HSC 2016 MX2 Integration Marathon (archive) (3 Viewers)

InteGrand · Apr 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
I don't want to be nitpicky, but isn't the technical term "cusp"?

I put the word corners in inverted commas because I was using it informally. I think for a cusp, the tangents should be equal for the two meeting branches (http://mathworld.wolfram.com/Cusp.html) (which is not the case for this function's graph).

leehuan · Apr 9, 2016

Re: MX2 2016 Integration Marathon

Oh right. I keep missing that 1-sin(2x) = perfect square result. But yeah the sign function I understand.

And lol I know what a cusp is. All good

leehuan · Apr 9, 2016

Re: MX2 2016 Integration Marathon

$\int{{x}^{2015}{e}^{{x}^{1008}}\,$d$x}$

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\sin { x } \cos ^{ 2 }{ x }\,\text{d}x }$

InteGrand · Apr 10, 2016

Re: MX2 2016 Integration Marathon

Additional Q:

$\noindent Find $\int \sqrt{\frac{6-x}{x-3}}\text{ d}x$.$

Paradoxica · Apr 11, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
Additional Q:

$\noindent Find $\int \sqrt{\frac{6-x}{x-3}}\text{ d}x$.$

I'm thinking of using the elliptic substitution...

$x = 6\cos^2{\theta}+3\sin^2{\theta}$

Paradoxica · Apr 11, 2016

Re: MX2 2016 Integration Marathon

$$\noindent a) Simplify $\text{sgn}(x) |x| \\ $b) $ \text{sgn}(\cos{x}-\cos{y}) \int \sqrt{(1-\cos{(x-y)})(1-\cos{(x+y)})}\,$d$x$

Drsoccerball · Apr 11, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
Additional Q:

$\noindent Find $\int \sqrt{\frac{6-x}{x-3}}\text{ d}x$.$

$u=x-\frac{9}{2}$

KingOfActing · Apr 12, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent Given $n$ is a non-negative integer, find the recurrence relation for $\\ I_n = \int_{-\infty}^\infty \frac{\text{d}x}{(x^2+1)^{n+1}} \\ $Hence, prove the following statement using induction:$ \\ \int_{-\infty}^\infty \frac{\text{d}x}{(x^2+1)^{n+1}} = \frac{(2n)!\,\pi}{2^{2n}(n!)^2}$

$$By first making the substitution $ x = \tan{\theta} $ and shifting index to $n+1$ we obtain $ I_{n+1} = 2\int_0^{\frac{\pi}{2}}\cos^{2n+2} \theta\, d \theta\\$Now, through one application of integration by parts: $\\ I_{n+1} = (2n+1)I_n - (2n+1)I_{n+1} \Longleftrightarrow I_{n+1} = \frac{2n+1}{2n+2}I_n\\$To prove the closed form by induction we first verify the base case $n = 0$ trivially. Assume the result holds for $n=k$, then for $n = k+1$ : $\\I_{k+1} = \frac{2k+1}{2k+2}I_k = \frac{2k+1}{2k+2}\frac{(2k)!\pi}{2^{2k}(k!)^2} = \frac{(2k+1)!\pi}{2^{2k+1}(k+1)!k!} =\frac{(2(k+1))!\pi}{2^{2(k+1)}(k+1)!^2} $ where in the last step we multiplied by $\frac{2k+2}{2k+2}$. Hence $I_n = \frac{(2n)!\pi}{2^{2n}(n!)^2}$

DatAtarLyfe · Apr 12, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
Additional Q:

$\noindent Find $\int \sqrt{\frac{6-x}{x-3}}\text{ d}x$.$

KingOfActing · Apr 12, 2016

Re: MX2 2016 Integration Marathon

This one is dedicated to Paradoxica because he asked me to:

