What is the integral from 0 to k of 1/2 (e^x-1) ?

The integral from 0 to k of 1/2 (e^x-1) is 1/2 (e^k-1-k)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to k of 1/2 (e^x-1)

\int_{0}^{k} \frac{1}{2} (e^{x} - 1) d x

\frac{1}{2} (e^{k} - 1 - k)

Solution steps

\int_{0}^{k} \frac{1}{2} (e^{x} - 1) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{k} e^{x} - 1 d x

Apply the Sum Rule:

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

= \frac{1}{2} (\int_{0}^{k} e^{x} d x - \int_{0}^{k} 1 d x)

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

\int_{0}^{k} 1 d x = k

= \frac{1}{2} (e^{k} - 1 - k)

\int x csc^{2} (3 x) d x

\frac{\partial}{\partial y} (tan (7 + 2 x^{2} y^{3} z^{2}))

\frac{d}{d x} (cosh (2 x))

\int - \frac{1}{e ^{x} ( 8 + e ^{- x} ) ^{2}} d x

x \to 0 + lim (1 4^{x} + 2)

What is the integral from 0 to k of 1/2 (e^x-1) ?
The integral from 0 to k of 1/2 (e^x-1) is 1/2 (e^k-1-k)