What is the integral from-infinity to-9 of 2e^{4x} ?

The integral from-infinity to-9 of 2e^{4x} is 1/(2e^{36)}

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integral from-infinity to-9 of 2e^{4x}

\int_{- \infty}^{- 9} 2 e^{4 x} d x

\frac{1}{2 e ^{36}}

Solution steps

\int_{- \infty}^{- 9} 2 e^{4 x} d x

Compute the indefinite integral:

\int 2 e^{4 x} d x = \frac{1}{2} e^{4 x} + C

Compute the boundaries:

\int_{- \infty}^{- 9} 2 e^{4 x} d x = \frac{1}{2 e ^{36}} - 0

= \frac{1}{2 e ^{36}} - 0

Simplify

= \frac{1}{2 e ^{36}}

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What is the integral from-infinity to-9 of 2e^{4x} ?
The integral from-infinity to-9 of 2e^{4x} is 1/(2e^{36)}