Integral where 5 is at top, 1 at bottom, of x^5-x^-1/x^2dx

[salu34]

I am in need of help for an integral question:

Integral where 5 is at the top and 1 is at the bottom of the integral sign x^5-x^-1/x^2 dx

Any help would be appreciated.

 

[PAULK]

Re: Integral help

I am in need of help for an integral question:

Integral where 5 is at the top and 1 is at the bottom of the integral sign x^5-x^-1/x^2 dx

Any help would be appreciated.

..........................
Is this it?

{5 x^-1
| [ x^5 - ----- ] dx
}1 x^2

If so, you have:

{5
| [ x^5 - x^-3 ] dx
}1

which should be routine. But maybe you left out something. [Parentheses, maybe?]

 

[galactus]

Re: Integral help

A little LaTex would make it easier to decipher.

Is it

\(\displaystyle \int_{1}^{5}\frac{x^{5}-x^{-1}}{x^{2}}dx=\int_{1}^{5}\left[x^{3}-\frac{1}{x^{3}}\right]dx\)

 

[salu34]

Re: Integral help

yes thats it. How did you get it to be x^3 - 1/x^3?

 

[galactus]

Re: Integral help

Just the exponent laws. With all due respect, you should have those down pat since you're in calculus.