The 2012 STEP Results Discussion Thread

Original post by Dog4444

Actually I tried part i:
I'm not sure, is it sufficient?

Spoiler

You could sketch the graph of (x+1)^(k-1) in the cases when k=-1,0,1 and show by shading the region between 0 and 1 in each case that in the cases when k=-1,1 the area under the curve is approximately the same as in the case of k=0. Hence the areas are approximately equal when k is approximately 0. I doubt they'll require more than a sketch proof.

Reply 1324

Original post by shamika

Apart from your last line, yes. Read the last part of (i) again.

(There isn't a huge amount to this question, and it probably ranks amongst the easiest STEP questions ever.)

Misspelled it.

Spoiler

Is it ok? It sounds too vague for me.

And part ii

Spoiler

Not sure what to do from here, if m is zero than it's obviously 0.5.
EDIT: something wrong with Latex.

(edited 12 years ago)

Reply 1325

Rahul.S

14

Original post by Dog4444

Misspelled it.

Spoiler

Is it ok? It sounds too vague for me.

And part ii

Spoiler

Not sure what to do from here, if m is zero than it's obviously 0.5.
EDIT: something wrong with Latex.

btw whats with the name dog when your avatar is of a bird.......CONTRADICTION DWARG!

Reply 1326

Original post by Dog4444

Misspelled it.

Spoiler

Is it ok? It sounds too vague for me.

And part ii

Spoiler

Not sure what to do from here, if m is zero than it's obviously 0.5.
EDIT: something wrong with Latex.

The second term is partially wrong, because I think you accidentally took (x+1) to be (x+2), it should be 2^(m+2) instead of 3^(m+2) and 1^(m+2)=1 instead of 2^(m+2).

Reply 1327

Farhan.Hanif93

18

Original post by Dog4444

Misspelled it.

Spoiler

Is it ok? It sounds too vague for me.

And part ii

Spoiler

Not sure what to do from here, if m is zero than it's obviously 0.5.
EDIT: something wrong with Latex.

You need to be careful of the cases where the exponent of any linear term is -1.

Parts and substitution is a very hammer-fisted approach. I would observe that

x(x+1)^m = (x+1)^{m+1} - (x+1)^m

.

(edited 12 years ago)

Reply 1328

Original post by Aurum

The second term is partially wrong, because I think you accidentally took (x+1) to be (x+2), it should be 2^(m+2) instead of 3^(m+2) and 1^(m+2)=1 instead of 2^(m+2).

I might be wrong, I do these questions in Paint. :facepalm2:

.
But, how did you integrate

Unparseable latex formula:

[br]\int_0^1 \! \frac{(x+1)^{m+1}}{m+1} \[br]

?
If you substitute u=x+1, your boundaries become 2 and 1. Or I'm wrong?

Original post by Rahul.S

btw whats with the name dog when your avatar is of a bird.......CONTRADICTION DWARG!

Well... you're not cannabis00 as well. :colone:

(edited 12 years ago)

Reply 1329

Original post by Dog4444

I might be wrong, I do these questions in Paint. :facepalm2:

.
But, how did you integrate

Unparseable latex formula:

[br]\int_0^1 \! \frac{(x+1)^{m+1}}{m+1} \[br]

?
If you substitute u=x+1, your boundaries become 2 and 1. Or I'm wrong?

Well... you're not cannabis00 as well. :colone:

When you used the substiution u=x+1 you forgot to replace x+1 by u in the integrand.

(edited 12 years ago)

Reply 1330

Rahul.S

14

Original post by Dog4444

Well... you're not cannabis00 as well. :colone:

might have to make a new account :colone:

Reply 1331

Original post by Dog4444

Misspelled it.

Spoiler

Is it ok? It sounds too vague for me.

And part ii

Spoiler

Not sure what to do from here, if m is zero than it's obviously 0.5.
EDIT: something wrong with Latex.

There's one more error, in your first term it should be just 2^(m+1) not 2^(m+1)-1, since when you plug 0 (the lower limit) x[(x+1)^(m+1)]/(m+1) becomes 0.

Also you can use parts if you want, but like Farhan said you need to realise that it won't be integrated in the same way when m=-1, since with 1/(x+1) the standard way of integrating (ax+b)^m doesn't work when m=-1.

(edited 12 years ago)

Reply 1332

shamika

16

Original post by Farhan.Hanif93

You need to be careful of the cases where the exponent of any linear term is -1.

Parts and substitution is a very hammer-fisted approach. I would observe that

x(x+1)^m = (x+1)^{m+1} - (x+1)^m

.

That's the easiest way sure, but I bet in exam conditions most people go with an integration by parts (or a substitution), simply because they require less 'thinking'.

@Aurum: you'll need to be careful if m=-2 as well.

(edited 12 years ago)

Reply 1333

Original post by Aurum

When you used the substiution u=x+1 you forgot to replace x+1 by u in the integrand.

[br]\int_0^1 \! \frac{(x+1)^{m+1}}{m+1} = \frac{1}{m+1} . \int_1^2 \! {u^{m+1}}

And you get:

[br]\displaystyle[br]\frac{(x+1)^{m+2}} {(m+2)(m+1)}[br]

And what's wrong with Latex?

Reply 1334

Original post by shamika

That's the easiest way sure, but I bet in exam conditions most people go with an integration by parts (or a substitution), simply because they require less 'thinking'.

@Aurum: you'll need to be careful if m=-2 as well.

Thanks for pointing that out, I overlooked the second integral after integrating by parts.

Reply 1335

Original post by Dog4444

[br]\int_0^1 \! \frac{(x+1)^{m+1}}{m+1} = \frac{1}{m+1} . \int_1^2 \! {u^{m+1}}

And you get:

[br]\displaystyle[br]\frac{(x+1)^{m+2}} {(m+2)(m+1)}[br]

And what's wrong with Latex?

If you want to change u back to x+1 after you've integrated, you'll have to change the limits too. You can use 1 and 2 with u or you can use 0 and 1 with x+1.

Reply 1336

Original post by Farhan.Hanif93

You need to be careful of the cases where the exponent of any linear term is -1.

Parts and substitution is a very hammer-fisted approach. I would observe that

x(x+1)^m = (x+1)^{m+1} - (x+1)^m

.

Are we able to integrate it, but not to consider m=-1, or we can't integrate it like that at all, because the denominator might be zero?

Reply 1337