$We begin by considering the contour integral $C$ around the pole at $ z = i$, made up from a line contour from $ -R$ to $R$, and a semi-circle contour $\gamma$ in the top half-plane of radius $R$, as $R \to \infty$.$\\\oint _C \frac{dz}{(z^2+1)^{n+1}} = \int_{-R}^R \frac{dz}{(z^2+1)^{n+1}} + \int_\gamma\frac{dz}{(z^2+1)^{n+1}}\\ \int_{-R}^R \frac{dz}{(z^2+1)^{n+1}} = \oint _C \frac{dz}{(z^2+1)^{n+1}} - \int_\gamma\frac{dz}{(z^2+1)^{n+1}}\\$The line segment contour is the integral we wish to evaluate. First we will deal with the contour integral (with an application of Cauchy's Integral formula):$ $\\\oint _C \frac{dz}{(z^2+1)^{n+1}} = \oint _C \frac{\frac{dz}{(z+i)^{n+1}}}{(z-i)^{n+1}} = \frac{2\pi i}{n!}\frac{d^n}{dz^n}\left.\left( \frac{1}{(z+i)^{n+1}}\right)\right|_{z=i} \\= \frac{2\pi i}{n!} \frac{(-1)^n(2n)!}{n!(2i)^{2n+1}}$ (Can be proved by induction.)$\\ = \frac{2(2n)!\pi i (-1)^n}{(n!)^22^{2n+1}i^{2n}} = \frac{(2n)!\pi}{2^{2n}(n!)^2}\\$Finally, we apply the estimation lemma to the semi-circle contour:$\\ \left|\int_\gamma f(z)\,dz \right| \leq ML $ where $M$ is the maximum of $f(z)$ on $\gamma$ and $L$ is the length of $\gamma$.$\\L = \pi R $ trivially by the geometry of the contour. For the maximum of the function, we apply the triangle inequality when $|z| = R$: $\\|z|^2 = |z^2| = |z^2 + 1 - 1| \leq |z^2 + 1| + 1\\$ $Hence: $\\|z^2+1| \geq R^2 - 1 > 0 \;\;\;$(as $R > 1$)$\\\left|\frac{1}{(z^2+1)^{n+1}}\right| \leq \frac{1}{(R^2 - 1)^{n+1}}\\$And so $\left|\int_\gamma f(z)\,dz \right| \leq \frac{\pi R}{(R^2 - 1)^{n+1}}\\$Taking the limit as $R \to \infty \implies \int_\gamma f(z)\,dz \to 0$ by the squeeze theorem.$\\$Putting all these results together: $ \int_{-\infty}^{\infty}\frac{1}{(x^2+1)^{n+1}}\,dx = \frac{(2n)!\pi}{2^{2n}(n!)^2}\\$ Q.E.D.$

Paradoxica · Apr 12, 2016

Re: MX2 2016 Integration Marathon

KingOfActing said:
This one is dedicated to Paradoxica because he asked me to:

$We begin by considering the contour integral $C$ around the pole at $ z = i$, made up from a line contour from $ -R$ to $R$, and a semi-circle contour $\gamma$ in the top half-plane of radius $R$, as $R \to \infty$.$\\\oint _C \frac{dz}{(z^2+1)^{n+1}} = \int_{-R}^R \frac{dz}{(z^2+1)^{n+1}} + \int_\gamma\frac{dz}{(z^2+1)^{n+1}}\\ \int_{-R}^R \frac{dz}{(z^2+1)^{n+1}} = \oint _C \frac{dz}{(z^2+1)^{n+1}} - \int_\gamma\frac{dz}{(z^2+1)^{n+1}}\\$The line segment contour is the integral we wish to evaluate. First we will deal with the contour integral:$ $\\\oint _C \frac{dz}{(z^2+1)^{n+1}} = \oint _C \frac{\frac{dz}{(z+i)^{n+1}}}{(z-i)^{n+1}} = \frac{2\pi i}{n!}\frac{d^n}{dz^n}\left.\left( \frac{1}{(z+i)^{n+1}}\right)\right|_{z=i} \\= \frac{2\pi i}{n!} \frac{(-1)^n(2n)!}{n!(2i)^{2n+1}}$ (Can be proved by induction.)$\\ = \frac{2(2n)!\pi i (-1)^n}{(n!)^22^{2n+1}i^{2n}} = \frac{(2n)!\pi}{2^{2n}(n!)^2}\\$Finally, we apply the estimation lemma to the semi-circle contour:$\\ \left|\int_\gamma f(z)\,dz \right| \leq ML $ where $M$ is the maximum of $f(z)$ on $\gamma$ and $L$ is the length of $\gamma$.$\\L = \pi R $ trivially by the geometry of the contour. For the maximum of the function, we apply the triangle inequality when $|z| = R$: $\\|z|^2 = |z^2| = |z^2 + 1 - 1| \leq |z^2 + 1| + 1\\$ $Hence: $\\|z^2+1| \geq R^2 - 1 > 0 \;\;\;$(as $R > 1$)$\\\left|\frac{1}{(z^2+1)^{n+1}}\right| \leq \frac{1}{(R^2 - 1)^{n+1}}\\$And so $\left|\int_\gamma f(z)\,dz \right| \leq \frac{\pi R}{(R^2 - 1)^{n+1}}\\$Taking the limit as $R \to \infty \implies \int_\gamma f(z)\,dz \to 0$ by the squeeze theorem.$\\$Putting all these results together: $ \int_{-\infty}^{\infty}\frac{1}{(x^2+1)^{n+1}}\,dx = \frac{(2n)!\pi}{2^{2n}(n!)^2}\\$ Q.E.D.$

"You must spread some Reputation around before giving it to KingOfActing again."

leehuan · Apr 12, 2016

Re: MX2 2016 Integration Marathon

50 IBPs later

https://www.youtube.com/watch?v=-reFBJ4R9iA

leehuan said:
$\int{{x}^{2015}{e}^{{x}^{1008}}\,$d$x}$

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\sin { x } \cos ^{ 2 }{ x }\,\text{d}x }$

