What is the integral from 1 to 2 of 4x^3+6/(x^2) ?

The integral from 1 to 2 of 4x^3+6/(x^2) is 18

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 4x^3+6/(x^2)

\int_{1}^{2} 4 x^{3} + \frac{6}{x ^{2}} d x

18

Solution steps

\int_{1}^{2} 4 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} 4 x^{3} d x + \int_{1}^{2} \frac{6}{x ^{2}} d x

\int_{1}^{2} 4 x^{3} d x = 15

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

= 15 + 3

Simplify

= 18

\int_{0}^{x} sin (2 t^{2}) d t

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

\int_{- 2}^{2} \frac{x ^{3}}{2} d x

\int_{0}^{\frac{π}{3}} 3 sin (x) d x

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

What is the integral from 1 to 2 of 4x^3+6/(x^2) ?
The integral from 1 to 2 of 4x^3+6/(x^2) is 18