What is the integral from 0 to 1 of (9x)/(e^{2x)} ?

The integral from 0 to 1 of (9x)/(e^{2x)} is 9(-3/(4e^2)+1/4)

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integral from 0 to 1 of (9x)/(e^{2x)}

\int_{0}^{1} \frac{9 x}{e ^{2 x}} d x

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

Decimal Notation

1.33648 \dots

Solution steps

\int_{0}^{1} \frac{9 x}{e ^{2 x}} d x

Take the constant out:

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

= 9 \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}

= 9 [- \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}

= 9 [- \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}

= 9 (- \frac{3}{4 e ^{2}} + \frac{1}{4})

\frac{d}{d x} (1 + \frac{3}{x})

\int \frac{1}{50 + t} d t

f (x) = 3 t^{4} - 7 t^{3} + 2

x \to 1 lim (sin (3 p i x))

What is the integral from 0 to 1 of (9x)/(e^{2x)} ?
The integral from 0 to 1 of (9x)/(e^{2x)} is 9(-3/(4e^2)+1/4)