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Math Help - How did they simplify this?

  1. #1
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    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.
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  2. #2
    Forum Admin topsquark's Avatar
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    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
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  3. #3
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    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!
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  4. #4
    Forum Admin topsquark's Avatar
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    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
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