What is the integral from 0 to 1 of-(5y)/(e^{2y)} ?

The integral from 0 to 1 of-(5y)/(e^{2y)} is -5(-3/(4e^2)+1/4)

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integral from 0 to 1 of-(5y)/(e^{2y)}

\int_{0}^{1} - \frac{5 y}{e ^{2 y}} d y

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

Decimal Notation

- 0.74249 \dots

Solution steps

\int_{0}^{1} - \frac{5 y}{e ^{2 y}} d y

Take the constant out:

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

= - 5 \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}

= - 5 [- \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}

= - 5 [- \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}

= - 5 (- \frac{3}{4 e ^{2}} + \frac{1}{4})

- x^{3} + x, x = - 1, y = - \frac{1}{2}

\int \frac{5 x + 1}{x ^{2} + 4 x + 5} d x

\int_{0}^{5} \frac{2}{x ^{2}} d x

\frac{d y}{d t} = - \frac{y}{t} - 7

\frac{d}{d x} (\frac{2}{( 1 - x ) ^{3}})

What is the integral from 0 to 1 of-(5y)/(e^{2y)} ?
The integral from 0 to 1 of-(5y)/(e^{2y)} is -5(-3/(4e^2)+1/4)