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

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

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integral of (1-x)^n

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

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

Solution steps

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

Apply u-substitution

: \int - u^{n} d u

= \int - u^{n} d u

Take the constant out:

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

= - \int u^{n} d u

Apply the Power Rule

: \frac{u ^{n + 1}}{n + 1}

= - \frac{u ^{n + 1}}{n + 1}

Substitute back

u = 1 - x

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

Add a constant to the solution

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

n = 0 \sum \infty e^{- 5 n}

\int_{0}^{2} 0.09375 (4 - x^{2}) d x

\int sec (6 w) tan (6 w) d w

\int 2 x - x d x

y = x + 5, at x = 11

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