What is the integral of (1-x)^{n-1} ?

The integral of (1-x)^{n-1} is -((1-x)^n)/n+C

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integral of (1-x)^{n-1}

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

- \frac{( 1 - x ) ^{n}}{n} + C

Solution steps

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

Apply u-substitution

: \int - u^{n - 1} d u

= \int - u^{n - 1} d u

Take the constant out:

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

= - \int u^{n - 1} d u

Apply the Power Rule

: \frac{u ^{n}}{n}

= - \frac{u ^{n}}{n}

Substitute back

u = 1 - x

= - \frac{( 1 - x ) ^{n}}{n}

Add a constant to the solution

= - \frac{( 1 - x ) ^{n}}{n} + C

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