How to integrate this?

Blaser211 · Jul 3, 2020

√y (1+y)dy

Dr.Peterson · Jul 4, 2020

Write it as powers of y, starting with y^(1/2) and expanding.

Singleton · Jul 4, 2020

Distribute.

Singleton · Jul 4, 2020

Singleton said:
Distribute.

Oh! where does the square root sign end?

Deleted member 4993 · Jul 4, 2020

Blaser211 said:
√y (1+y)dy

Is it:

\(\displaystyle \sqrt{y} * (y + 1)\)

or

\(\displaystyle \sqrt{y * (y + 1)}\)

Blaser211 · Jul 5, 2020

\(\displaystyle \sqrt{y} * (y + 1)\)

Blaser211 · Jul 5, 2020

y^1/2 (1+y) dy
=y^1/2 (dy) +2y^1/2 (dy)
Is this correct?

Deleted member 4993 · Jul 5, 2020

Blaser211 said:
y^1/2 (1+y) dy
=y^1/2 (dy) +2y^1/2 (dy)
Is this correct?.........................No

y^1/2 (1+y) dy

=y^(1/2) dy + y^(3/2) dy

Dr.Peterson · Jul 5, 2020

Blaser211 said:
y^1/2 (1+y) dy
=y^1/2 (dy) +2y^1/2 (dy)
Is this correct?

No. Take it slowly:

y^(1/2) (1+y) = y^(1/2)*1 + y^(1/2)*y = y^(1/2) + y^(1/2)*y^1 = y^(1/2) + y^(1/2 + 1) = y^(1/2) + y^(3/2)

Steven G · Jul 5, 2020

Blaser211 said:
y^1/2 (1+y) dy
=y^1/2 (dy) +2y^1/2 (dy)
Is this correct?

You are in calculus and do not know that only y^1/2*2 gives you 2y^1/2. So y^1/2*y can't be 2y^1/2