What is the integral from 0 to 1 of (7y)/(e^{2y)} ?

The integral from 0 to 1 of (7y)/(e^{2y)} is 7(-3/(4e^2)+1/4)

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integral from 0 to 1 of (7y)/(e^{2y)}

\int_{0}^{1} \frac{7 y}{e ^{2 y}} d y

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

Decimal Notation

1.03948 \dots

Solution steps

\int_{0}^{1} \frac{7 y}{e ^{2 y}} d y

Take the constant out:

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

= 7 \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}

= 7 [- \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}

= 7 [- \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}

= 7 (- \frac{3}{4 e ^{2}} + \frac{1}{4})

- 2 y^{^{'}} + \frac{2 y}{x} = x^{4} y^{3}

n = 1 \sum \infty \frac{( n ^{n} )}{n !}

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

\int sin (2 x) cos (2 x) tan (2 x) d x

2 sin^{3} (t)

What is the integral from 0 to 1 of (7y)/(e^{2y)} ?
The integral from 0 to 1 of (7y)/(e^{2y)} is 7(-3/(4e^2)+1/4)