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

The integral from 1 to 2 of ((x-4)^2)/(x^5) is 43/24

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{43}{24}

Decimal Notation

1.79166 \dots

Solution steps

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

Expand

\frac{( x - 4 ) ^{2}}{x ^{5}} : \frac{1}{x ^{3}} - \frac{8}{x ^{4}} + \frac{16}{x ^{5}}

= \int_{1}^{2} \frac{1}{x ^{3}} - \frac{8}{x ^{4}} + \frac{16}{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_{1}^{2} \frac{1}{x ^{3}} d x - \int_{1}^{2} \frac{8}{x ^{4}} d x + \int_{1}^{2} \frac{16}{x ^{5}} d x

\int_{1}^{2} \frac{1}{x ^{3}} d x = \frac{3}{8}

\int_{1}^{2} \frac{8}{x ^{4}} d x = \frac{7}{3}

\int_{1}^{2} \frac{16}{x ^{5}} d x = \frac{15}{4}

= \frac{3}{8} - \frac{7}{3} + \frac{15}{4}

Simplify

\frac{3}{8} - \frac{7}{3} + \frac{15}{4} : \frac{43}{24}

= \frac{43}{24}

(\frac{11}{8}, - \frac{1}{2}), (\frac{1}{4}, \frac{1}{4})

x \to 0 lim (\frac{sin ( 3 )}{2 x})

y = 3 x^{2}, (- 4, 48)

\int \frac{x ^{2} + 3}{x ^{2} + 1} d x

x \to - 3 lim (3 x + 9)

What is the integral from 1 to 2 of ((x-4)^2)/(x^5) ?
The integral from 1 to 2 of ((x-4)^2)/(x^5) is 43/24