What is the integral from 0 to 2 of (4x+5)e^{-x} ?

The integral from 0 to 2 of (4x+5)e^{-x} is -(17)/(e^2)+9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of (4x+5)e^{-x}

\int_{0}^{2} (4 x + 5) e^{- x} d x

- \frac{17}{e ^{2}} + 9

Decimal Notation

6.69930 \dots

Solution steps

\int_{0}^{2} (4 x + 5) e^{- x} d x

Multiply by the conjugate of

4 x + 5 : \frac{16 x ^{2} - 25}{4 x - 5}

= \int_{0}^{2} \frac{16 x ^{2} - 25}{4 x - 5} e^{- x} d x

Find undefined (singularity) points:

x = \frac{5}{4}

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{\frac{5}{4}} \frac{16 x ^{2} - 25}{4 x - 5} e^{- x} d x + \int_{\frac{5}{4}}^{2} \frac{16 x ^{2} - 25}{4 x - 5} e^{- x} d x

\int_{0}^{\frac{5}{4}} \frac{16 x ^{2} - 25}{4 x - 5} e^{- x} d x = - \frac{14}{e ^{\frac{5}{4}}} + 9

\int_{\frac{5}{4}}^{2} \frac{16 x ^{2} - 25}{4 x - 5} e^{- x} d x = - \frac{17 e ^{\frac{5}{4}} - 14 e ^{2}}{e ^{\frac{13}{4}}}

= - \frac{14}{e ^{\frac{5}{4}}} + 9 - \frac{17 e ^{\frac{5}{4}} - 14 e ^{2}}{e ^{\frac{13}{4}}}

Simplify

- \frac{14}{e ^{\frac{5}{4}}} + 9 - \frac{17 e ^{\frac{5}{4}} - 14 e ^{2}}{e ^{\frac{13}{4}}} : - \frac{17}{e ^{2}} + 9

= - \frac{17}{e ^{2}} + 9

\frac{d}{d x} (\frac{2 x ^{2} + 2 x + 4}{x})

y^{^{'}} = \frac{( x y + 2 y )}{( x + 3 )}

\int (4 x) d x

\frac{d y}{d t} = \frac{2 t}{1 + t ^{2}} y + \frac{2}{1 + t ^{2}}

\frac{d}{d x} (9 - 2 x)

What is the integral from 0 to 2 of (4x+5)e^{-x} ?
The integral from 0 to 2 of (4x+5)e^{-x} is -(17)/(e^2)+9