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How to integrate y^2

How to integrate y^2 with respect to x?

I know that the result is x*y^2 but why?

int(y^2)dx=int(1*y^2)dx

set:
u=y^2

dv/dx=1

find du/dx and v

du/dx=2y*dy/dx

v=x

use the integration by parts formula

I=uv - int(v*du/dx)

I=x*y^2 - int(x*(2y*dy/dx))dx

Seems to me like one awkward equation.... or maybe it has something to do with me having only 3 hours of sleep today?

Step by step explanation greatly appreciated :smile:
y^2 is just a constant.

Say if you had to integrate INT 9.dx, it would be 9x+c, well its the same principal with y^2.
Reply 2
Avatar for TheBigRC
TheBigRC
8
y^2 is a constant unless y is a function of x

y2∫ dx=xy2y^2\displaystyle\int\ dx=xy^2

maybe?
Reply 3
Avatar for Slumpy
Slumpy
16
Lover's Knot
y^2 is just a constant.

Say if you had to integrate INT 9.dx, it would be 9x+c, well its the same principal with y^2.


This is quite probably wrong, assuming that in the above context, y is a function of x, which would seem reasonable.

OP-I'd say it depends on y I think, but I'm a bit sleep deprived also:p:

Edit-OP, are you given that the answer is xy^2?
If so, the above are both correct, and y^2 is just a constant.
Reply 4
Avatar for Moa
Moa
OP
13
Lover's Knot
y^2 is just a constant.

Say if you had to integrate INT 9.dx, it would be 9x+c, well its the same principal with y^2.


Here it is also ok to do this?

d^2/dx^2= int(x^3 - y^2)dx
d^2/dx^2= int(x^3)-(y^2)dx
d^2/dx^2= int(x^3)-x*y^2
Moa
Here it is also ok to do this?

d^2/dx^2= int(x^3 - y^2)dx
d^2/dx^2= int(x^3)-(y^2)dx
d^2/dx^2= int(x^3)-x*y^2

Yes that's fine, but don't forget the +c.
Reply 6
Avatar for Slumpy
Slumpy
16
Moa
Here it is also ok to do this?

d^2/dx^2= int(x^3 - y^2)dx
d^2/dx^2= int(x^3)-(y^2)dx
d^2/dx^2= int(x^3)-x*y^2


d^2/dx^2 is not a valid 'thing' as it were, it needs to be d^2 something by dx^2.
Reply 7
Avatar for Moa
Moa
OP
13
Slumpy
d^2/dx^2 is not a valid 'thing' as it were, it needs to be d^2 something by dx^2.


yup missed "y" when I was copying from my notes :biggrin:
Reply 8
Avatar for Slumpy
Slumpy
16
Moa
yup missed "y" when I was copying from my notes :biggrin:


Based on what you've written above, where are you getting that the answer to integrating y^2 is xy^2 from? The book?

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