KingOfActing · Apr 12, 2016

Re: MX2 2016 Integration Marathon

$\int x^{2015}e^{x^{1008}}\,dx = \frac{1}{1008}\int ue^u\,du = \frac{1}{1008}(ue^u - e^u) + C = \frac{1}{1008}x^{1008}e^{x^{1008}} - \frac{1}{1008}e^{x^{1008}} + C$

abecina · Apr 15, 2016

Re: MX2 2016 Integration Marathon

$\\ $Find a reduction formula for the integral ${ I }_{ n }=\int {x^{p-2}(1-x^p)^n}\textrm{d}x$

leehuan · Apr 15, 2016

Re: MX2 2016 Integration Marathon

Couldn't be bothered checking/simplifying my answer

$\left(p-1\right){I}_{n}={x}^{p-1}{\left(1-{x}^{p}\right)}^{n}-p{I}_{n}+p{I}_{n-1}$

DatAtarLyfe · Apr 15, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Couldn't be bothered checking/simplifying my answer

$\left(p-1\right){I}_{n}={x}^{p-1}{\left(1-{x}^{p}\right)}^{n}-p{I}_{n}+p{I}_{n-1}$

correct me if im wrong, but i think you're missing a few n's in front of the ${I}_{n}$ and ${I}_{n-1}$

leehuan · Apr 15, 2016

Re: MX2 2016 Integration Marathon

DatAtarLyfe said:
correct me if im wrong, but i think you're missing a few n's in front of the ${I}_{n}$ and ${I}_{n-1}$

Yes I did. (Just thinking back to my working out which I left at home it's like oh right.)

Paradoxica · Apr 15, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
$\int{{x}^{2015}{e}^{{x}^{1008}}\,$d$x}$

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\sin { x } \cos ^{ 2 }{ x }\,\text{d}x }$

$$\noindent$\frac{1}{4}\int_{0}^{\frac{\pi}{2}}e^x({4\sin{x}-4\sin^3{x}})$d$x =\frac{1}{4}\int_0^\frac{\pi}{2}e^x(\sin{3x}+\sin{x})$d$x \\$d$(e^x\sin{3x})=e^x(\sin{3x}+3\cos{3x})$d$x,\,$d$(e^x\cos{3x})=e^x(\cos{3x}-3{\sin{3x}})$d$x \\ $d$(e^x\sin{x})=e^x(\sin{x}+\cos{x})$d$x,\,$d$(e^x\cos{x})=e^x(\cos{x}-\sin{x})$d$x \\ $d$(e^x\sin{3x})-3$d$(e^x\cos{3x})=10e^x\sin{3x}$d$x\\5$d$(e^x\sin{x})-5$d$(e^x\cos{x})=10e^x\sin{x}$d$x \\40I = \left[e^x(\sin{3x}-3\cos{3x}+5\sin{x}-5\cos{x}) \right]_0^\frac{\pi}{2}=(-1+5)e^\frac{\pi}{2}-(-3-5)\\I =\frac{e^\frac{\pi}{2}+2}{10}$

DatAtarLyfe · Apr 16, 2016

Re: MX2 2016 Integration Marathon

I had a dream about integration and volumes and gabriels horn....too much maffs

leehuan · Apr 16, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent$\frac{1}{4}\int_{0}^{\frac{\pi}{2}}e^x({4\sin{x}-4\sin^3{x}})$d$x =\frac{1}{4}\int_0^\frac{\pi}{2}e^x(\sin{3x}+\sin{x})$d$x \\$d$(e^x\sin{3x})=e^x(\sin{3x}+3\cos{3x})$d$x,\,$d$(e^x\cos{3x})=e^x(\cos{3x}-3{\sin{3x}})$d$x \\ $d$(e^x\sin{x})=e^x(\sin{x}+\cos{x})$d$x,\,$d$(e^x\cos{x})=e^x(\cos{x}-\sin{x})$d$x \\ $d$(e^x\sin{3x})-3$d$(e^x\cos{3x})=10e^x\sin{3x}$d$x\\5$d$(e^x\sin{x})-5$d$(e^x\cos{x})=10e^x\sin{x}$d$x \\40I = \left[e^x(\sin{3x}-3\cos{3x}+5\sin{x}-5\cos{x}) \right]_0^\frac{\pi}{2}=(-1+5)e^\frac{\pi}{2}-(-3-5)\\I =\frac{e^\frac{\pi}{2}+2}{10}$

I got this question off an integration bee btw

DatAtarLyfe said:
I had a dream about integration and volumes and gabriels horn....too much maffs

Are you ok

HSC 2016 MX2 Integration Marathon (archive) (3 Viewers)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

-insert title here-

-insert title here-

Well-Known Member

lukewarm mess

Booty Connoisseur

lukewarm mess

-insert title here-

Well-Known Member

lukewarm mess

Member

Well-Known Member

Booty Connoisseur

Well-Known Member

-insert title here-

Booty Connoisseur

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 3)