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

The integral from 0 to 1 of y/(e^{4y)} is -5/(16e^4)+1/16

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

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

- \frac{5}{16 e ^{4}} + \frac{1}{16}

Decimal Notation

0.05677 \dots

Solution steps

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

Apply Integration By Parts

: [- \frac{1}{4} e^{- 4 y} y - \int - \frac{1}{4} e^{- 4 y} d y]_{0}^{1}

= [- \frac{1}{4} e^{- 4 y} y - \int - \frac{1}{4} e^{- 4 y} d y]_{0}^{1}

\int - \frac{1}{4} e^{- 4 y} d y = \frac{1}{16} e^{- 4 y}

= [- \frac{1}{4} e^{- 4 y} y - \frac{1}{16} e^{- 4 y}]_{0}^{1}

Compute the boundaries:

- \frac{5}{16 e ^{4}} + \frac{1}{16}

= - \frac{5}{16 e ^{4}} + \frac{1}{16}

\frac{d}{d x} (x x + 6)

x \to \infty lim (- \frac{1}{x} + 1)

f (x) = ln (\frac{4 + x ^{2}}{x})

f (x) = 20 x, (4, 40)

y^{^{''}} + 5 y^{^{'}} - 2 y = 0

What is the integral from 0 to 1 of y/(e^{4y)} ?
The integral from 0 to 1 of y/(e^{4y)} is -5/(16e^4)+1/16