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

Question

2188 views around the world

You can reuse this answer
Creative Commons License

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))| )#