What is the integral from 0 to 1 of e^{3x}-e^x ?

The integral from 0 to 1 of e^{3x}-e^x is (e^3-1)/3-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 e^{3x}-e^x

\int_{0}^{1} e^{3 x} - e^{x} d x

\frac{e ^{3} - 1}{3} - e + 1

Decimal Notation

4.64356 \dots

Solution steps

\int_{0}^{1} e^{3 x} - e^{x} 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

= \int_{0}^{1} e^{3 x} d x - \int_{0}^{1} e^{x} d x

\int_{0}^{1} e^{3 x} d x = \frac{e ^{3} - 1}{3}

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

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

Simplify

= \frac{e ^{3} - 1}{3} - e + 1

\int \frac{y + 1}{y ^{2} + 2 y - 4} d y

\int 9 x^{4} d x

x \to 3 - lim (c x^{2} + 2 x)

x^{^{'}} = \frac{1}{x ^{2}}

\frac{d}{d x} (x + \frac{x ^{2}}{2})

What is the integral from 0 to 1 of e^{3x}-e^x ?
The integral from 0 to 1 of e^{3x}-e^x is (e^3-1)/3-e+1