What is the integral of (x/2)^n ?

The integral of (x/2)^n is (x^{n+1})/(2^n(n+1))+C

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integral of (x/2)^n

\int (\frac{x}{2})^{n} d x

\frac{x ^{n + 1}}{2 ^{n} ( n + 1 )} + C

Solution steps

\int (\frac{x}{2})^{n} d x

Apply exponent rule

= \int \frac{x ^{n}}{2 ^{n}} d x

Take the constant out:

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

= \frac{1}{2 ^{n}} \cdot \int x^{n} d x

Apply the Power Rule

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

= \frac{1}{2 ^{n}} \cdot \frac{x ^{n + 1}}{n + 1}

Simplify

\frac{1}{2 ^{n}} \cdot \frac{x ^{n + 1}}{n + 1} : \frac{x ^{n + 1}}{2 ^{n} ( n + 1 )}

= \frac{x ^{n + 1}}{2 ^{n} ( n + 1 )}

Add a constant to the solution

= \frac{x ^{n + 1}}{2 ^{n} ( n + 1 )} + C

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