What is the integral from-1 to-4 of (2e^x+5) ?

The integral from-1 to-4 of (2e^x+5) is -2/e+2/(e^4)-15

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to-4 of (2e^x+5)

\int_{- 1}^{- 4} (2 e^{x} + 5) d x

- \frac{2}{e} + \frac{2}{e ^{4}} - 15

Decimal Notation

- 15.69912 \dots

Solution steps

\int_{- 1}^{- 4} 2 e^{x} + 5 d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 4}^{- 1} 2 e^{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_{- 4}^{- 1} 2 e^{x} d x + \int_{- 4}^{- 1} 5 d x)

\int_{- 4}^{- 1} 2 e^{x} d x = 2 (\frac{1}{e} - \frac{1}{e ^{4}})

\int_{- 4}^{- 1} 5 d x = 15

= - (2 (\frac{1}{e} - \frac{1}{e ^{4}}) + 15)

Simplify

- (2 (\frac{1}{e} - \frac{1}{e ^{4}}) + 15) : - \frac{2}{e} + \frac{2}{e ^{4}} - 15

= - \frac{2}{e} + \frac{2}{e ^{4}} - 15

\int_{x}^{x} x d x

\int_{- 1}^{x} x + 1 d x

\int_{2}^{4} (x)^{2} d x

\int_{- π}^{0} (π + t) cos (n t) d t

\int_{500}^{10 x + 500} \frac{1}{75000} d y

What is the integral from-1 to-4 of (2e^x+5) ?
The integral from-1 to-4 of (2e^x+5) is -2/e+2/(e^4)-15