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

The integral from 0 to 1 of 3e^{-4sqrt(x)} is -(3(-e^4+5))/(8e^4)

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integral from 0 to 1 of 3e^{-4sqrt(x)}

\int_{0}^{1} 3 e^{- 4 x} d x

- \frac{3 ( - e ^{4} + 5 )}{8 e ^{4}}

Decimal

0.34065 \dots

Solution steps

\int_{0}^{1} 3 e^{- 4 x} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{1} e^{- 4 x} d x

Apply u-substitution

: \int_{0}^{1} e^{- 4 u} \cdot 2 u d u

= 3 \cdot \int_{0}^{1} e^{- 4 u} \cdot 2 u d u

Take the constant out:

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

= 3 \cdot 2 \cdot \int_{0}^{1} e^{- 4 u} u d u

Apply u-substitution

: \int_{0}^{- 4} \frac{e ^{v} v}{16} d v

= 3 \cdot 2 \cdot \int_{0}^{- 4} \frac{e ^{v} v}{16} d v

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

= 3 \cdot 2 (- \int_{- 4}^{0} \frac{e ^{v} v}{16} d v)

Take the constant out:

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

= 3 \cdot 2 (- \frac{1}{16} \cdot \int_{- 4}^{0} e^{v} v d v)

Apply Integration By Parts

: [e^{v} v - \int e^{v} d v]_{- 4}^{0}

= 3 \cdot 2 (- \frac{1}{16} [e^{v} v - \int e^{v} d v]_{- 4}^{0})

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

= 3 \cdot 2 (- \frac{1}{16} [e^{v} v - e^{v}]_{- 4}^{0})

Simplify

3 \cdot 2 (- \frac{1}{16} [e^{v} v - e^{v}]_{- 4}^{0}) : - \frac{3}{8} [e^{v} v - e^{v}]_{- 4}^{0}

= - \frac{3}{8} [e^{v} v - e^{v}]_{- 4}^{0}

Compute the boundaries:

- 1 + \frac{5}{e ^{4}}

= - \frac{3}{8} (- 1 + \frac{5}{e ^{4}})

Simplify

= - \frac{3 ( - e ^{4} + 5 )}{8 e ^{4}}

\frac{\partial ^{2}}{\partial x \partial y} (x arcsin (\frac{y}{x}))

\frac{d y}{d t} = e^{4 t + 5 y}

y = ln (π)

\frac{d}{d x} (\frac{4 x ^{2} + 4 x + 8}{x})

\int x (ln (x))^{2} d x

What is the integral from 0 to 1 of 3e^{-4sqrt(x)} ?
The integral from 0 to 1 of 3e^{-4sqrt(x)} is -(3(-e^4+5))/(8e^4)