The discussion revolves around evaluating the integral of the square root of 3x with respect to x, a topic within calculus. Participants are exploring the correct approach to simplifying and solving the integral.
The discussion is active, with participants providing insights and asking clarifying questions. Some guidance has been offered regarding the manipulation of constants in integrals, but there is no explicit consensus on the final evaluation of the integral.
Participants are navigating the nuances of integral calculus, particularly in relation to constants and radical expressions. There is an indication of differing interpretations of the integral's setup.
I'm taking this to meanNawz said:Homework Statement
Evaluate:
Integral sign: Square root of 3x times dx
Nawz said:Homework Equations
The Attempt at a Solution
integral of 3x^1/2 dx
(3x^(3/2)) / (3/2)
so I multiplied it by 2/3 and got:
(2/3x^(3/2)) +C
but the answer is 2 times the square root of 3 divided by 3 then x^3/2.. ? I don't know how they kept the square root of 3.
Mark44 said:I'm taking this to mean
[tex]\int \sqrt{3x}dx[/tex]
The simplest way to approach this is to bring out a factor of sqrt(3), and rewriting the integral using exponents, rather than radicals.
[tex]\sqrt{3}\int x^{1/2}~dx[/tex]
Can you carry out the rest of this?