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

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

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integral from 0 to 1 of (-4y)/(e^{2y)}

\int_{0}^{1} \frac{- 4 y}{e ^{2 y}} d y

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

Decimal Notation

- 0.59399 \dots

Solution steps

\int_{0}^{1} \frac{- 4 y}{e ^{2 y}} d y

Take the constant out:

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

= - 4 \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}

= - 4 [- \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}

= - 4 [- \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}

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

Simplify

= \frac{3}{e ^{2}} - 1

\frac{\partial}{\partial x} (6 x^{2} y + 9 x y^{3})

\frac{\partial}{\partial x} (- x y (12 - x - y))

\frac{\partial}{\partial t} (ln (x + t^{2}))

\frac{d}{d x} (\frac{7 + 8 x - 4 x}{x})

\int_{3}^{6} 2 π x (- \frac{3}{2} x + 9) d x

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