HSC 2015 MX2 Integration Marathon (archive) (2 Viewers)

VBN2470 · Apr 12, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Find $\int_{1}^{\infty} \frac{\text{ d}x}{x{\sqrt{x^2+2x-1}}}$$

EASIER METHOD:

$Use the substitution $u = \frac{1}{x}$ to get $\int_{0}^{1} \frac{\text{ d}u}{\sqrt{2-(u-1)^2}}$ which becomes easier to evaluate.$

Drsoccerball · Apr 12, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
$so$
$\int{a}^{b}$
$like$
$\frac{c}{d}$
$this$
Thanks!

Like this -.-:

Ekman said:
Next Question:

$\int_{0}^{\frac{\pi}{4}} \frac{\sec(x)}{2+\sin^2(x)} dx$

$$ Let u = $ sinx$

$du= cosx dx$

$\int{\frac{du}{(1-u^2)(2+u^2)}}$

$use partial fractions from here on...$

Look at my Latex and notice i started a new tex eveytime i needed a new line and how i left a gap between them aswell..

Drsoccerball · Apr 12, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
EASIER METHOD:

$Use the substitution $u = \frac{1}{x}$ to get $\int_{0}^{1} \frac{\text{ d}u}{\sqrt{2-(u-1)^2}}$ which becomes easier to evaluate.$

I tried this substitution got halfway and said "ceebs"

Drsoccerball · Apr 12, 2015

Re: MX2 2015 Integration Marathon

$I_n = \int_0^{\frac{\pi}{4}}{\frac{1-cos(2nx)}{sin(2x)}}$

$i) $ Show that for r $ > $ 0$

$a) I_{2r+1} -I_{2r-1}=\frac{1-(-1)^r}{2r}$

$b) I_{2r}-I_{2r-2}= \frac{1}{2r-1}$

Ekman · Apr 12, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$I_n = \int_0^{\frac{\pi}{4}}{\frac{1-cos(2nx)}{sin(2x)}}$

$i) $ Show that for r $ > $ 0$

$a) I_{2r+1} -I_{2r-1}=\frac{1-(-1)^r}{2r}$

$b) I_{2r}-I_{2r-2}= \frac{1}{2r-1}$

$a) I_{2r+1} -I_{2r-1}=\int_0^{\frac{\pi}{4}}{\frac{1-cos(4rx+2x)}{sin(2x)}} - \int_0^{\frac{\pi}{4}}{\frac{1-cos(4rx-2x)}{sin(2x)}}$

$\int_0^{\frac{\pi}{4}}{\frac{cos(4rx-2x)-cos(4rx+2x)}{sin(2x)}} = \int_0^{\frac{\pi}{4}}{\frac{2sin(4rx)sin(2x)}{sin(2x)}}$

$= \left[-\frac{cos(4rx)}{2r} \right]_{0}^{\frac{\pi}{4}} = \frac{1-(-1)^r}{2r}$

$b) I_n = \int_0^{\frac{\pi}{4}}{\frac{1-cos(2nx)}{sin(2x)}} = \int_0^{\frac{\pi}{4}}{\frac{sin^{2}(nx)}{sin(x)cos(x)}}$

$I_{2r}-I_{2r-2}=\int_0^{\frac{\pi}{4}}{\frac{sin^{2}(2rx)}{sin(x)cos(x)}} - {\frac{sin^{2}((2r-2)x)}{sin(x)cos(x)}}$

$=\int_0^{\frac{\pi}{4}}{\frac{(sin(2rx)-sin((2r-2)x))\times(sin(2rx)+sin((2r-2)x)) }{sin(x)cos(x)}}$

$=\int_0^{\frac{\pi}{4}} \frac{4sin(2rx-x)cos(x)\times \cos(2rx-x)sin(x)}{sin(x)cos(x)} = \int_0^{\frac{\pi}{4}} 2\sin(2x(2r-1))$

$=\left[-\frac{\cos(2x(2r-1))}{2r-1} \right]_{0}^{\frac{\pi}{4}} = \frac{1}{2r-1}$

Drsoccerball · Apr 13, 2015

Re: MX2 2015 Integration Marathon

NEW QUESTION :

