What is the integral from 1 to 4 of (5e^{-x})/2 ?

The integral from 1 to 4 of (5e^{-x})/2 is (5(e^3-1))/(2e^4)

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integral from 1 to 4 of (5e^{-x})/2

\int_{1}^{4} \frac{5 e ^{- x}}{2} d x

\frac{5 ( e ^{3} - 1 )}{2 e ^{4}}

Decimal

0.87390 \dots

Solution steps

\int_{1}^{4} \frac{5 e ^{- x}}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{5}{2} \cdot \int_{1}^{4} e^{- x} d x

Apply u-substitution

: \int_{- 1}^{- 4} - e^{u} d u

= \frac{5}{2} \cdot \int_{- 1}^{- 4} - e^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{5}{2} (- \int_{- 4}^{- 1} - e^{u} d u)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{5}{2} (- (- \int_{- 4}^{- 1} e^{u} d u))

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{5}{2} (- (- [e^{u}]_{- 4}^{- 1}))

Simplify

= \frac{5}{2} [e^{u}]_{- 4}^{- 1}

Compute the boundaries:

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

= \frac{5}{2} (\frac{1}{e} - \frac{1}{e ^{4}})

Simplify

= \frac{5 ( e ^{3} - 1 )}{2 e ^{4}}

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What is the integral from 1 to 4 of (5e^{-x})/2 ?
The integral from 1 to 4 of (5e^{-x})/2 is (5(e^3-1))/(2e^4)