What is the integral from 0 to 1 of 5e^{-3sqrt(x)} ?

The integral from 0 to 1 of 5e^{-3sqrt(x)} is -(10(-e^3+4))/(9e^3)

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integral from 0 to 1 of 5e^{-3sqrt(x)}

\int_{0}^{1} 5 e^{- 3 x} d x

- \frac{10 ( - e ^{3} + 4 )}{9 e ^{3}}

Decimal Notation

0.88983 \dots

Solution steps

\int_{0}^{1} 5 e^{- 3 x} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} e^{- 3 x} d x

Apply u-substitution

: \int_{0}^{1} e^{- 3 u} \cdot 2 u d u

= 5 \cdot \int_{0}^{1} e^{- 3 u} \cdot 2 u d u

Take the constant out:

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

= 5 \cdot 2 \cdot \int_{0}^{1} e^{- 3 u} u d u

Apply u-substitution

: \int_{0}^{- 3} \frac{e ^{v} v}{9} d v

= 5 \cdot 2 \cdot \int_{0}^{- 3} \frac{e ^{v} v}{9} d v

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

= 5 \cdot 2 (- \int_{- 3}^{0} \frac{e ^{v} v}{9} d v)

Take the constant out:

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

= 5 \cdot 2 (- \frac{1}{9} \cdot \int_{- 3}^{0} e^{v} v d v)

Apply Integration By Parts

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

= 5 \cdot 2 (- \frac{1}{9} [e^{v} v - \int e^{v} d v]_{- 3}^{0})

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

= 5 \cdot 2 (- \frac{1}{9} [e^{v} v - e^{v}]_{- 3}^{0})

Simplify

5 \cdot 2 (- \frac{1}{9} [e^{v} v - e^{v}]_{- 3}^{0}) : - \frac{10}{9} [e^{v} v - e^{v}]_{- 3}^{0}

= - \frac{10}{9} [e^{v} v - e^{v}]_{- 3}^{0}

Compute the boundaries:

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

= - \frac{10}{9} (- 1 + \frac{4}{e ^{3}})

Simplify

= - \frac{10 ( - e ^{3} + 4 )}{9 e ^{3}}

\int_{7}^{9} (x^{2} + 2 x - 4) d x

\frac{d r}{d θ} = 10 - 6 r

\int \frac{tan ( z )}{cos ^{2} ( z )} d z

x \to 1 + lim (\frac{4 x - 3}{x - 1})

f (x) = ln (- x)

What is the integral from 0 to 1 of 5e^{-3sqrt(x)} ?
The integral from 0 to 1 of 5e^{-3sqrt(x)} is -(10(-e^3+4))/(9e^3)