What is the integral from 1 to-1 of 3e^x ?

The integral from 1 to-1 of 3e^x is -3(e-1/e)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to-1 of 3e^x

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

- 3 (e - \frac{1}{e})

Decimal Notation

- 7.05120 \dots

Solution steps

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

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

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

Take the constant out:

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

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

Use the common integral:

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

= - 3 [e^{x}]_{- 1}^{1}

Compute the boundaries:

e - \frac{1}{e}

= - 3 (e - \frac{1}{e})

t^{2} y^{^{''}} + 2 t y^{^{'}} - 12 y = 0

f (x) = 9 x^{\frac{7}{5}} - 5 x^{2} + 1 0^{4}

3 x + sin (5 x)

y = \frac{x}{4 x + 7}

f (x) = ln (cos (x))

What is the integral from 1 to-1 of 3e^x ?
The integral from 1 to-1 of 3e^x is -3(e-1/e)