That’s pretty easy you just have to basically y=mx+b
If f(x) has an inverse on [a, b], then integrating by parts (take u = f(x) and dv = dx), we can show

Let
. Compute the inverse:
![f\left(f^{-1}(x)\right) = \sqrt{1 + f^{-1}(x)^3} = x \implies f^{-1}(x) = \sqrt[3]{x^2-1}](https://tex.z-dn.net/?f=f%5Cleft%28f%5E%7B-1%7D%28x%29%5Cright%29%20%3D%20%5Csqrt%7B1%20%2B%20f%5E%7B-1%7D%28x%29%5E3%7D%20%3D%20x%20%5Cimplies%20f%5E%7B-1%7D%28x%29%20%3D%20%5Csqrt%5B3%5D%7Bx%5E2-1%7D)
and we immediately notice that
.
So, we can write the given integral as

Splitting up terms and replacing
in the first integral, we get

The diameter fits all the way across the circle, and the radius fits half way across the circle, thus making the radius half the length of the diameter so the ratio for diameter/radius would be 2/1 or 1/.5
Answer:
AAS
Step-by-step explanation:
the ansewer