Find the value of p if $\displaystyle \int_{1}^{\infty} \frac{2}{x^2},dx=3$$\displaystyle \int_{1}^{p} \frac{2}{x^2}dx$
many thanks!
and where can i find the codes for the [tex] function?
how does the 3 play a role in this equation?
Just be easy on me as this is my first week of intergration.
thanks for you help.
I did:
$\displaystyle \int_{1}^{\infty} \frac{2}{x^2} dx$
$\displaystyle =[\frac{-2}{x}]_{1}^{\infty}$
$\displaystyle =0-(-2)$
$\displaystyle =2 $
So do you divide by 3:
$\displaystyle \frac{2}{3} = $ $\displaystyle \int_{1}^{p} \frac{2}{x^2} dx$
$\displaystyle = [\frac{-2}{x}]_{1}^{p}$
$\displaystyle = \frac{-2}{p}-(-2)=\frac{2}{3}$
$\displaystyle = p=\frac{-2}{\frac{2}{3}-2}$
$\displaystyle p = 1.5 $