What is the integral from 0 to 12 of e^{-sqrt(x)} ?

The integral from 0 to 12 of e^{-sqrt(x)} is -2(-1-(-2sqrt(3)-1)/(e^{2sqrt(3))})

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integral from 0 to 12 of e^{-sqrt(x)}

\int_{0}^{12} e^{- x} d x

- 2 (- 1 - \frac{- 2 3 - 1}{e ^{23}})

Decimal Notation

1.72053 \dots

Solution steps

\int_{0}^{12} e^{- x} d x

Apply u-substitution

: \int_{0}^{- 23} 2 e^{u} u d u

= \int_{0}^{- 23} 2 e^{u} u d u

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

= - \int_{- 23}^{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_{- 23}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 23}^{0}

= - 2 [e^{u} u - \int e^{u} d u]_{- 23}^{0}

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

= - 2 [e^{u} u - e^{u}]_{- 23}^{0}

Compute the boundaries:

- 1 - \frac{- 2 3 - 1}{e ^{23}}

= - 2 (- 1 - \frac{- 2 3 - 1}{e ^{23}})

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

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

\int_{0}^{1} 2 e^{\frac{x}{7}} + 5 d x

\int_{0}^{t} sin (x) \cdot sin (t - x) d x

\int_{0}^{4} x 25 - x^{2} d x

What is the integral from 0 to 12 of e^{-sqrt(x)} ?
The integral from 0 to 12 of e^{-sqrt(x)} is -2(-1-(-2sqrt(3)-1)/(e^{2sqrt(3))})