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

Popular >

integral from 0 to 1 of-12e^x

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

- 12 (e - 1)

Decimal Notation

- 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)

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

ln (x), 6

\frac{d}{d x} (8 + 3 sin (x))

\int_{0}^{1} x^{\frac{1}{2}} d x

n = 1 \sum \infty \frac{1}{2 ^{n} n !}

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