# partially solved line integral

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• December 14th 2010, 03:28 PM
manyarrows
partially solved line integral
I have gotten a line integral this far and can't figure out how to integrate it. I solved it on an integral calc, which gave me the correct value, I just don't know how to do the steps.

$\int^1_0 2t^3 \sqrt[2]{1+4t^2+4t^4} dt$
• December 14th 2010, 04:30 PM
dwsmith
$\displaystyle 1+4t^2+4t^4=(2t^2+1)^2$

$\displaystyle \int_0^1 (2t^3*(2t^2+1))dt$

Spoiler:
$\displaystyle \int_0^1 (4t^5+2t^3)dt=\left[\frac{2t^6}{3}+\frac{t^4}{2}\right]_0^1=\frac{2}{3}+\frac{1}{2}\right)=\frac{7}{6}$
• December 14th 2010, 04:54 PM
matheagle
It looks like you have two 2's, instead of one.
• December 14th 2010, 08:16 PM
matheagle
ok, you killed that extra 2, but the upper bound was a 1 not a 2.
you have a thing with 2's?
• December 14th 2010, 08:21 PM
dwsmith
You can never have enough 2's.