Let $f:[a,b] \to \mathbb{R}$ be a continuous differentiable function. The curve $c_f:[a,b] \to \mathbb{R}^2,\; t\mapsto \bigl(t,f(t)\bigr)$ is the parametrization of the graph of $f$. The arc length of the graph of $f$ is $$ L(c_f) = \int_{a}^{b} \! \sqrt { 1 + [f'(t)]^2 } \, dt. $$
In an exercise it is explicitly said, that I should calculate the integrals
a) $\displaystyle \int_{0}^{1} x^2 \, dx$
b) $\displaystyle \int_{0}^{1} |2x| \, dx$
using the formula for the arc length $L$. I don't know how I can use the formula for $L$ to solve an integral and why I should do that? A part of the exercise is also to explain the geometrical meaning.
A hint is to use hyperbolic functions.
Those integrals are not complicated to solve:
a) $\displaystyle \int_{0}^{1}x^2 \, dx = \biggl[\frac{x^3}{3}\biggr]_0^1 = \frac13$
b) $\displaystyle \int_{0}^{1} |2x| \, dx = 2\int_{0}^{1} |x| \, dx = 2\int_{0}^{1}x \, dx = 2 \biggl[\frac{x^2}{2}\biggr]_0^1 = 1$
I'm confused about this exercise. Does the exercise make any sense to you?
Edit - Corrections and additional information based on the comments :
1)More context: I actually derived the formula of $L(c_f)$ in a previous exercise from the general formula for a continuous differentiable curve $L(c)= \int_{a}^{b} \! || c'(t) || \, dt$. We derived this formula in our lectures, we started with the length of a polygon chain...We have done nothing further on this subject.
2)The exact phrasing of the exercise is: In the previous exercise, the formula of the arc length of the graph of a function $f$ was derived $L(c_f) = \int_{a}^{b} \! \sqrt { 1 + [f'(t)]^2 } \, dt$. With this in mind, calculate the integrals:
a) $\displaystyle \int_{0}^{1} x^2 \, dx$
b) $\displaystyle \int_{0}^{1} |2x| \, dx$ and explain the geometric meaning
Edit 2
- Perhaps the integrals should not be considered individually, there is another integral in the problem:
c) $\displaystyle \int_{0}^{1} \sqrt{1+4x^2} \, dx$
I haven't written this until now as it seemed irrelevant to me, but maybe that was a mistake.