What is the integral from 0 to 1 of 6e^{-6sqrt(x)} ?

The integral from 0 to 1 of 6e^{-6sqrt(x)} is -(-e^6+7)/(3e^6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 1 of 6e^{-6sqrt(x)}

\int_{0}^{1} 6 e^{- 6 x} d x

- \frac{- e ^{6} + 7}{3 e ^{6}}

Decimal

0.32754 \dots

Solution steps

\int_{0}^{1} 6 e^{- 6 x} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} e^{- 6 x} d x

Apply u-substitution

= 6 \cdot \int_{0}^{1} e^{- 6 u} \cdot 2 u d u

Take the constant out:

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

= 6 \cdot 2 \cdot \int_{0}^{1} e^{- 6 u} u d u

Apply u-substitution

= 6 \cdot 2 \cdot \int_{0}^{- 6} \frac{e ^{v} v}{36} d v

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

= 6 \cdot 2 (- \int_{- 6}^{0} \frac{e ^{v} v}{36} d v)

Take the constant out:

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

= 6 \cdot 2 (- \frac{1}{36} \cdot \int_{- 6}^{0} e^{v} v d v)

Apply Integration By Parts

= 6 \cdot 2 (- \frac{1}{36} [e^{v} v - \int e^{v} d v]_{- 6}^{0})

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

= 6 \cdot 2 (- \frac{1}{36} [e^{v} v - e^{v}]_{- 6}^{0})

Simplify

6 \cdot 2 (- \frac{1}{36} [e^{v} v - e^{v}]_{- 6}^{0}) : - \frac{1}{3} [e^{v} v - e^{v}]_{- 6}^{0}

= - \frac{1}{3} [e^{v} v - e^{v}]_{- 6}^{0}

Compute the boundaries:

- 1 + \frac{7}{e ^{6}}

= - \frac{1}{3} (- 1 + \frac{7}{e ^{6}})

Simplify

= - \frac{- e ^{6} + 7}{3 e ^{6}}

What is the integral from 0 to 1 of 6e^{-6sqrt(x)} ?
The integral from 0 to 1 of 6e^{-6sqrt(x)} is -(-e^6+7)/(3e^6)