Evaluate the following integral ?

Question

∫ xe^-2x dx with an upper boundary of 1 and a lower boundary of 0

Answer 1

I = ∫ x e^(-2x) dx

I = u v - ∫ v (du/dx) dx

u = x , dv/dx = e^(-2x)

du/dx = 1 , v = (-1/2) e^(-2x)

I = x (-1/2) e^(-2x) + (1/2) ∫  e^(- 2x) dx

I = x (-1/2) e^(-2x) - (1/4) ∫  e^(- 2x) dx-----limits 0 to 1

I = (-1/2) e^(-2) - (1/4) e^(-1) + 1/4

I = - 1 / (2e²) - 1 / (4e) + 1/4

Answer 2

∫ x e^(-2x) dx =

this is a typical by parts integral:

let x = u → dx = du

e^(-2x) dx = dv → (-1/2)e^(-2x) = v

thus, integrating by parts, you get:

∫ u dv = u v - ∫ v du →

∫ x e^(-2x) dx = (-1/2)x e^(-2x) - ∫ (-1/2)e^(-2x) dx =

(-1/2)x e^(-2x) + (1/2) ∫ e^(-2x) dx =

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

[(-1/2)e^(-2x)] [x + (1/2)]  + C

that is the antiderivative;

thus:

1

∫ x e^(-2x) dx = { {(-1/2)e^[-2(1)]} [1+ (1/2)] } - { {(-1/2)e^[-2(0)]} [0 + (1/2)] }

0

[(-1/2)e^(-2) (3/2)] + [(1/2)e^0)] (1/2) =

(-3/4)e^(-2) + (1/4)

I hope it helps...

Bye!

Answer 3

this is integration by parts

let u = x, dv = e^-2xdx, so du = dx and v = (-1/2)e^-2x

plug em into the formula...you can do this