What is the integral of ((1-z)^5)/(10) ?

The integral of ((1-z)^5)/(10) is -1/60 (1-z)^6+C

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integral of ((1-z)^5)/(10)

integral

\frac{( 1 - z ) ^{5}}{10}

- \frac{1}{60} (1 - z)^{6} + C

Solution steps

\int \frac{( 1 - z ) ^{5}}{10} d z

Take the constant out:

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

= \frac{1}{10} \cdot \int (1 - z)^{5} d z

Apply u-substitution

: \int - u^{5} d u

= \frac{1}{10} \cdot \int - u^{5} d u

Take the constant out:

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

= \frac{1}{10} (- \int u^{5} d u)

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{1}{10} (- \frac{u ^{6}}{6})

Substitute back

u = 1 - z

= \frac{1}{10} (- \frac{( 1 - z ) ^{6}}{6})

Simplify

\frac{1}{10} (- \frac{( 1 - z ) ^{6}}{6}) : - \frac{1}{60} (1 - z)^{6}

= - \frac{1}{60} (1 - z)^{6}

Add a constant to the solution

= - \frac{1}{60} (1 - z)^{6} + C

\frac{sin ^{3} ( x )}{cos ^{2} ( x )}

- 3 t^{2}

\frac{x}{12}

1 - 2 x

(x + 1)^{3}

What is the integral of ((1-z)^5)/(10) ?
The integral of ((1-z)^5)/(10) is -1/60 (1-z)^6+C