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

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

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

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

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

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

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

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

= \frac{4}{3} - 0

Simplify

= \frac{4}{3}

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a^{2} - x^{2}

h (x) = \frac{2 x}{sin ( x )}

\frac{d}{d x} ((x^{2} + x) (6 - x))

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