What is the integral of k(1-y)y^n ?

The integral of k(1-y)y^n is k((y^{n+1})/(n+1)-(y^{n+2})/(n+2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of k(1-y)y^n

\int k (1 - y) y^{n} d y

k (\frac{y ^{n + 1}}{n + 1} - \frac{y ^{n + 2}}{n + 2}) + C

Solution steps

\int k (1 - y) y^{n} d y

Take the constant out:

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

= k \cdot \int (1 - y) y^{n} d y

Expand

(1 - y) y^{n} : y^{n} - y^{1 + n}

= k \cdot \int y^{n} - y^{1 + n} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= k (\int y^{n} d y - \int y^{1 + n} d y)

\int y^{n} d y = \frac{y ^{n + 1}}{n + 1}

\int y^{1 + n} d y = \frac{y ^{n + 2}}{n + 2}

= k (\frac{y ^{n + 1}}{n + 1} - \frac{y ^{n + 2}}{n + 2})

Add a constant to the solution

= k (\frac{y ^{n + 1}}{n + 1} - \frac{y ^{n + 2}}{n + 2}) + C

\int 2 x y - 1 d x

\int x (x - 3)^{2} (x + 1) d x

\int x (x^{\frac{3}{2}} - 10)^{4} d x

\int x^{(- 1)} d x

\int \frac{x + 3}{x ^{2} + 4 x + 5} d x

What is the integral of k(1-y)y^n ?
The integral of k(1-y)y^n is k((y^{n+1})/(n+1)-(y^{n+2})/(n+2))+C