What is the integral from 0 to 1 of (2t-1)^3 ?

The integral from 0 to 1 of (2t-1)^3 is 0

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integral from 0 to 1 of (2t-1)^3

\int_{0}^{1} (2 t - 1)^{3} d t

0

Solution steps

\int_{0}^{1} (2 t - 1)^{3} d t

Apply u-substitution

: \int_{- 1}^{1} \frac{u ^{3}}{2} d u

= \int_{- 1}^{1} \frac{u ^{3}}{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^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{- 1}^{1}

= \frac{1}{2} [\frac{u ^{4}}{4}]_{- 1}^{1}

Compute the boundaries:

0

= \frac{1}{2} \cdot 0

Simplify

= 0

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What is the integral from 0 to 1 of (2t-1)^3 ?
The integral from 0 to 1 of (2t-1)^3 is 0