What is the integral from 0 to 1 of (-10y)/(e^{2y)} ?

The integral from 0 to 1 of (-10y)/(e^{2y)} is -10(-3/(4e^2)+1/4)

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integral from 0 to 1 of (-10y)/(e^{2y)}

\int_{0}^{1} \frac{- 10 y}{e ^{2 y}} d y

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

Decimal

- 1.48498 \dots

Solution steps

\int_{0}^{1} \frac{- 10 y}{e ^{2 y}} d y

Take the constant out:

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

= - 10 \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}

= - 10 [- \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}

= - 10 [- \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}

= - 10 (- \frac{3}{4 e ^{2}} + \frac{1}{4})

x \to 3 + lim (\frac{( x - 8 )}{x - 3})

x \to 0 + lim ((x) (e^{\frac{1}{x}}))

y = \frac{x ^{2}}{4 + x}

n = 0 \sum \infty \frac{n}{2 n + 1}

f (x) = \frac{64 x}{x ^{2} + 64}

What is the integral from 0 to 1 of (-10y)/(e^{2y)} ?
The integral from 0 to 1 of (-10y)/(e^{2y)} is -10(-3/(4e^2)+1/4)