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

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

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

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

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

Decimal Notation

- 0.74249 \dots

Solution steps

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

Take the constant out:

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

= - 5 \cdot \int_{0}^{1} \frac{x}{e ^{2 x}} d x

Apply Integration By Parts

: [- \frac{1}{2} e^{- 2 x} x - \int - \frac{1}{2} e^{- 2 x} d x]_{0}^{1}

= - 5 [- \frac{1}{2} e^{- 2 x} x - \int - \frac{1}{2} e^{- 2 x} d x]_{0}^{1}

\int - \frac{1}{2} e^{- 2 x} d x = \frac{1}{4} e^{- 2 x}

= - 5 [- \frac{1}{2} e^{- 2 x} x - \frac{1}{4} e^{- 2 x}]_{0}^{1}

Compute the boundaries:

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

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

\frac{d}{d x} (16 (x + 2)^{3})

\frac{2}{5}

f (x) = \frac{3 x}{7 x ^{2} + 1}

\frac{\partial}{\partial x} (3 y - \frac{x}{y ^{2}})

2 - 0.5 e^{- 10 t}

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