What is the integral from 0 to infinity of 14e^{-7x} ?

The integral from 0 to infinity of 14e^{-7x} is 2

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integral from 0 to infinity of 14e^{-7x}

\int_{0}^{\infty} 14 e^{- 7 x} d x

2

Solution steps

\int_{0}^{\infty} 14 e^{- 7 x} d x

Compute the indefinite integral:

\int 14 e^{- 7 x} d x = - 2 e^{- 7 x} + C

Compute the boundaries:

\int_{0}^{\infty} 14 e^{- 7 x} d x = 0 - (- 2)

= 0 - (- 2)

Simplify

= 2

8 x - 3

x \to 0 lim (x cot (5 x))

\int (2 x - 5) (x - 3)^{9} d x

(y^{2} + 16) d x - (\frac{1}{( 3 x + 5 ) ^{2}}) d y = 0

\int \frac{12}{12 + e ^{x}} d x

What is the integral from 0 to infinity of 14e^{-7x} ?
The integral from 0 to infinity of 14e^{-7x} is 2