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

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

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integral from 0 to 1 of (3x)/(e^{2x)}

\int_{0}^{1} \frac{3 x}{e ^{2 x}} d x

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

Solution steps

\int_{0}^{1} \frac{3 x}{e ^{2 x}} d x

Take the constant out:

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

= 3 \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}

= 3 [- \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}

= 3 [- \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}

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

sin (t) H (t - π)

x^{3}, 0, 2

\frac{\partial}{\partial s} (- s^{2} t^{2})

\frac{d}{d x} (cot (x) sec (x))

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