What is the integral from 1 to 2 of (24x^3+6/(x^2)) ?

The integral from 1 to 2 of (24x^3+6/(x^2)) is 93

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to 2 of (24x^3+6/(x^2))

\int_{1}^{2} (24 x^{3} + \frac{6}{x ^{2}}) d x

93

Solution steps

\int_{1}^{2} 24 x^{3} + \frac{6}{x ^{2}} 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} 24 x^{3} d x + \int_{1}^{2} \frac{6}{x ^{2}} d x

\int_{1}^{2} 24 x^{3} d x = 90

\int_{1}^{2} \frac{6}{x ^{2}} d x = 3

= 90 + 3

Simplify

= 93

What is the integral from 1 to 2 of (24x^3+6/(x^2)) ?
The integral from 1 to 2 of (24x^3+6/(x^2)) is 93