By using the arc length formula, we will see that the length of the curve is L = 1.48
<h3>
How to use the arc length formula?</h3>
Here we have the curve:
y = 2 - x^2 with 0 ≤ x ≤ 1
And we want to find the length of the curve.
The arc length formula for a curve y in the interval [x₁, x₂] is given by:
![L =\int\limits^{x_2}_{x_1} {\sqrt{1 + \frac{dy}{dx}^2} } } \, dx](https://tex.z-dn.net/?f=L%20%20%3D%5Cint%5Climits%5E%7Bx_2%7D_%7Bx_1%7D%20%7B%5Csqrt%7B1%20%2B%20%5Cfrac%7Bdy%7D%7Bdx%7D%5E2%7D%20%7D%20%7D%20%5C%2C%20dx)
For our curve, we have:
dy/dx = -2x
And the interval is [0, 1]
Replacing that we get:
![L =\int\limits^{1}_{0} {\sqrt{1 + (-2x)^2} } } \, dx\\\\L \int\limits^{1}_{0} {\sqrt{1 + 4x^2} } } \, dx\\](https://tex.z-dn.net/?f=L%20%20%3D%5Cint%5Climits%5E%7B1%7D_%7B0%7D%20%7B%5Csqrt%7B1%20%2B%20%28-2x%29%5E2%7D%20%7D%20%7D%20%5C%2C%20dx%5C%5C%5C%5CL%20%20%5Cint%5Climits%5E%7B1%7D_%7B0%7D%20%7B%5Csqrt%7B1%20%2B%204x%5E2%7D%20%7D%20%7D%20%5C%2C%20dx%5C%5C)
This integral is not trivial, using a table you can see that this is equal to:
![L = (\frac{Arsinh(2x)}{4} + \frac{x*\sqrt{4x^2 + 1} }{2})](https://tex.z-dn.net/?f=L%20%3D%20%28%5Cfrac%7BArsinh%282x%29%7D%7B4%7D%20%20%2B%20%5Cfrac%7Bx%2A%5Csqrt%7B4x%5E2%20%2B%201%7D%20%7D%7B2%7D%29)
evaluated from x = 1 to x = 0, when we do that, we will get:
![L = \frac{Arsinh(2) + 2*\sqrt{5} }{4} = 1.48](https://tex.z-dn.net/?f=L%20%3D%20%5Cfrac%7BArsinh%282%29%20%2B%202%2A%5Csqrt%7B5%7D%20%7D%7B4%7D%20%20%3D%201.48)
That is the length of the curve.
If you want to learn more about curve length, you can read:
brainly.com/question/2005046