What is the integral from-1 to 0 of 8(e^{1/x})/(x^3) ?

The integral from-1 to 0 of 8(e^{1/x})/(x^3) is -(16)/e

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 0 of 8(e^{1/x})/(x^3)

\int_{- 1}^{0} 8 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

- \frac{16}{e}

Decimal Notation

- 5.88607 \dots

Solution steps

\int_{- 1}^{0} 8 \cdot \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

Take the constant out:

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

= 8 \cdot \int_{- 1}^{0} \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

Apply Integration By Parts

: [\frac{e ^{\frac{1}{x}}}{x ^{2}} - \int - \frac{e ^{\frac{1}{x}} ( 3 x + 1 )}{x ^{4}} d x]_{- 1}^{0}

= 8 [\frac{e ^{\frac{1}{x}}}{x ^{2}} - \int - \frac{e ^{\frac{1}{x}} ( 3 x + 1 )}{x ^{4}} d x]_{- 1}^{0}

\int - \frac{e ^{\frac{1}{x}} ( 3 x + 1 )}{x ^{4}} d x = \frac{3 e ^{\frac{1}{x}}}{x} - e^{\frac{1}{x}} + \frac{e ^{\frac{1}{x}}}{x ^{2}} - \frac{2 e ^{\frac{1}{x}}}{x}

= 8 [\frac{e ^{\frac{1}{x}}}{x ^{2}} - (\frac{3 e ^{\frac{1}{x}}}{x} - e^{\frac{1}{x}} + \frac{e ^{\frac{1}{x}}}{x ^{2}} - \frac{2 e ^{\frac{1}{x}}}{x})]_{- 1}^{0}

Simplify

8 [\frac{e ^{\frac{1}{x}}}{x ^{2}} - (\frac{3 e ^{\frac{1}{x}}}{x} - e^{\frac{1}{x}} + \frac{e ^{\frac{1}{x}}}{x ^{2}} - \frac{2 e ^{\frac{1}{x}}}{x})]_{- 1}^{0} : 8 [\frac{- e ^{\frac{1}{x}} + e ^{\frac{1}{x}} x}{x}]_{- 1}^{0}

= 8 [\frac{- e ^{\frac{1}{x}} + e ^{\frac{1}{x}} x}{x}]_{- 1}^{0}

Compute the boundaries:

- \frac{2}{e}

= 8 (- \frac{2}{e})

Simplify

= - \frac{16}{e}

y^{^{'}} = - \frac{x y}{x ^{2} + y ^{2}}

\int \frac{x ^{2}}{1 + x ^{2}} d x

x y^{^{'}} + 4 y = 8 x^{4}, y (1) = 2

x^{2} + 3 x + 9, at x = 4

\frac{d}{d x} ((x + \frac{1}{x})^{2} - \frac{5}{3 x})

What is the integral from-1 to 0 of 8(e^{1/x})/(x^3) ?
The integral from-1 to 0 of 8(e^{1/x})/(x^3) is -(16)/e