What is the integral from 1 to infinity of 4e^{-4x} ?

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

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integral from 1 to infinity of 4e^{-4x}

\int_{1}^{\infty} 4 e^{- 4 x} d x

\frac{1}{e ^{4}}

Decimal Notation

0.01831 \dots

Solution steps

\int_{1}^{\infty} 4 e^{- 4 x} d x

Compute the indefinite integral:

\int 4 e^{- 4 x} d x = - e^{- 4 x} + C

Compute the boundaries:

\int_{1}^{\infty} 4 e^{- 4 x} d x = 0 - (- \frac{1}{e ^{4}})

= 0 - (- \frac{1}{e ^{4}})

Simplify

= \frac{1}{e ^{4}}

\int_{3}^{2} 1

\int_{\frac{1}{2}}^{1} (x^{- 3} - 15) d x

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\int_{0}^{1} 3 - 4 x^{2} + 3 x^{\frac{3}{4}} d x

\int_{- 4}^{4} x^{2} + y^{2} d x

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