The discussion focuses on the application of the substitution rule for integrals, specifically for the integral $$\int_{0}^{2} r\sqrt{5-\sqrt{4-r^2}} dr$$. The initial substitution of $$u=4-r^2$$ was deemed incorrect due to improper handling of the differential. A more effective substitution is proposed: $$u = \sqrt{4 - r^2}$$, leading to the transformed integral $$\int_3^5 (5 - u)\sqrt{u} du$$, which simplifies the evaluation process significantly.
PREREQUISITESStudents and educators in calculus, particularly those focusing on integral calculus and substitution techniques, as well as anyone seeking to improve their skills in simplifying complex integrands.
This doesn't look very promising, and besides, it doesn't look like you did your substitution correctly. If ##u = 4 - r^2##, what is du? You can't simply replace dr by du.Arnoldjavs3 said:Homework Statement
$$\int_{0}^{2} r\sqrt{5-\sqrt{4-r^2}} dr$$
Homework Equations
The Attempt at a Solution
would i substitute ##u=4-r^2##?
After of which I would input into the integral and get:
$$\int_{0}^{2} \sqrt{5-\sqrt{u}}du$$
As I said, this doesn't seem very promising. I would try this substitution: ##u = \sqrt{4 - r^2}##, or equivalently, ##u^2 = 4 - r^2##.Arnoldjavs3 said:What would I do here? Do I just work inside the radical(so 5r - 2/3u^3/2)? Or do I have to account that it's inside a radical when solving?