x\left (x^{2}-8 \right )^{8}}{16}$

Check: Hang on a minute, I can’t differentiate this without expanding, or using the product rule…

Therefore: … I’m stuck

Teacher: For the time being …

However, having seen further examples of the chain rule, students can successfully guess and check increasingly complicated integrals:

$\dpi{80} \fn_jvn \small \int 3\sin 8xdx=$?

$\dpi{80} \fn_jvn \small \int 4\sin 2x\cos ^3{}2xdx=$?

(and it’s interesting to ask them to compare this with $\dpi{80} \fn_jvn \small \int \cos ^3{}2xdx$)

Now for the payoff! Every time I teach this next bit I am struck by how quickly most students see what’s happening – they pick up the idea much faster than I did when I was at school and was taught from the textbook “the formal method”.

So, guessing and checking for grown-ups, as my first Head of Department called it (and I’m sure he took it from his previous Head of Department!), works like this (and happens in the scheme after the product rule has been explored and mastered!):

Question: $\dpi{80} \fn_jvn \small \int x\sin 3xdx=$?

Guess: I’ll try $\dpi{80} \fn_jvn \small x\cos 3x$ and see what happens

Check: The derivative of this is $\dpi{80} \fn_jvn \small -3x\sin 3x+\cos 3x$

Thought: Hmmm. Tricky. One step at a time. Let's sort the first term out.

New guess: $\dpi{80} \fn_jvn \small -\frac{x\cos 3x}{3}$

Check: This differentiates to $\dpi{80} \fn_jvn \small x\sin 3x-\frac{\cos 3x}{3}$

Thought: Now I need to include in my guess an extra term which, when differentiated, cancels the second term, which I don’t want

Guess again: $\dpi{80} \fn_jvn \small -\frac{x\cos 3x}{3}+\frac{\sin3x }{9}$

Check: This now differentiates to $\dpi{80} \fn_jvn \small x\sin 3x-\frac{\cos 3x}{3}+\frac{\cos3x }{3}$

Therefore: $\dpi{80} \fn_jvn \small \int x\sin 3xdx=-\frac{x\cos 3x}{3}+\frac{\sin3x }{3}+c$

Getting the first guess may take a while, and it is important to praise, and explore, the alternative suggestion:

Question:$\dpi{80} \fn_jvn \small \int x\sin 3xdx=$?

Guess: I'll try $\dpi{80} \fn_jvn \small \frac{x^{2}\sin3x}{2}$ and see what happens

Check: $\dpi{80} \fn_jvn \small x\sin3x+\frac{3x^{2}\cos3x}{2}$

Thought: Now I need to subtract something but I have no idea what, because $\dpi{80} \fn_jvn \small \int \frac{3x^{2}\cos3x}{2}dx$ is a harder question than the one I started with...

Therefore: Back to the drawing board!

By the time the students have tried two or three of these they should be able to conclude that, with this type of question, there are two “natural” guesses, but one ends up with a second integral that is easier than the original, whereas the other produces a second integral that is harder than the original: clearly the skill is in picking the better first guess.

Once they’re feeling confident, do get them to try$\dpi{80} \fn_jvn \small \int x^{2}e^{4x}dx$, and $\dpi{80} \fn_jvn \small \int \ln xdx$ is an important example to cover at this point.

By the second or third lesson on this method, having also looked at definite integrals, students need to generalise:

Question: $\dpi{80} \fn_jvn \small \int {uv}'dx$

Guess: $\dpi{80} \fn_jvn \small uv$

Check: This has derivative $\dpi{80} \fn_jvn \small {u}'v+u{v}'$

Thought: This is what I want, with an extra term, so I need to subtract from the guess a term which, when differentiated, gives $\dpi{80} \fn_jvn \small {u}'v$

New guess: $\dpi{80} \fn_jvn \small {u}'v-\int {u}'vdx$

Check: $\dpi{80} \fn_jvn \small {u}'v+{uv}'-{u}'v$

Therefore: $\dpi{80} \fn_jvn \small \int {uv}'dx=uv-\int {u}'vdx+c$

Hence, informal guessing and checking for grown-ups becomes traditional integration by parts, but specialising before generalising helps enormously. Repeatedly applying an “out of the blue” formula will probably give students procedural fluency, but their conceptual understanding is very unlikely to develop too. I’m reasonably sure that the so-called explanation which my sixth form self was given started with the last line of the above argument, and it took me a long while to understand what on earth was happening!

Image credit
Page header by takomabibelot (adapted), some rights reserved

 Add to your NCETM favourites Remove from your NCETM favourites Add a note on this item Recommend to a friend Comment on this item Send to printer Request a reminder of this item Cancel a reminder of this item

18 June 2015 11:34
Hi RobertWilne. I agree and believe that this approach would lead to more successful exam outcomes. The thought process of checking using inverses is also applicable to other topic areas.