What is the integral from 0 to 1 of (3x^3-x^2+2x-4) ?

The integral from 0 to 1 of (3x^3-x^2+2x-4) is -31/12

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integral from 0 to 1 of (3x^3-x^2+2x-4)

\int_{0}^{1} (3 x^{3} - x^{2} + 2 x - 4) d x

- \frac{31}{12}

Decimal Notation

- 2.58333 \dots

Solution steps

\int_{0}^{1} 3 x^{3} - x^{2} + 2 x - 4 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} 3 x^{3} d x - \int_{0}^{1} x^{2} d x + \int_{0}^{1} 2 x d x - \int_{0}^{1} 4 d x

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

\int_{0}^{1} x^{2} d x = \frac{1}{3}

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

\int_{0}^{1} 4 d x = 4

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

Simplify

\frac{3}{4} - \frac{1}{3} + 1 - 4 : - \frac{31}{12}

= - \frac{31}{12}

\int_{0}^{1} \frac{32}{9} (x y)^{\frac{1}{3}} d y

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What is the integral from 0 to 1 of (3x^3-x^2+2x-4) ?
The integral from 0 to 1 of (3x^3-x^2+2x-4) is -31/12