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

Answer 1

Christopher Allers

#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 tutor.hix, so there is no point of repeating it here.

https://tutor.hix.ai

# 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))| )#

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Answer 2

Isaiah Allen

To find the length of the curve described by the integral (\int_0^x \sqrt{t^2 + 2t} , dt) for (0 \leq x \leq 10), you would need to evaluate the integral from 0 to x and then substitute the upper limit x into the result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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