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

The integral from 0 to 1 of y/(e^{9y)} is -(10)/(81e^9)+1/81

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

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

- \frac{10}{81 e ^{9}} + \frac{1}{81}

Decimal

0.01233 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{1}{9} e^{- 9 y} d y = \frac{1}{81} e^{- 9 y}

= [- \frac{1}{9} e^{- 9 y} y - \frac{1}{81} e^{- 9 y}]_{0}^{1}

Compute the boundaries:

- \frac{10}{81 e ^{9}} + \frac{1}{81}

= - \frac{10}{81 e ^{9}} + \frac{1}{81}

10 (t + 1) y^{^{'}} - 8 y = 16 t

y = 2 + 4 x^{2} - 2 x^{3}

\int - 3 d x

\int \frac{1}{x x ^{4} - 4} d x

y^{^{''}} + 5 y = 4 t^{4}, y (0) = 0, y^{^{'}} (0) = 0

What is the integral from 0 to 1 of y/(e^{9y)} ?
The integral from 0 to 1 of y/(e^{9y)} is -(10)/(81e^9)+1/81