Integral

[jkeynes]

Hello,

I am working on a question at the moment and I am stumped. I am attempting find the integral of the following equation:

(Not sure how to type the equation clearly here - I will attempt to use words. Here is a link to the formula written correctly in Wolfram Alpha if this is not clear enough: http://www.wolframalpha.com/input/?i=int+(2/5)+-+1.6/(2x^2)^(1/3)&a=*C.int-_*IntegralsWord-)

2/5 - 1.6 / (3rd root of (2x^2))

I am not sure what method of integration to use. At first I thought that the (3rd root of (2x^2)) was the same as 2x^(2/3) - and thus would just use the power rule.

Any help would be appreciated!

- John Keynes

 

[galactus]

Before I proceed, is this your integral?

\(\displaystyle \int\left[\frac{2}{5}-\frac{1.6}{\sqrt[3]{2x^{2}}}\right]dx\)

 

[jkeynes]

Yes! Thank you very much for your reply. However, I believe have solved it!

2x / 5 - 4.8x^(1/3) / 2^(1/3)

May I ask what you use to write equations in their proper forms? It would be much easier when asking for help to post something like that. And to show the steps I took to calculate my answer.

- John Keynes

 

[galactus]

Yep, that's it.

Except, to be anal, write everything in fractions instead of mixing up fractions and decimals. Looks better that way.

Just a thought.

You could also write it as:

\(\displaystyle \frac{2x}{5}-\frac{12}{5}\cdot (4x)^{\frac{1}{3}}\) to compact things a bit.