What is the integral from 0 to 1 of (-12e^x) ?

The integral from 0 to 1 of (-12e^x) is -12(e-1)

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 (-12e^x)

\int_{0}^{1} (- 12 e^{x}) d x

- 12 (e - 1)

Decimal

- 20.61938 \dots

Solution steps

\int_{0}^{1} - 12 e^{x} d x

Take the constant out:

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

= - 12 \cdot \int_{0}^{1} e^{x} d x

Use the common integral:

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

= - 12 [e^{x}]_{0}^{1}

Compute the boundaries:

e - 1

= - 12 (e - 1)

x \to \infty lim (\frac{- 1}{x})

\frac{d}{d x} (x^{9} f (x))

\frac{d}{d x} (c e^{\frac{x}{6}})

\int 4 x^{2} - sin (x) + C d x

\int \frac{1}{x ^{3}} + x d x

What is the integral from 0 to 1 of (-12e^x) ?
The integral from 0 to 1 of (-12e^x) is -12(e-1)