1. ## Integration

Find the value of p if $\int_{1}^{\infty} \frac{2}{x^2},dx=3$ $\int_{1}^{p} \frac{2}{x^2}dx$

many thanks!

and where can i find the codes for the [tex] function?

2. Originally Posted by BabyMilo
Find the value of p if $\int_{1}^{\infty} \frac{2}{x^2},dx=3$ $\int_{1}^{p} \frac{2}{x^2}dx$

many thanks!

and where can i find the codes for the [tex] function?
what value have you determined for

$\int_{1}^{\infty} \frac{2}{x^2} \, dx$

?

3. Originally Posted by BabyMilo
Find the value of p if $\int_{1}^{\infty} \frac{2}{x^2},dx=3$ $\int_{1}^{p} \frac{2}{x^2}dx$

many thanks!

and where can i find the codes for the [tex] function?

what has you confused about the problem? Evaluate the left hand side. Then evaluate the right hand side, plugging in P as your upper limit, and solve for P

4. Originally Posted by skeeter
what value have you determined for

$\int_{1}^{\infty} \frac{2}{x^2} \, dx$

?
2

5. Originally Posted by artvandalay11
what has you confused about the problem? Evaluate the left hand side. Then evaluate the right hand side, plugging in P as your upper limit, and solve for P
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:

$\int_{1}^{\infty} \frac{2}{x^2} dx$

$=[\frac{-2}{x}]_{1}^{\infty}$

$=0-(-2)$

$=2$

6. So do you divide by 3:

$\frac{2}{3} =$ $\int_{1}^{p} \frac{2}{x^2} dx$

$= [\frac{-2}{x}]_{1}^{p}$

$= \frac{-2}{p}-(-2)=\frac{2}{3}$

$= p=\frac{-2}{\frac{2}{3}-2}$

$p = 1.5$

7. I thought you had to sub in the 3 in to $x$
O.o
thanks!

8. lol, I love when you pose questions and then answer them, good job