What is the integral of (a-by)^4 ?

The integral of (a-by)^4 is -((a-by)^5)/(5b)+C

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integral of (a-by)^4

\int (a - b y)^{4} d y

- \frac{( a - b y ) ^{5}}{5 b} + C

Solution steps

\int (a - b y)^{4} d y

Apply u-substitution

: \int - \frac{u ^{4}}{b} d u

= \int - \frac{u ^{4}}{b} d u

Take the constant out:

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

= - \frac{1}{b} \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= - \frac{1}{b} \cdot \frac{u ^{5}}{5}

Substitute back

u = a - b y

= - \frac{1}{b} \cdot \frac{( a - b y ) ^{5}}{5}

Simplify

- \frac{1}{b} \cdot \frac{( a - b y ) ^{5}}{5} : - \frac{( a - b y ) ^{5}}{5 b}

= - \frac{( a - b y ) ^{5}}{5 b}

Add a constant to the solution

= - \frac{( a - b y ) ^{5}}{5 b} + C

x \to 5 + lim (x^{2} + 2 x - 3)

\frac{\partial}{\partial y} (sin (2 y + x) sin (x))

n = 0 \sum \infty 3 n - 1

\int \frac{12}{1 - cos ( 3 x )} d x

\frac{\partial}{\partial y} (4 x^{3} - 4 x)

What is the integral of (a-by)^4 ?
The integral of (a-by)^4 is -((a-by)^5)/(5b)+C