What is the integral from 0 to 1 of (2x-1)^6 ?

The integral from 0 to 1 of (2x-1)^6 is 1/7

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integral from 0 to 1 of (2x-1)^6

\int_{0}^{1} (2 x - 1)^{6} d x

\frac{1}{7}

Decimal

0.14285 \dots

Solution steps

\int_{0}^{1} (2 x - 1)^{6} d x

Apply u-substitution

= \int_{- 1}^{1} \frac{u ^{6}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- 1}^{1} u^{6} d u

Apply the Power Rule

= \frac{1}{2} [\frac{u ^{7}}{7}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{7}

= \frac{1}{2} \cdot \frac{2}{7}

Simplify

= \frac{1}{7}

What is the integral from 0 to 1 of (2x-1)^6 ?
The integral from 0 to 1 of (2x-1)^6 is 1/7