$\int{sin2x cos3x}dx$

HINT: Dont expand out solution

swagswagyoloswag · Apr 13, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
NEW QUESTION :

$\int{sin2x cos3x}dx$

HINT: Dont expand out solution

$\int \frac{1}{2}[sin(5x) - sin(x)]dx$

$\frac{1}{2} [cos(x) - \frac{1}{5}cos(5x)] + C$

braintic · Apr 14, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
$\int \frac{1}{2}[sin(5x) - sin(x)]dx$

$\frac{1}{2} [cos(x) - \frac{1}{5}cos(5x)] + C$

I just want to make sure everyone realises that in an exam you can't just quote the formula you have used.
The result would have to be derived.

swagswagyoloswag · Apr 14, 2015

Re: MX2 2015 Integration Marathon

braintic said:
I just want to make sure everyone realises that in an exam you can't just quote the formula you have used.
The result would have to be derived.

Yep. For this one,
sin(a+b) = sinacosb + sinbcosa
sin(a-b) = sinacosb - sinbcosa
Subtracting, sin(a+b) - sin(a-b) = 2sinbcosa
Sub b = 2x and a = 3x

swagswagyoloswag · Apr 14, 2015

Re: MX2 2015 Integration Marathon

$$Explain why, for \alpha >0$ , \int_{0}^{1} x^{\alpha}e^{x}dx < \frac{3}{\alpha +1}$

InteGrand · Apr 14, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
$$Explain why, for \alpha >0$ , \int_{0}^{1} x^{\alpha}e^{x}dx < \frac{3}{\alpha +1}$

Because e^x < 3 for all x in that domain of integration.

VBN2470 · Apr 14, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
$$Explain why, for \alpha >0$ , \int_{0}^{1} x^{\alpha}e^{x}dx < \frac{3}{\alpha +1}$

If I remember correctly, this q was also asked in the 2001 MX2 HSC Paper Q8 ...

Drsoccerball · Apr 14, 2015

Re: MX2 2015 Integration Marathon

$\int_0^1{\frac{dx}{(1+\sqrt{x}) \sqrt{x -x^2}}}dx$

swagswagyoloswag · Apr 15, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
If I remember correctly, this q was also asked in the 2001 MX2 HSC Paper Q8 ...

Asked this because i didn't know it (lol just one reasoning) and it was on a worksheet i had

FrankXie · Apr 15, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int_0^1{\frac{dx}{(1+\sqrt{x}) \sqrt{x -x^2}}}dx$

substitution $x=\sin^2\theta$

swagswagyoloswag · Apr 15, 2015

Re: MX2 2015 Integration Marathon

$$ Let $ I_n = \int_{0}^{1} (1-x^{2})^{n} dx$

$J_n = \int_{0}^{1} x^{2}(1-x^{2})^{n}dx$

$$ a) Show $ I_n = 2nJ_{n-1}$

$b) Hence show that I_n = \frac{2n}{2n+1}I_{n-1}$

$c) $ Show that $ J_n = I_n - I_{n+1} $ and hence deduce that $ J_n = \frac{1}{2n+3}I_n$

$d) $ Write down a reduction formulae $ for J_n $ in terms of $ J_{n-1}$

Drsoccerball · Apr 16, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
$$ Let $ I_n = \int_{0}^{1} (1-x^{2})^{n} dx$

$J_n = \int_{0}^{1} x^{2}(1-x^{2})^{n}dx$

$$ a) Show $ I_n = 2nJ_{n-1}$

$b) Hence show that I_n = \frac{2n}{2n+1}I_{n-1}$

$c) $ Show that $ J_n = I_n - I_{n+1} $ and hence deduce that $ J_n = \frac{1}{2n+3}I_n$

$d) $ Write down a reduction formulae $ for J_n $ in terms of $ J_{n-1}$

This seems as a combination of all recurrence questions posted on the forum

leehuan · Apr 16, 2015

Re: MX2 2015 Integration Marathon

I've seen this question before in a past paper I think...

