What is the integral from 0 to infinity of e^{5-2x} ?

The integral from 0 to infinity of e^{5-2x} is (e^5)/2

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integral from 0 to infinity of e^{5-2x}

\int_{0}^{\infty} e^{5 - 2 x} d x

\frac{e ^{5}}{2}

Decimal

74.20657 \dots

Solution steps

\int_{0}^{\infty} e^{5 - 2 x} d x

Compute the indefinite integral:

\int e^{5 - 2 x} d x = - \frac{1}{2} e^{5 - 2 x} + C

Compute the boundaries:

\int_{0}^{\infty} e^{5 - 2 x} d x = 0 - (- \frac{e ^{5}}{2})

= 0 - (- \frac{e ^{5}}{2})

Simplify

= \frac{e ^{5}}{2}

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What is the integral from 0 to infinity of e^{5-2x} ?
The integral from 0 to infinity of e^{5-2x} is (e^5)/2