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

The integral from 1 to 2 of (x-1)-(x-1)^3 is 1/4

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integral from 1 to 2 of (x-1)-(x-1)^3

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

\frac{1}{4}

Decimal Notation

0.25

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} x d x - \int_{1}^{2} 1 d x - \int_{1}^{2} (x - 1)^{3} d x

\int_{1}^{2} x d x = \frac{3}{2}

\int_{1}^{2} 1 d x = 1

\int_{1}^{2} (x - 1)^{3} d x = \frac{1}{4}

= \frac{3}{2} - 1 - \frac{1}{4}

Simplify

\frac{3}{2} - 1 - \frac{1}{4} : \frac{1}{4}

= \frac{1}{4}

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What is the integral from 1 to 2 of (x-1)-(x-1)^3 ?
The integral from 1 to 2 of (x-1)-(x-1)^3 is 1/4