What is the integral from 0 to 1 of (2x-1)^{100} ?

The integral from 0 to 1 of (2x-1)^{100} is 1/101

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integral from 0 to 1 of (2x-1)^{100}

\int_{0}^{1} (2 x - 1)^{100} d x

\frac{1}{101}

Solution steps

\int_{0}^{1} (2 x - 1)^{100} d x

Apply u-substitution

: \int_{- 1}^{1} \frac{u ^{100}}{2} d u

= \int_{- 1}^{1} \frac{u ^{100}}{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^{100} d u

Apply the Power Rule

: [\frac{u ^{101}}{101}]_{- 1}^{1}

= \frac{1}{2} [\frac{u ^{101}}{101}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{101}

= \frac{1}{2} \cdot \frac{2}{101}

Simplify

= \frac{1}{101}

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What is the integral from 0 to 1 of (2x-1)^{100} ?
The integral from 0 to 1 of (2x-1)^{100} is 1/101