What is the integral from-infinity to 0 of ze^{4z} ?

The integral from-infinity to 0 of ze^{4z} is -1/16

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integral from-infinity to 0 of ze^{4z}

\int_{- \infty}^{0} z e^{4 z} d z

- \frac{1}{16}

Decimal Notation

- 0.0625

Solution steps

\int_{- \infty}^{0} z e^{4 z} d z

Compute the indefinite integral:

\int z e^{4 z} d z = \frac{1}{16} (e^{4 z} \cdot 4 z - e^{4 z}) + C

Compute the boundaries:

\int_{- \infty}^{0} z e^{4 z} d z = - \frac{1}{16} - 0

= - \frac{1}{16} - 0

Simplify

= - \frac{1}{16}

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x + \frac{1}{x ^{2}}

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n = 2 \sum \infty \frac{6}{5 ^{n}}

\int cos^{4} (x s i) n^{3} x d x

What is the integral from-infinity to 0 of ze^{4z} ?
The integral from-infinity to 0 of ze^{4z} is -1/16