Edexcel FP3 - 27th June, 2016

Original post by A Slice of Pi

Given that

Unparseable latex formula:

\displaystyle I_n = \int_{0}^{2}x^n \sqrt{2x-x^2}\hspace{4pt}\mathrm{d}x, \hspace{15} n\geqslant0

(a) find

I_0

(b) Show that

\displaystyle I_n = \frac{2n+1}{n+2}I_{n-1}

I_n = \displaystyle\frac{(2n+1)! \pi}{n!(n+2)!2^n}

a) Write

\int_0^2 \sqrt{1-(x-1)^2} \, \mathrm{d}x

, then shift it over to get

I_0 = \int_{-1}^{1} \sqrt{1-x^2} \, \mathrm{d}x

where a straightforward

u=\sin x

sub sorts it out, gives

\frac{\pi}{2}

.

b) Write

x^n\sqrt{2x-x^2} = x^{n + 1/2} \sqrt{2-x}

and use IBP with

u = x^{n + 1/2}

and

\mathrm{d}v = \sqrt{2-x}

. The

uv

portion clearly disappears and we are left with:

Unparseable latex formula:

\displaystyle [br]\begin{align*}I_n &= \frac{(2n+1)}{3} \int_0^2 x^{n-1} \sqrt{2x-x^2} (2-x) \, \mathrm{d}x \\ \\ & \iff = 3I_n = 2(2n+1)I_{n-1} - (2n+1)I_n \\ \\ & \iff 2(n+2)I_n = 2(2n+1)I_{n-1} \\ \\ & \iff I_n = \frac{2n+1}{n+2}I_{n-1}\end{align*}

as required.

c) It follows straightforwardly that we end up with

I_n = \frac{2(2n+1)!!}{(n+2)!} I_0 = \frac{(2n+1)!}{2^{n-1}n!(n+2)!} \times \frac{\pi}{2} = \frac{(2n+1)! \pi}{n! (n+2)! 2^n}

using the well known double factorial identity.

Reply 362

Anon-

FP3 June 2009

this paper, question 7 c). shortest distance between two skew vectors. mark scheme says you have to use a formula (link), but i've never seen this formula and i can't see it in the formula booklet?

could someone please explain how you would find the solution?

Reply 363

Original post by A Slice of Pi

How did this turn out for you? Did you get the result?

Couldn't figure out a substitution for a, so I tried to complete the square but it hasn't worked as that :s-smilie:

. Guessing there's a sneaky little thing which I cannot be arsed waiting for myself to figure out. I moved onto b at this point.

As for b, I tried to do the method as I suggested last night, but this fails to work as I get

\int{\frac{x}{2-2x}} . \sqrt{u} du

for v. I tried to take a

x^\frac12

outside of the square root which surprisingly lead me quite far, but at the end I was stuck with a (2x-x^2) which I wouldn't be able to multiply out, as it would lead to

I_{n+1}

stuff.

Reply 364

Original post by Craig1998

Couldn't figure out a substitution for a, so I tried to complete the square but it hasn't worked as that :s-smilie:

\int{\frac{x}{2-2x}} . \sqrt{u} du

for v. I tried to take a

x^\frac12

outside of the square root which surprisingly lead me quite far, but at the end I was stuck with a (2x-x^2) which I wouldn't be able to multiply out, as it would lead to

I_{n+1}

stuff.

See two posts above for my solution.

Reply 365

Original post by Zacken

See two posts above for my solution.

Dammit. Forgot the power was 3/2 when I turned my piece of paper over. Thank you for figuring out the stupidness of part a which I'm going to be screwed over by if they ask something like it in the exam.

Reply 366

@A Slice of Pi where did you get the reduction integration from btw? Would those sorts of ones appear in FP3 (even as a really hard question).

Reply 367

Original post by Craig1998

@A Slice of Pi where did you get the reduction integration from btw? Would those sorts of ones appear in FP3 (even as a really hard question).

It seemed like a standard FP3 question to me. (and I think he made it up himself).

Reply 368

A Slice of Pi

Original post by Craig1998

@A Slice of Pi where did you get the reduction integration from btw? Would those sorts of ones appear in FP3 (even as a really hard question).

I based the part b question on a reduction formula I came across a few months ago, then put it into the 'Edexcel style'. It's one I could remember because the splitting of the integral wasn't obvious. I can't remember the source of the original question, which had a slightly different integral but the principle was the same. I'm not 100% sure whether something like this will come up. I'm with AQA for A-level so I don't know what level of difficulty the Edexcel FP3 papers go to.

Reply 369

Original post by Zacken

It seemed like a standard FP3 question to me. (and I think he made it up himself).

Most of the ones I've seen tend to be either:
Generic easy ones like

I_n = \int_\beta^\alpha x^n e^x dx

Trig ones like

I_n = \int_\beta^\alpha \cos^n x dx

Weird ones where the thought comes after the ibp

\int_\beta^\alpha x^n \sqrt{2-x^2} dx

- this was similar to the other one, which is why my first thought was to split it into

x . x^{n-1}

then do a substitution.

