What is the integral from 1 to 4 of 1/(x^2)-8/(x^3) ?

The integral from 1 to 4 of 1/(x^2)-8/(x^3) is -3

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 1/(x^2)-8/(x^3)

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

- 3

Solution steps

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

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

\int_{1}^{4} \frac{8}{x ^{3}} d x = \frac{15}{4}

= \frac{3}{4} - \frac{15}{4}

Simplify

\frac{3}{4} - \frac{15}{4} : - 3

= - 3

x^{^{'}} = x^{3}

\frac{1}{1 + cos ^{2} ( x )}

\int 4 x^{3} + 24 x - 1 d x

x \to 3 lim (\frac{3 - ∣ x ∣}{x + 3})

y^{^{''}} + 2 k y^{^{'}} + a k^{2} y = 0

What is the integral from 1 to 4 of 1/(x^2)-8/(x^3) ?
The integral from 1 to 4 of 1/(x^2)-8/(x^3) is -3