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

The integral from 0 to 1 of (x-1)^3-(x-1) is 1/4

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integral from 0 to 1 of (x-1)^3-(x-1)

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

\frac{1}{4}

Solution steps

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

Simplify

= \int_{0}^{1} (x - 1)^{3} - x + 1 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_{0}^{1} (x - 1)^{3} d x - \int_{0}^{1} x d x + \int_{0}^{1} 1 d x

\int_{0}^{1} (x - 1)^{3} d x = - \frac{1}{4}

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} 1 d x = 1

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

Simplify

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

= \frac{1}{4}

\frac{1}{s \cdot ( s ^{2} + \frac{9}{2} s + 15 )}

\int 2 - 1 (e - x + 2) d x

f (x) = 3 + 2 x^{3}

\frac{\partial}{\partial x} (2^{4 x} + 4 x^{2})

x^{2} - x y + 2 y - 2 = 0

What is the integral from 0 to 1 of (x-1)^3-(x-1) ?
The integral from 0 to 1 of (x-1)^3-(x-1) is 1/4