What is the integral from 4 to infinity of 9e^{-y/2} ?

The integral from 4 to infinity of 9e^{-y/2} is (18)/(e^2)

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integral from 4 to infinity of 9e^{-y/2}

\int_{4}^{\infty} 9 e^{- \frac{y}{2}} d y

\frac{18}{e ^{2}}

Decimal Notation

2.43603 \dots

Solution steps

\int_{4}^{\infty} 9 e^{- \frac{y}{2}} d y

Compute the indefinite integral:

\int 9 e^{- \frac{y}{2}} d y = - 18 e^{- \frac{y}{2}} + C

Compute the boundaries:

\int_{4}^{\infty} 9 e^{- \frac{y}{2}} d y = 0 - (- \frac{18}{e ^{2}})

= 0 - (- \frac{18}{e ^{2}})

Simplify

= \frac{18}{e ^{2}}

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\frac{\partial}{\partial x} (ln (x y^{4}))

y = 3 x e^{2 x}

\int - 1120 (1 + 4 t)^{- 3} d t

f (x) = 5 x^{2} - x^{3}

What is the integral from 4 to infinity of 9e^{-y/2} ?
The integral from 4 to infinity of 9e^{-y/2} is (18)/(e^2)