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

Answer:
Side YZ is the longest
Step-by-step explanation:
The longest side is across from the largest angle.
Angle X must be 80° because its exterior angle is 100° and they're supplementary. Angle Z is 60° for the same reason. Inside a triangle is always 180°, and since the measures of angle Z and X add up to 140°, so there's 40° left, meaning angle X is the largest.
Angle X is across from size YZ, so that's the longest
Answer:
D
Step-by-step explanation:
commission is what you get when you make sales
Answer:
Area: 181.5
Volume: 166.4
Step-by-step explanation:
Hope this helps.