Question

∫γ(xz2y+1)ds, where γ(t) =(lnt, t^2, 2t) and tϵ[1,e] please help

Answer 1

x(t)=lnt y(t)=t^2 z(t)= 2t
x'(t)= 1/t y'(t)=2t z'(t)=2

Answer 2

ds=\[\sqrt{\frac{ 1 }{ t^2 }+4t^2+4}dt \]

Answer 3

\[\int\limits_{1}^{e} \frac{ lnt*2t }{ 2t^2+1 }\sqrt{\frac{ 1 }{ t^2 }+4t^2+4} dt\]

Answer 4

and here i'am stuck. please help

Answer 5

The expression under the radical is a perfect square.

Answer 6

\[\frac{ 2t^2 +1}{ t }?\]

Answer 7

yep

Answer 8

thank you. i was writing 4 instead of 1 and i didn't noticed the perfect square. the answer is 2.

Answer 9

no problem