What is the integral of 6x(1-x) ?

The integral of 6x(1-x) is 6((x^2)/2-(x^3)/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6x(1-x)

\int 6 x (1 - x) d x

6 (\frac{x ^{2}}{2} - \frac{x ^{3}}{3}) + C

Solution steps

\int 6 x (1 - x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int x (1 - x) d x

Expand

x (1 - x) : x - x^{2}

= 6 \cdot \int x - 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

= 6 (\int x d x - \int x^{2} d x)

\int x d x = \frac{x ^{2}}{2}

\int x^{2} d x = \frac{x ^{3}}{3}

= 6 (\frac{x ^{2}}{2} - \frac{x ^{3}}{3})

Add a constant to the solution

= 6 (\frac{x ^{2}}{2} - \frac{x ^{3}}{3}) + C

t^{2} \cdot y^{^{'}} + t y = 7, y (1) = 7

\int 8 sin^{2} (x) cos (x) d x

y^{^{''}} + 2 y^{^{'}} + y = 3 x + 8

\frac{d}{d x} (a cos (2 x) + b cos (2 x))

\frac{\partial}{\partial x} (y - 5 x)

What is the integral of 6x(1-x) ?
The integral of 6x(1-x) is 6((x^2)/2-(x^3)/3)+C