Evaluate: Essentials of Calculus integral problem

Nawz
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Homework Statement



Evaluate:

Integral sign: Square root of 3x times dx



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.
 
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Nawz said:

Homework Statement



Evaluate:

Integral sign: Square root of 3x times dx
I'm taking this to mean
\int \sqrt{3x}dx

The simplest way to approach this is to bring out a factor of sqrt(3), and rewriting the integral using exponents, rather than radicals.

\sqrt{3}\int x^{1/2}~dx

Can you carry out the rest of this?

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
\int \sqrt{3x}dx

The simplest way to approach this is to bring out a factor of sqrt(3), and rewriting the integral using exponents, rather than radicals.

\sqrt{3}\int x^{1/2}~dx

Can you carry out the rest of this?

Wow. Yes that makes it much easier. When can you do this? Can you do it For all constants in front of an x? Is it recommended to do that?
 
Yes, you can always move a constant into or out of an integral.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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