Looking for some proof

I derived the following equality using a few methods that aren't all that complicated. I'm wondering if there is any way to show this is true other than the way I used to derive it.

\[\int _{ -1 }^{ 1 }{ \frac { 2{ x }^{ 3 }+2bx+a+1 }{ \sqrt { 2{ x }^{ 2 }+2\left[ a+b \right] x+{ \left[ a+1 \right] }^{ 2 }+{ \left[ b-1 \right] }^{ 2 } } } dx } =\int _{ 0 }^{ 2 }{ \frac { a+2-y }{ \sqrt { { y }^{ 2 }-2\left[ a+2 \right] y+{ \left[ a+2 \right] }^{ 2 }+{ b }^{ 2 } } } dy } \]

where aa and bb are real numbers.

Something about it makes me think there's a trick that will make them look the same.

If there isn't any other generally simple way to get this, then I might mess around with what I did to see what can come of it.

#Calculus

Note by Austin Antonacci
5 years, 4 months ago

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Comments

Show your proof :)

Aman Rajput - 5 years, 4 months ago

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I used a vector field based off of a gravitational field centered at zero:

F(x,y)=<−xx2+y2,−yx2+y2>F\left( x,y \right) =\left< \frac { -x }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } ,\frac { -y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } \right>

This field is conservative, so any closed path integral should result in zero. I decided to create an arbitrary path starting at (a,b)\left( a,b \right) that follows a parabola to (a+2,b)\left( a+2,b \right) then proceeds linearly back to (a,b)\left( a,b \right) . I set up the following parametric equations for these paths:

r1(t)=<t+a+1,t2+b−1>{ r }_{ 1 }\left( t \right) =\left< t+a+1,{ t }^{ 2 }+b-1 \right> for −1≤t≤1-1\le t\le 1

r2(t)=<a+2−t,b>{ r }_{ 2 }\left( t \right) =\left< a+2-t,b \right> for 0≤t≤20\le t\le 2

The line integrals give us

∫r1F(x,y)dr+∫r2F(x,y)dr=0\int _{ { r }_{ 1 } }^{ }{ F\left( x,y \right) dr } \quad +\quad \int _{ { r }_{ 2 } }^{ }{ F\left( x,y \right) dr } \quad =\quad 0

∫r1F(x,y)ds=−∫r2F(x,y)dr\int _{ { r }_{ 1 } }^{ }{ F\left( x,y \right) ds } \quad =\quad -\int _{ { r }_{ 2 } }^{ }{ F\left( x,y \right) dr }

From here, if I have copied everything down correctly, it should give the claim above. I might elaborate on the last steps later.

Austin Antonacci - 5 years, 4 months ago
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