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

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

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

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

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

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

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

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

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

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

= 0 - \frac{2}{3} + 0 + 2

Simplify

0 - \frac{2}{3} + 0 + 2 : \frac{4}{3}

= \frac{4}{3}

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