Improper Integrals

• May 28th 2007, 09:32 PM
darkblue
Improper Integrals

1. int {1/[radical(x-1)]} from 2 to infinity. 2 is the lower limit, infinity is the upper limit. i put it converges because i looked at the graph and i think it has a limit?

2. int {e^2x} dx from -infinity to 0

i put it diverges because the limit does not exist. i got [(e^2x)/2] 0 to N. then i put (e^0 / 2) - (e^2N / 2)
and then i looked at the graph of the integral.
• May 28th 2007, 09:54 PM
Jhevon
Quote:

Originally Posted by darkblue

1. int {1/[radical(x-1)]} from 2 to infinity. 2 is the lower limit, infinity is the upper limit. i put it converges because i looked at the graph and i think it has a limit?

this diverges. we only have a problem at x = 1, so since that is not between our limits of integration we dont have to split the integral up.

$\displaystyle \int_{2}^{ \infty} \frac {1}{ \sqrt {x - 1}} dx = \lim_{N \to \infty} \int_{2}^{N} (x - 1)^{- \frac {1}{2}}$

$\displaystyle = \lim_{N \to \infty} \left[ 2(x - 1)^{ \frac {1}{2}} \right]_{2}^{N}$

$\displaystyle = \lim_{N \to \infty} \left[ 2(N - 1)^{ \frac {1}{2}} - 2(2 - 1)^{ \frac {1}{2}} \right]$

$\displaystyle = \infty$

So the integral diverges

looking at the graph and seeing that is has a limit means nothing. look at the graph of $\displaystyle \frac {1}{x}$, it's limit is 0 as x goes to $\displaystyle \infty$, but its improper integral diverges

Quote:

2. int {e^2x} dx from -infinity to 0

i put it diverges because the limit does not exist. i got [(e^2x)/2] 0 to N. then i put (e^0 / 2) - (e^2N / 2)
and then i looked at the graph of the integral.
this converges.

$\displaystyle \int_{- \infty}^{0} e^{2x}dx = \lim_{N \to - \infty} \int_{N}^{0} e^{2x} dx$

$\displaystyle = \lim_{N \to - \infty} \left[ \frac {1}{2} e^{2x} \right]_{N}^{0}$

$\displaystyle = \lim_{N \to - \infty} \left[ \frac {1}{2} e^0 - \frac {1}{2} e^{N} \right]$

$\displaystyle = \frac {1}{2} - 0 < \infty$

Thus the integral converges

remember, when we raise something to a negative exponent, it's the same as putting 1 over the something to the positive exponent. So $\displaystyle e^{- \infty} = \frac {1}{e^{ \infty}} = \frac {1}{ \infty} = 0$