Calculus II Integration by parts

Question

Evaluate the integral: from 0 to 1 ∫(-2y)/(e^2y) dy

Answer 1

-2 integral [ y*dy/ e^2y ]

= -2 integral [ y * e^(-2y) dy ]

IBP:

U = y dV = e^(-2y) dy

dU = dy V = (-1/2)e^(-2y)

UV - integral [ V du] =

(-1/2)y e^(-2y) - integral [ (-1/2)e^(-2y) dy]

(-1/2) y e^(-2y) + (1/2) integral [ e^(-2y) dy

(-1/2) y e^(2y) + (1/2) (-1/2) e^(-2y)

(-1/2) y e^(2y) - (1/4) e^(-2y)

However, the ORIGINAL integral is multiplied by -2

PRIOR to the IBP. So the anti-derivative is:

y e^(-2y) + 1/2 e^(-2y)

check by differentiation:

(-2)y e^(-2y) + e^(-2y) - e^(-2y)

(-2)y e^(-2y)

which is the original function!

therefore the anti-derivative is correct!

limit as y->0: y e^(-2y) + 1/2 e^(-2y)

0 * 1 + 1/2 * 1 = 1/2

limit as y->1: 1 * 1/e^2 + 1/2 * 1/ e^2

= 1/e^2 [ 1+ 1/2]

= (3/2) (1/e^2)

subtracting the limits:

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