What is the integral from 0 to 1 of 7e^{-7sqrt(x)} ?

The integral from 0 to 1 of 7e^{-7sqrt(x)} is -(2(-e^7+8))/(7e^7)

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integral from 0 to 1 of 7e^{-7sqrt(x)}

\int_{0}^{1} 7 e^{- 7 x} d x

- \frac{2 ( - e ^{7} + 8 )}{7 e ^{7}}

Decimal

0.28362 \dots

Solution steps

\int_{0}^{1} 7 e^{- 7 x} d x

Take the constant out:

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

= 7 \cdot \int_{0}^{1} e^{- 7 x} d x

Apply u-substitution

: \int_{0}^{1} e^{- 7 u} \cdot 2 u d u

= 7 \cdot \int_{0}^{1} e^{- 7 u} \cdot 2 u d u

Take the constant out:

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

= 7 \cdot 2 \cdot \int_{0}^{1} e^{- 7 u} u d u

Apply u-substitution

: \int_{0}^{- 7} \frac{e ^{v} v}{49} d v

= 7 \cdot 2 \cdot \int_{0}^{- 7} \frac{e ^{v} v}{49} d v

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

= 7 \cdot 2 (- \int_{- 7}^{0} \frac{e ^{v} v}{49} d v)

Take the constant out:

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

= 7 \cdot 2 (- \frac{1}{49} \cdot \int_{- 7}^{0} e^{v} v d v)

Apply Integration By Parts

: [e^{v} v - \int e^{v} d v]_{- 7}^{0}

= 7 \cdot 2 (- \frac{1}{49} [e^{v} v - \int e^{v} d v]_{- 7}^{0})

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

= 7 \cdot 2 (- \frac{1}{49} [e^{v} v - e^{v}]_{- 7}^{0})

Simplify

7 \cdot 2 (- \frac{1}{49} [e^{v} v - e^{v}]_{- 7}^{0}) : - \frac{2}{7} [e^{v} v - e^{v}]_{- 7}^{0}

= - \frac{2}{7} [e^{v} v - e^{v}]_{- 7}^{0}

Compute the boundaries:

- 1 + \frac{8}{e ^{7}}

= - \frac{2}{7} (- 1 + \frac{8}{e ^{7}})

Simplify

= - \frac{2 ( - e ^{7} + 8 )}{7 e ^{7}}

x \to 5 lim (\frac{x + 2}{x - 4})

\int 0.1 t d t

\frac{\partial}{\partial x} (x \cdot ln (y))

\frac{d}{d x} (x^{2} + x^{2} + 1)

y = 2 e^{- 4 x^{2}}

What is the integral from 0 to 1 of 7e^{-7sqrt(x)} ?
The integral from 0 to 1 of 7e^{-7sqrt(x)} is -(2(-e^7+8))/(7e^7)