$a)\quad { I }_{ n }\quad =\quad \int _{ 0 }^{ 1 }{ { \left( 1-{ x }^{ 2 } \right) }^{ n }dx\quad } \\ ={ \left[ { x\left( 1-{ x }^{ 2 } \right) }^{ n } \right] }_{ 0 }^{ 1 }-\int _{ 0 }^{ 1 }{ { x\cdot \left( -2x \right) n\left( 1-{ x }^{ 2 } \right) }^{ n-1 }dx\quad } \\ =\left( { 1\left( 1-{ 1 }^{ 2 } \right) }^{ n }-{ 0\left( 1-0^{ 2 } \right) }^{ n } \right) \quad +\quad 2n\int _{ 0 }^{ 1 }{ { x }^{ 2 } } { \left( 1-{ x }^{ 2 } \right) }^{ n-1 }dx\\ =2n{ J }_{ n-1 }$

$b)\quad However:\\ 2n\int _{ 0 }^{ 1 }{ { x }^{ 2 } } { \left( 1-{ x }^{ 2 } \right) }^{ n-1 }dx\\ =-2n\int _{ 0 }^{ 1 }{ -{ x }^{ 2 } } { \left( 1-{ x }^{ 2 } \right) }^{ n-1 }dx\\ =-2n\left( \int _{ 0 }^{ 1 }{ \left( 1-{ x }^{ 2 } \right) } { \left( 1-{ x }^{ 2 } \right) }^{ n-1 }dx\quad -\quad \int _{ 0 }^{ 1 }{ 1 } { \left( 1-{ x }^{ 2 } \right) }^{ n-1 }dx \right) \\ =-2n\int _{ 0 }^{ 1 }{ { \left( 1-{ x }^{ 2 } \right) }^{ n } } dx\quad +\quad 2n\int _{ 0 }^{ 1 }{ { \left( 1-{ x }^{ 2 } \right) }^{ n-1 } } dx$
$\therefore { I }_{ n }=-2n{ I }_{ n }+2n{ I }_{ n-1 }\\ (2n+1){ I }_{ n }=2n{ I }_{ n-1 }\\ { I }_{ n }=\frac { 2n }{ 2n+1 } { I }_{ n-1 }$

$c)\quad { J }_{ n }=\int _{ 0 }^{ 1 }{ { x }^{ 2 }{ \left( 1-{ x }^{ 2 } \right) }^{ n }dx } \\ =-\int _{ 0 }^{ 1 }{ { \left( 1-{ x }^{ 2 } \right) }{ \left( 1-{ x }^{ 2 } \right) }^{ n }dx } +\int _{ 0 }^{ 1 }{ { 1 }{ \left( 1-{ x }^{ 2 } \right) }^{ n }dx } \\ =\int _{ 0 }^{ 1 }{ { \left( 1-{ x }^{ 2 } \right) }^{ n }dx } -\int _{ 0 }^{ 1 }{ { \left( 1-{ x }^{ 2 } \right) }^{ n+1 }dx } \\ ={ I }_{ n }-{ I }_{ n+1 }$

$From\quad b)\quad { I }_{ n+1 }=\frac { 2(n+1) }{ 2(n+1)+1 } { I }_{ n }\\ =\frac { 2n+2 }{ 2n+3 } { I }_{ n }\\ \therefore { J }_{ n }=\frac { 2n+3 }{ 2n+3 } { I }_{ n }-\frac { 2n+2 }{ 2n+3 } { I }_{ n }\\ =\frac { 1 }{ 2n+3 } { I }_{ n }$

$d){ J }_{ n }=\frac { 2n }{ 2n+3 } { J }_{ n-1 }\quad \quad using\quad a)$

Ekman · Apr 16, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
This seems as a combination of all recurrence questions posted on the forum

It looks like the first bit of the 2010 q8 integration question...

Sy123 · Apr 16, 2015

Re: MX2 2015 Integration Marathon

$\\ $i) Explain why$ \ \int_0^a f(x) \ dx = \int_0^a f(a-x) \ dx \\ \\ $ii) Hence find$ \ \int_0^{\pi/4} \ln (1 + \tan x) \ dx$

HSC 2015 MX2 Integration Marathon (archive) (2 Viewers)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Member

Member

Well-Known Member

Well-Known Member

Well-Known Member

Member

Active Member

Member

Well-Known Member

Well-Known Member

Well-Known Member

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 2)