Integrating a Polynomial with Fractional and Negative Indices

[_Mayday_]

I have a bit of a problem with this question, I will do my best to offer an answer. I think the problem is not with the differentiation but with my indices.

Here is the initial formula.

$$\int (6x + 2 + x^{-\frac{1}{2}}) dx$$

Here is my attempt

$$\frac{6x^2}{2} + 2x + \frac{x\frac{1}{2}}{\frac{1}{2}}$$

$$3x^2 + 2x + \frac{\sqrt\frac{1}{2}}{\frac{1}{2}} + c$$


Where have I gone wrong? I'll bet it's the whole indices thing!

Thanks for any help

 

[Tedjn]

Do you mean:

$$\frac{\sqrt{x}}{\frac{1}{2}} = 2\sqrt{x}$$

I don't see how the x disappeared in the last term of your solution.

 

[dynamicsolo]

Here is my attempt

$$\frac{6x^2}{2} + 2x + \frac{x\frac{1}{2}}{\frac{1}{2}}$$

$$3x^2 + 2x + \frac{\sqrt\frac{1}{2}}{\frac{1}{2}} + c$$

The problem is in the last term: that should be

$$\frac{x^{\frac{1}{2}}}{\frac{1}{2}}$$ ,

which is to say, x to the one-half power divided by one-half. X to the one-half power is the square root of x, but in any case, what you did in the next line was to drop the "x". You want to say

$$\frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 2 {x^{\frac{1}{2}}$$ .

 

[_Mayday_]

$$\frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 2 {x^{\frac{1}{2}}$$ .

Ah ok, why does it do that then? I don't understand that could you take me through it please.

Thanks to both of you so far

 

[Tedjn]

Do you mean, why is dividing by 1/2 equal to multiplying by 2?

 

[_Mayday_]

Well, how does this work?

$$\frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 2 {x^{\frac{1}{2}}$$ .

 

[dynamicsolo]

Well, how does this work?

$$\frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 2 {x^{\frac{1}{2}}$$ .

You want to "eliminate" the 1/2 in the denominator of this quotient, which means making it 1, so that you end up with the numerator divided by 1, which is just the numerator. You can multiply the numerator and denominator by 2, which is like multiplying your quotient by 2/2 = 1, which means the value of the fraction is unchanged. You will end up with

$$\frac{2x^{\frac{1}{2}}}{1} = 2 {x^{\frac{1}{2}} = 2\sqrt{x}$$

 

[_Mayday_]

You want to "eliminate" the 1/2 in the denominator of this quotient, which means making it 1, so that you end up with the numerator divided by 1, which is just the numerator. You can multiply the numerator and denominator by 2, which is like multiplying your quotient by 2/2 = 1, which means the value of the fraction is unchanged. You will end up with

$$\frac{2x^{\frac{1}{2}}}{1} = 2 {x^{\frac{1}{2}} = 2\sqrt{x}$$

Brilliant! Thanks for your time you two! Funny how trivial these things seem in hind sight! :shy:

Thanks Again!

 

[_Mayday_]

So then my final equation will be:

$$3x^2 + 2x + 2x^{\frac{1}{2}} + c$$

 

[dynamicsolo]

So then my final equation will be:

$$3x^2 + 2x + 2x^{\frac{1}{2}} + c$$

If you want to check it, differentiate it term-by-term: you'll get your original integrand back again... (It's correct!)