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

##### 1 Answer
Dec 27, 2017

$\frac{1}{2} \left(11 \sqrt{120} - \ln | \left(11 + \sqrt{120}\right) |\right)$

#### Explanation:

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

Let $x = t + 1 \Rightarrow \mathrm{dx} = \mathrm{dt}$

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.

https://socratic.org/questions/how-do-you-evaluate-the-integral-int-sqrt-x-2-1-dx:

$\int \sqrt{{x}^{2} - 1} \mathrm{dx} = \frac{x}{2} \sqrt{{x}^{2} - 1} - \frac{1}{2} \ln | x + \sqrt{{x}^{2} - 1} | + C$

${\int}_{1}^{11} \sqrt{{x}^{2} - 1} \mathrm{dx} = {\left(\frac{x}{2} \sqrt{{x}^{2} - 1} - \frac{1}{2} \ln | x + \sqrt{{x}^{2} - 1} |\right)}_{1}^{11}$

$= \frac{1}{2} \left(11 \sqrt{120} - 0 - \ln | \frac{11 + \sqrt{120}}{1} |\right)$

$= \frac{1}{2} \left(11 \sqrt{120} - \ln | \left(11 + \sqrt{120}\right) |\right)$