They seem to stop after that and don't really have ones that require a lot of thought before the ibp. Not in the past papers I've seen anyway.

The only other one's I have seen are one's that are like

I_{m,n}

, but weve never been taught them so I assume they are not on the spec.

Reply 370

Original post by Zacken

a) Write

\int_0^2 \sqrt{1-(x-1)^2} \, \mathrm{d}x

, then shift it over to get

I_0 = \int_{-1}^{1} \sqrt{1-x^2} \, \mathrm{d}x

where a straightforward

u=\sin x

sub sorts it out, gives

\frac{\pi}{2}

.

b) Write

x^n\sqrt{2x-x^2} = x^{n + 1/2} \sqrt{2-x}

and use IBP with

u = x^{n + 1/2}

and

\mathrm{d}v = \sqrt{2-x}

. The

uv

portion clearly disappears and we are left with:

Unparseable latex formula:

as required.

c) It follows straightforwardly that we end up with

I_n = \frac{2(2n+1)!!}{(n+2)!} I_0 = \frac{(2n+1)!}{2^{n-1}n!(n+2)!} \times \frac{\pi}{2} = \frac{(2n+1)! \pi}{n! (n+2)! 2^n}

using the well known double factorial identity.

My method was about 7 steps longer even though it is fundamentally just like yours :biggrin:

Wish I knew those tricks like shifting the limits and the double factorial thing.

(edited 7 years ago)

Reply 371

Original post by oShahpo

My method was about 7 steps longer even though it is fundamentally just like yours :biggrin:

Wish I knew those tricks like shifting the limits and the double factorial thing.

If it helps, "shifting the limits" is just me sneakily using the sub

x \mapsto x-1

u=x-1

in my head, nothing fancy there. :tongue:

And then, the double factorial thing isn't much work at all, it's just:

\displaystyle (2n+1)!! = (2n+1)(2n-1)\cdots (1) = \frac{(2n+1)!}{(2n)(2n-2)(2n-4)\cdots(2)} = \frac{(2n+1)!}{2^n n!}

So it's not as impressive as you think. :tongue:

Reply 372

Original post by Zacken

If it helps, "shifting the limits" is just me sneakily using the sub

x \mapsto x-1

u=x-1

in my head, nothing fancy there. :tongue:

And then, the double factorial thing isn't much work at all, it's just:

\displaystyle (2n+1)!! = (2n+1)(2n-1)\cdots (1) = \frac{(2n+1)!}{(2n)(2n-2)(2n-4)\cdots(2)} = \frac{(2n+1)!}{2^n n!}

So it's not as impressive as you think. :tongue:

What I did is that I multiplied both numerator and denominator by 2^n (n)!, it's just the double factorial thing looked more elegant :biggrin:

Where does the factor of 2 in the numerator come from btw?

(edited 7 years ago)

Reply 373

Original post by Zacken

a) Write

\int_0^2 \sqrt{1-(x-1)^2} \, \mathrm{d}x

, then shift it over to get

I_0 = \int_{-1}^{1} \sqrt{1-x^2} \, \mathrm{d}x

where a straightforward

u=\sin x

sub sorts it out, gives

\frac{\pi}{2}

.

b) Write

x^n\sqrt{2x-x^2} = x^{n + 1/2} \sqrt{2-x}

and use IBP with

u = x^{n + 1/2}

and

\mathrm{d}v = \sqrt{2-x}

. The

uv

portion clearly disappears and we are left with:

Unparseable latex formula:

as required.

c) It follows straightforwardly that we end up with

I_n = \frac{2(2n+1)!!}{(n+2)!} I_0 = \frac{(2n+1)!}{2^{n-1}n!(n+2)!} \times \frac{\pi}{2} = \frac{(2n+1)! \pi}{n! (n+2)! 2^n}

using the well known double factorial identity.

Hey Zack, where does the factor of 2 come from in C?

Reply 374

username1162972

Original post by A Slice of Pi

What method have you chosen for (b)?

One second

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Reply 375

Original post by oShahpo

Hey Zack, where does the factor of 2 come from in C?

We require

n \geq 1

so the

(n+2)!

in the denominator doesn't contain the

(0+2) = 2

terms. But I wrote it as a factorial and multiplied the numerator by 2 to account for the extra insertion of 2 in the denominator from the factorial.

i.e: the denominator should really just be

(n+2)(n+1) \cdots (3)

but I write

(n+2)! = (n+2)(n+1) \cdots (2)(1)

so had to multiply the numerator by 2 to compensate for that.

Reply 376

username1162972

Original post by A Slice of Pi

What method have you chosen for (b)?

ImageUploadedByStudent Room1464821712.246474.jpg

My way for this nice reduction question.

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Reply 377