What is the integral from 0 to 1 of x/(e^{7x)} ?

The integral from 0 to 1 of x/(e^{7x)} is -8/(49e^7)+1/49

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integral from 0 to 1 of x/(e^{7x)}

\int_{0}^{1} \frac{x}{e ^{7 x}} d x

- \frac{8}{49 e ^{7}} + \frac{1}{49}

Decimal Notation

0.02025 \dots

Solution steps

\int_{0}^{1} \frac{x}{e ^{7 x}} d x

Apply Integration By Parts

: [- \frac{1}{7} e^{- 7 x} x - \int - \frac{1}{7} e^{- 7 x} d x]_{0}^{1}

= [- \frac{1}{7} e^{- 7 x} x - \int - \frac{1}{7} e^{- 7 x} d x]_{0}^{1}

\int - \frac{1}{7} e^{- 7 x} d x = \frac{1}{49} e^{- 7 x}

= [- \frac{1}{7} e^{- 7 x} x - \frac{1}{49} e^{- 7 x}]_{0}^{1}

Compute the boundaries:

- \frac{8}{49 e ^{7}} + \frac{1}{49}

= - \frac{8}{49 e ^{7}} + \frac{1}{49}

cos (arctan (2 x))

\frac{\partial}{\partial x} (z x)

\frac{d}{d t} (sin (t) (1 - sin (t)))

\frac{d y}{d x} + 2 = x

x \to 0 lim (+ sin (x) ln (x))

What is the integral from 0 to 1 of x/(e^{7x)} ?
The integral from 0 to 1 of x/(e^{7x)} is -8/(49e^7)+1/49