Find the length of the curve.
-$$y = \int _ { 1 } ^ { x } \sqrt { t ^ { 2 } - 1 } d t , 3 \leq x \leq 6$$
A) 18
B) $\frac { 27 } { 2 }$
C) 9
D) 27

Question

Quizplus · Accepted Answer

The length of the curve can be found using the formula for arc length: $\int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$. Here, $\frac{dy}{dx} = \sqrt{x^2 - 1}$. Thus, the arc length is $\int_{3}^{6} \sqrt{1 + (x^2 - 1)} \, dx = \int_{3}^{6} x \, dx$, which evaluates to $\frac{27}{2}$.

Find the Length of the Curve y=∫1xt2−1dt,3≤x≤6y = \int _ { 1 } ^ { x } \sqrt { t ^ { 2 } - 1 } d t , 3 \leq x \leq 6y=∫1x​t2−1​dt,3≤x≤6

Find the Length of the Curve $y = \int _ { 1 } ^ { x } \sqrt { t ^ { 2 } - 1 } d t , 3 \leq x \leq 6$