homeworkd help--Calc?

Question

Find the exact value of the Integral of X^3*e^(-2x)dx from 2 to infinity. NO DECIMAL ANSWERS

Help me get started.

Answer 1

Indefinite integral is: Int [x^3*e^(-2x)] dx

Integrating by parts
Int [x^3*e^(-2x)] dx
= - {x^3*e^(-2x)}/2 - Int [- {3x^2*e^(-2x)}/2] dx
= - {x^3*e^(-2x)}/2 + 1.5*Int [x^2*e^(-2x)] dx ________ (1)

Again integrating by parts
Int [x^2*e^(-2x)] dx
= - {x^2*e^(-2x)}/2 - Int [{2x*e^(-2x)}/2] dx
= - {x^2*e^(-2x)}/2 - Int [x*e^(-2x)] dx _________ (2)

Again integrating by parts
Int [x*e^(-2x)] dx
= - {x*e^(-2x)}/2 - Int [1*e^(-2x)/(-2)] dx
= - {x*e^(-2x)}/2 + 0.5*Int [e^(-2x)] dx
= - {x*e^(-2x)}/2 - 0.25*e^(-2x) ________(3)

Imposing the limits definite integral in (3) becomes
= - [- {2*e^(-2*2)}/2 - 0.25*e^(-2*2)] [since e^(-inf) = 0]
= e^(-4) + 0.25*e^(-4)
= 1.25*e^(-4)

Substituting this value in (2) and also imposing the limits there
= - [- {2^2*e^(-2*2)}/2] - 1.25*e^(-4)
= 2e^(-4) - 1.25*e^(-4)
= 0.75*e^(-4)

Substituting this value in (1) and also imposing the limits there
= - [- {2^3*e^(-2*2)}/2] + 1.5*0.75*e^(-4)
= 4e^(-4) + (9/8)e^(-4)
= 5.125e^(-4)
= (9/8)*e^(-4)

Answer 2

to get you started use integration by parts several times
= -1/2x^3e^-2x +1/2 Int 3x^2*e^-2x dx
To find the Intx^2e^-2x do the same

Answer 3

int( x^3*e^(-2x) )dx
Integrating by parts:
int(x^3 d(e^(-2x)/(-2))
= x^3 e^(-2x) / (-2) + (3x^2/2) int( e^(-2x)) dx
Integrating 2nd term by parts again:
x^3e^(-2x) / (-2) - (3/2) int(x^2*e^(-2x)) dx
and so on, reducing the power of x in the product by one each time.

Answer 4

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