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

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

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

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

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

Decimal

0.29699 \dots

Solution steps

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

Take the constant out:

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

= 2 \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}

= 2 [- \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}

= 2 [- \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}

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

y = 2 x^{3} + 3 x^{2} - 12 x + 7

\frac{d}{d x} (e^{2 x} (2 x + 1))

h \to 0 lim (\frac{49 + h - 7}{h})

2 \cdot sin (3 t)

y = e^{3 x}, y = e^{- 3 x}, x = - 1

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