What is the integral from 0 to 1 of (e^{2x}-5) ?

The integral from 0 to 1 of (e^{2x}-5) is (e^2-1)/2-5

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^{2x}-5)

\int_{0}^{1} (e^{2 x} - 5) d x

\frac{e ^{2} - 1}{2} - 5

Decimal Notation

- 1.80547 \dots

Solution steps

\int_{0}^{1} e^{2 x} - 5 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^{2 x} d x - \int_{0}^{1} 5 d x

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

\int_{0}^{1} 5 d x = 5

= \frac{e ^{2} - 1}{2} - 5

\int_{0}^{y} 2 e^{- 2 x} d x

\int_{1}^{\infty} \frac{1}{2 x + 6} d x

\int_{1}^{2} \frac{x ^{6} + 1}{2 x ^{3}} d x

\int_{0}^{t} e^{- t + x} cos (t) d x

\int_{1}^{4} \frac{1}{x ^{2}} ln (6 x) d x

What is the integral from 0 to 1 of (e^{2x}-5) ?
The integral from 0 to 1 of (e^{2x}-5) is (e^2-1)/2-5