What is the integral from-6 to 6 of sqrt(2+2e^{2t)} ?

The integral from-6 to 6 of sqrt(2+2e^{2t)} is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from-6 to 6 of sqrt(2+2e^{2t)}

\int_{- 6}^{6} 2 + 2 e^{2 t} d t

- 2 \frac{ln ( 1 + e ^{12} + 1 ) - ln ( ( 1 + e ^{12} - 1 ) ( 1 + \frac{1}{e ^{12}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{12}} - 1 )}{2} - 1 + e^{12} + 1 + \frac{1}{e ^{12}}

Decimal

578.58404 \dots

Solution steps

\int_{- 6}^{6} 2 + 2 e^{2 t} d t

2 + 2 e^{2 t} = 2 (1 + \frac{2 e ^{2 t}}{2})

= \int_{- 6}^{6} 2 (1 + \frac{2 e ^{2 t}}{2}) d t

Simplify

2 (1 + \frac{2 e ^{2 t}}{2}) : 2 1 + e^{2 t}

= \int_{- 6}^{6} 2 1 + e^{2 t} d t

Take the constant out:

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

= 2 \cdot \int_{- 6}^{6} 1 + e^{2 t} d t

Apply u-substitution

= 2 \cdot \int_{\frac{1}{e ^{12}}}^{e^{12}} \frac{1 + u}{2 u} d u

Take the constant out:

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

= 2 \frac{1}{2} \cdot \int_{\frac{1}{e ^{12}}}^{e^{12}} \frac{1 + u}{u} d u

Apply u-substitution

= 2 \frac{1}{2} \cdot \int_{1 + \frac{1}{e ^{12}}}^{1 + e^{12}} \frac{2 v ^{2}}{v ^{2} - 1} d v

Take the constant out:

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

= 2 \frac{1}{2} \cdot 2 \cdot \int_{1 + \frac{1}{e ^{12}}}^{1 + e^{12}} \frac{v ^{2}}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= 2 \frac{1}{2} \cdot 2 \cdot \int_{1 + \frac{1}{e ^{12}}}^{1 + e^{12}} \frac{v ^{2}}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 2 \frac{1}{2} \cdot 2 (- \int_{1 + \frac{1}{e ^{12}}}^{1 + e^{12}} \frac{v ^{2}}{- v ^{2} + 1} d v)

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

= 2 \frac{1}{2} \cdot 2 (- \int_{1 + \frac{1}{e ^{12}}}^{1 + e^{12}} \frac{1}{- v ^{2} + 1} - 1 d v)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \frac{1}{2} \cdot 2 (- (\int_{1 + \frac{1}{e ^{12}}}^{1 + e^{12}} \frac{1}{- v ^{2} + 1} d v - \int_{1 + \frac{1}{e ^{12}}}^{1 + e^{12}} 1 d v))

\int_{1 + \frac{1}{e ^{12}}}^{1 + e^{12}} \frac{1}{- v ^{2} + 1} d v = \frac{ln ( 1 + e ^{12} + 1 ) - ln ( ( 1 + e ^{12} - 1 ) ( 1 + \frac{1}{e ^{12}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{12}} - 1 )}{2}

\int_{1 + \frac{1}{e ^{12}}}^{1 + e^{12}} 1 d v = 1 + e^{12} - 1 + \frac{1}{e ^{12}}

= 2 \frac{1}{2} \cdot 2 - \frac{ln ( 1 + e ^{12} + 1 ) - ln ( ( 1 + e ^{12} - 1 ) ( 1 + \frac{1}{e ^{12}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{12}} - 1 )}{2} - (1 + e^{12} - 1 + \frac{1}{e ^{12}})

Simplify

2 \frac{1}{2} \cdot 2 - \frac{ln ( 1 + e ^{12} + 1 ) - ln ( ( 1 + e ^{12} - 1 ) ( 1 + \frac{1}{e ^{12}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{12}} - 1 )}{2} - (1 + e^{12} - 1 + \frac{1}{e ^{12}}) : - 2 \frac{ln ( 1 + e ^{12} + 1 ) - ln ( ( 1 + e ^{12} - 1 ) ( 1 + \frac{1}{e ^{12}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{12}} - 1 )}{2} - 1 + e^{12} + 1 + \frac{1}{e ^{12}}

= - 2 \frac{ln ( 1 + e ^{12} + 1 ) - ln ( ( 1 + e ^{12} - 1 ) ( 1 + \frac{1}{e ^{12}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{12}} - 1 )}{2} - 1 + e^{12} + 1 + \frac{1}{e ^{12}}

What is the integral from-6 to 6 of sqrt(2+2e^{2t)} ?
The integral from-6 to 6 of sqrt(2+2e^{2t)} is