# How did they simplify this?

• Jul 31st 2006, 05:59 PM
c_323_h
How did they simplify this?
This: $2\pi\int_1^4 \sqrt{y}\sqrt{1+\frac{1}{4y}}dy$

To get: $\pi\int_1^4\sqrt{4y+1}dy$

I thought they found the common denominator under the radical, distributed the $\sqrt{y}$, then divided by 2. Apparently I'm wrong unless I did the algebra incorrectly.
• Jul 31st 2006, 06:18 PM
topsquark
Quote:

Originally Posted by c_323_h
This: $2\pi\int_1^4 \sqrt{y}\sqrt{1+\frac{1}{4y}}dy$

To get: $\pi\int_1^4\sqrt{4y+1}dy$

I thought they found the common denominator under the radical, distributed the $\sqrt{y}$, then divided by 2. Apparently I'm wrong unless I did the algebra incorrectly.

It's simpler to bring the 2 inside the square root:

$2\pi\int_1^4 \sqrt{y}\sqrt{1+\frac{1}{4y}}dy$

$\pi\int_1^4 2\sqrt{y}\sqrt{1+\frac{1}{4y}}dy$

$\pi\int_1^4 \sqrt{4y}\sqrt{1+\frac{1}{4y}}dy$

$\pi\int_1^4 \sqrt{4y+1}dy$

-Dan
• Jul 31st 2006, 06:22 PM
c_323_h
Quote:

Originally Posted by topsquark
It's simpler to bring the 2 inside the square root:

$2\pi\int_1^4 \sqrt{y}\sqrt{1+\frac{1}{4y}}dy$

$\pi\int_1^4 2\sqrt{y}\sqrt{1+\frac{1}{4y}}dy$

$\pi\int_1^4 \sqrt{4y}\sqrt{1+\frac{1}{4y}}dy$

$\pi\int_1^4 \sqrt{4y+1}dy$

-Dan

ugh...why do they keep doing this to me? math beats me up sometimes:( jk, Thanks!
• Jul 31st 2006, 06:25 PM
topsquark
Quote:

Originally Posted by c_323_h
ugh...why do they keep doing this to me? math beats me up sometimes:( jk, Thanks!

Don't worry about it. It happens to the best of us. ;)

-Dan