What is the integral from 2 to infinity of e^{-3p} ?

The integral from 2 to infinity of e^{-3p} is 1/(3e^6)

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integral from 2 to infinity of e^{-3p}

\int_{2}^{\infty} e^{- 3 p} d p

\frac{1}{3 e ^{6}}

Decimal

0.00082 \dots

Solution steps

\int_{2}^{\infty} e^{- 3 p} d p

Compute the indefinite integral:

\int e^{- 3 p} d p = - \frac{1}{3} e^{- 3 p} + C

Compute the boundaries:

\int_{2}^{\infty} e^{- 3 p} d p = 0 - (- \frac{1}{3 e ^{6}})

= 0 - (- \frac{1}{3 e ^{6}})

Simplify

= \frac{1}{3 e ^{6}}

\int \frac{1}{e ^{π}} d x

\int_{0}^{\frac{π}{3}} x sin (2 x) d x

\frac{1}{( s + 1 ) ^{3}}

n = 1 \sum \infty \frac{2}{n ^{\frac{3}{2}}}

x y^{2} y^{^{'}} = x + 1

What is the integral from 2 to infinity of e^{-3p} ?
The integral from 2 to infinity of e^{-3p} is 1/(3e^6)