What is the integral from 0 to 9 of e^{-sqrt(x)} ?

The integral from 0 to 9 of e^{-sqrt(x)} is -2(-1+4/(e^3))

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integral from 0 to 9 of e^{-sqrt(x)}

\int_{0}^{9} e^{- x} d x

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

Decimal Notation

1.60170 \dots

Solution steps

\int_{0}^{9} e^{- x} d x

Apply u-substitution

: \int_{0}^{- 3} 2 e^{u} u d u

= \int_{0}^{- 3} 2 e^{u} u d u

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

= - \int_{- 3}^{0} 2 e^{u} u d u

Take the constant out:

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

= - 2 \cdot \int_{- 3}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 3}^{0}

= - 2 [e^{u} u - \int e^{u} d u]_{- 3}^{0}

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

= - 2 [e^{u} u - e^{u}]_{- 3}^{0}

Compute the boundaries:

- 1 + \frac{4}{e ^{3}}

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

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What is the integral from 0 to 9 of e^{-sqrt(x)} ?
The integral from 0 to 9 of e^{-sqrt(x)} is -2(-1+4/(e^3))