What is the integral from 1 to 3 of (-x^3+6x^2+7x-5) ?

The integral from 1 to 3 of (-x^3+6x^2+7x-5) is 50

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to 3 of (-x^3+6x^2+7x-5)

\int_{1}^{3} (- x^{3} + 6 x^{2} + 7 x - 5) d x

50

Solution steps

\int_{1}^{3} - x^{3} + 6 x^{2} + 7 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}^{3} x^{3} d x + \int_{1}^{3} 6 x^{2} d x + \int_{1}^{3} 7 x d x - \int_{1}^{3} 5 d x

\int_{1}^{3} x^{3} d x = 20

\int_{1}^{3} 6 x^{2} d x = 52

\int_{1}^{3} 7 x d x = 28

\int_{1}^{3} 5 d x = 10

= - 20 + 52 + 28 - 10

Simplify

= 50

What is the integral from 1 to 3 of (-x^3+6x^2+7x-5) ?
The integral from 1 to 3 of (-x^3+6x^2+7x-5) is 50