How did they simplify this?

**c_323_h** · July 31st 2006, 04:59 PM

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.

**topsquark** · July 31st 2006, 05:18 PM

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

**c_323_h** · July 31st 2006, 05:22 PM

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!

**topsquark** · July 31st 2006, 05:25 PM

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

Thread: How did they simplify this?

LinkBack

Thread Tools

Display

How did they simplify this?

Similar Math Help Forum Discussions

simplify #2

Please simplify

Can you simplify this further?

Simplify

Search Tags