I need some assistance

Question

For what value of k, if any is ∫₀ ^∞ kxe ^-2x dx = 1?

A. 1/4

B 1

C. 4

D. There is no such value of k

Answer 1

Let U = -2x

then dU = -2 dx

(-1/2) dU = dx

AND

(-1/2)U = x

k factors out of the integral and moves up front

the integral becomes:

(-1/2)U e^(U) dU

the (-1/2)*(-1/2)=1/4 factors out of the integral and moves up front...

integrating by parts...

M = u dN =e^U dU

dM = dU N = e^U

the integral becomes:

U * e^U - integral [ e^U du] =

U * e^U - e^u =

e^u [ U-1]

{perhaps this i.b.p is in the table somewhere, and I

just "reinvented the wheel" ... oh well}

the indefinite integral is:

e^(-2x) [ -2x-1] = F(x)

= [-2x-1]/e^(2x)

------------------------------

checks by differentiation, product rule:

e^(-2x) [-2] + [-2x-1] * e^(-2x) *(-2)

-2* e^(-2x)[1 + -2x-1]

-2 * e^(-2x)(-2x) =

e^(-2x)*(4x)

however, there was a factor of 1/4 that was factored, so

the derivative of the anti-derivative is x * e^(-2x),

which is the original integrand, so yes it checks...

---------------------------------------------------

limit as x--> of F(x) = [ -2(0)-1] / e^(-2*0)

= (-1)/1 = -1

limit as x--> infinity of F(x) is zero, as the

exponential in the denominator shall DOMINATE!

subtracting these limits: 0 - -1 = 0+1 = 1

so the final limit, is k*1 = k

I would bet on option B, k=1

Answer 2

Link to Desmos graph referenced in video:

desmos.com/calculator/ep1eigt8oh