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

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

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

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

\frac{3}{4}

Decimal Notation

0.75

Solution steps

\int_{0}^{1} x^{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^{3} d x - \int_{0}^{1} x d x + \int_{0}^{1} 1 d x

\int_{0}^{1} x^{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{3}{4}

= \frac{3}{4}

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

\int_{0}^{3} r d r

\int_{0}^{1} \frac{x}{y} d x

\int_{0}^{\frac{π}{2}} 2 sin^{2} (2 x) d x

\int_{0}^{1} r e^{- r^{2}} d r

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