Set up an integral for the length of the curve.
-$$y = x ^ { 4 } , 0 \leq x \leq 1$$
A) $\int _ { 0 } ^ { 1 } \sqrt { 1 + 4 x ^ { 3 } } d x$
B) $\int _ { 0 } ^ { 1 } \sqrt { 1 + 16 x ^ { 8 } } d x$
C) $\int _ { 0 } ^ { 1 } \sqrt { 1 + 4 x ^ { 6 } } d x$
D) $\int _ { 0 } ^ { 1 } \sqrt { 1 + 16 x ^ { 6 } } d x$

Question

Quizplus · Accepted Answer

The formula for the length of a curve $ y = f(x) $ is $\int \sqrt{1 + (f'(x))^2} \, dx$. Here, $ f(x) = x^4 $, so $ f'(x) = 4x^3 $. Therefore, $(f'(x))^2 = 16x^6$, leading to the integral $\int _ { 0 } ^ { 1 } \sqrt { 1 + 16 x ^ { 6 } } d x$.

Set Up an Integral for the Length of the Curve y=x4,0≤x≤1y = x ^ { 4 } , 0 \leq x \leq 1y=x4,0≤x≤1

Set Up an Integral for the Length of the Curve $y = x ^ { 4 } , 0 \leq x \leq 1$