How do you find the lengths of the curve $y=intsqrt(t^2+2t)dt$ from [0,x] for the interval $0<=x<=10$ ?

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Answer 1 · 2017-12-27T22:27:11

$1/2(11sqrt(120)-ln|(11+sqrt(120))| )$

$y= int_0^10 sqrt(t^2+2t)dt = int_0^10 sqrt((t+1)^2-1)dt ^$

Let $x=t+1 rArr dx = dt$

$y= int_1^11 sqrt(x^2-1)dx ^$

The solution to this integration already posted elsewhere in socratic, so there is no point of repeating it here.

$int sqrt(x^2-1)dx = x/2sqrt(x^2-1)-1/2ln|x+sqrt(x^2-1)|+C$

$int_1^11sqrt(x^2-1)dx = (x/2sqrt(x^2-1)-1/2ln|x+sqrt(x^2-1)|)_1^11$

$=1/2(11sqrt(120) - 0 -ln|(11+sqrt(120))/1| )$

$=1/2(11sqrt(120)-ln|(11+sqrt(120))| )$

How do you find the lengths of the curve y=intsqrt(t^2+2t)dty=intsqrt(t^2+2t)dt from [0,x] for the interval 0<=x<=100<=x<=10?