Question

Find the value of m so that Integral from m to infinity e^(-x/2)dx = 1/2

Answer 1

\[\int\limits_{m}^{\infty} e^(-x/2) dx =1/2\]

Answer 2

i guess the anti derivative is \(-2e^{-\frac{x}{2}}\)

Answer 3

as x goes to infinity this goes to zero right?

Answer 4

so your real job is to set
\[\frac{1}{2}e^{-\frac{x}{2}}=\frac{1}{2}\] and solve for \(x\)

Answer 5

make sense? and is it clear what to do now?

Answer 6

How did you get the 1/2 e^-x/2

Answer 7

the first time or the second time?

Answer 8

second time

Answer 9

first you want
\[\int e^{-\frac{x}{2}}dx\] the anti - derivative. i get by a mental u - sub
\[-\frac{1}{2}e^{-\frac{x}{2}}\] so far so good?

Answer 10

Yup!

Answer 11

then to compute
\[\int_ m^{\infty}e^{-\frac{x}{2}}dx\] you take the anti derivative you compute
\[\lim_{t\to \infty}-\frac{1}{2}e^{-\frac{x}{2}}-\left (-\frac{1}{2}e^{\frac{m}{2}}\right)\]

Answer 12

now the limit is part is clearly zero because
\(e^{-\frac{x}{2}}\) goes to zero lickety split

Answer 13

so you are left with
\[\frac{1}{2}e^{-\frac{m}{2}}\] and you want that to be \(\frac{1}{2}\) so write
\[\frac{1}{2}e^{-\frac{m}{2}}=\frac{1}{2}\] and solve for \(m\)

Answer 14

k?

Answer 15

Sweet thanks so much!

Answer 16

yw