What is the integral from 0 to 10 of e^{-sqrt(x)} ?

The integral from 0 to 10 of e^{-sqrt(x)} is -2(-1-(-sqrt(10)-1)/(e^{sqrt(10))})

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integral from 0 to 10 of e^{-sqrt(x)}

\int_{0}^{10} e^{- x} d x

- 2 (- 1 - \frac{- 10 - 1}{e ^{10}})

Solution steps

\int_{0}^{10} e^{- x} d x

Apply u-substitution

: \int_{0}^{- 10} 2 e^{u} u d u

= \int_{0}^{- 10} 2 e^{u} u d u

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

= - \int_{- 10}^{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_{- 10}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 10}^{0}

= - 2 [e^{u} u - \int e^{u} d u]_{- 10}^{0}

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

= - 2 [e^{u} u - e^{u}]_{- 10}^{0}

Compute the boundaries:

- 1 - \frac{- 10 - 1}{e ^{10}}

= - 2 (- 1 - \frac{- 10 - 1}{e ^{10}})

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

\int_{1}^{3} e^{x} - \frac{1}{x} d x

\int_{0}^{1} e^{5 x - 4} d x

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

\int_{0}^{5} (3 x - 8) d x

What is the integral from 0 to 10 of e^{-sqrt(x)} ?
The integral from 0 to 10 of e^{-sqrt(x)} is -2(-1-(-sqrt(10)-1)/(e^{sqrt(10))})