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Math question

M

MikeMike

Lifer
ive never learned integration by parts, but i need to do it for this stat hmwk.

Evaluate 1(funkysign over)0 x*e^(-3x)dx

now how the hell do you do integration by parts.

i dont need the answer, just some hints and help.

thanks.

and:

Solve for K
DX/X(3x^2 + Kx) |x-1 = 1

i have even less clue on this one.

same deal as above.
 
Oh geez, my calc 2 teacher who can't speak english "taught" us this today. I'll be using that google link, no doubt. Thanks!
 
That's the expectation for exponential function. Are you sure you evaluate from 0 to 1 and not 0 to infinity?

x = u e^(-3x) = dv
dx = du e^(-3x)/-3 = v

uv - integral of vdu

x*e^(-3x)/-3 - int of e^(-3x)/-3

x*e^(-3x)/-3 + e^(-3x)/9 evaluates from 0 to 1

 
Originally posted by: MIKEMIKE
now, it might have seemes so easily, but what is "integral" mean??? im really honestly confused on this ******.
Click to expand...

You said you didn't now what integration by parts was, as if you knew calculus... and now you're asking what an integral is 😕
 
something ain't right, are you sure it is not Evaluate infinity(funkysign over)0 x*3*e^(-3x)dx?

The standard form for exponential distribution PDF is ae^(-ax).
 
Originally posted by: UncleWai
something ain't right, are you sure it is not Evaluate infinity(funkysign over)0 x*3*e^(-3x)dx?

The standard form for exponential distribution PDF is ae^(-ax).
Click to expand...

positive its a 1. it may be almost 1am, but it a 1
 
Originally posted by: MIKEMIKE
Originally posted by: UncleWai
something ain't right, are you sure it is not Evaluate infinity(funkysign over)0 x*3*e^(-3x)dx?

The standard form for exponential distribution PDF is ae^(-ax).
Click to expand...

positive its a 1. it may be almost 1am, but it a 1
Click to expand...

There should be a 3* next to x* i think, otherwise it's not a pdf?
You try plug in the number, you get negative number.
 
Originally posted by: UncleWai
Originally posted by: MIKEMIKE
Originally posted by: UncleWai
something ain't right, are you sure it is not Evaluate infinity(funkysign over)0 x*3*e^(-3x)dx?

The standard form for exponential distribution PDF is ae^(-ax).
Click to expand...

positive its a 1. it may be almost 1am, but it a 1
Click to expand...

There should be a 3* next to x* i think, otherwise it's not a pdf?
You try plug in the number, you get negative number.
Click to expand...

nope X*e^-3x dx
 
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