Question

need help with improper integral...

Answer 1

\[\int\limits_{2}^{\infty}e ^{-au}du=4\] Find a such that it equals 4

Answer 2

what's the thing at the bottom of the integral sign, its a little hard to read

Answer 3

2 to infinity

Answer 4

e to the -a x u

Answer 5

need help

Answer 6

i solved and now have a = 4e^2a

Answer 7

oih wait nvm im an idiot

Answer 8

i have 1/4 = ae^(2a)

Answer 9

\[a=-0.176\]This returns the integral value of 3.995 (my value is rounded so I'm guessing it's about right)

Answer 10

I don't think your method is quite right there QED, maybe check your working?

Answer 11

\[= \lim_{\alpha \rightarrow \infty} \int\limits_{2}^{\alpha} e^{-au} du = \lim_{\alpha \rightarrow \infty} -\frac{1}{a} [ e^{-a \alpha} - e^{-2a} ] = -\frac{1}{a}[ 0 -e^{-2a} ]\]

Answer 12

Because \[\lim_{x \rightarrow \infty} e^{-x} = 0 \]

Answer 13

So, \[\frac{1}{a} e^{-2a} = 4 \]

Answer 14

You need to guess and check to approximate solns to that.