What is the integral from 3 to infinity of e^{-7x} ?

The integral from 3 to infinity of e^{-7x} is 1/(7e^{21)}

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integral from 3 to infinity of e^{-7x}

\int_{3}^{\infty} e^{- 7 x} d x

\frac{1}{7 e ^{21}}

Decimal

1.08322 E - 10

Solution steps

\int_{3}^{\infty} e^{- 7 x} d x

Compute the indefinite integral:

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

Compute the boundaries:

\int_{3}^{\infty} e^{- 7 x} d x = 0 - (- \frac{1}{7 e ^{21}})

= 0 - (- \frac{1}{7 e ^{21}})

Simplify

= \frac{1}{7 e ^{21}}

x \to 7 - lim (\frac{4}{( x - 7 ) ^{2}})

y = - 6 x + k

\int_{0}^{1} 1 (x^{2} + 1) e^{- x} d x

\frac{5}{x ^{5}} + \frac{3}{x ^{2}}

- 3 x^{2} - 5 x

What is the integral from 3 to infinity of e^{-7x} ?
The integral from 3 to infinity of e^{-7x} is 1/(7e^{21)}