What is the integral from 0 to 1 of (9y)/(e^{2y)} ?

The integral from 0 to 1 of (9y)/(e^{2y)} is 9(-3/(4e^2)+1/4)

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integral from 0 to 1 of (9y)/(e^{2y)}

\int_{0}^{1} \frac{9 y}{e ^{2 y}} d y

9 (- \frac{3}{4 e ^{2}} + \frac{1}{4})

Decimal

1.33648 \dots

Solution steps

\int_{0}^{1} \frac{9 y}{e ^{2 y}} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int_{0}^{1} \frac{y}{e ^{2 y}} d y

Apply Integration By Parts

: [- \frac{1}{2} e^{- 2 y} y - \int - \frac{1}{2} e^{- 2 y} d y]_{0}^{1}

= 9 [- \frac{1}{2} e^{- 2 y} y - \int - \frac{1}{2} e^{- 2 y} d y]_{0}^{1}

\int - \frac{1}{2} e^{- 2 y} d y = \frac{1}{4} e^{- 2 y}

= 9 [- \frac{1}{2} e^{- 2 y} y - \frac{1}{4} e^{- 2 y}]_{0}^{1}

Compute the boundaries:

- \frac{3}{4 e ^{2}} + \frac{1}{4}

= 9 (- \frac{3}{4 e ^{2}} + \frac{1}{4})

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What is the integral from 0 to 1 of (9y)/(e^{2y)} ?
The integral from 0 to 1 of (9y)/(e^{2y)} is 9(-3/(4e^2)+1/4)