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

The integral from-2 to 1 of (3x^3-2x^2+x) is -75/4

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

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

- \frac{75}{4}

Decimal Notation

- 18.75

Solution steps

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

\int_{- 2}^{1} 3 x^{3} d x = - \frac{45}{4}

\int_{- 2}^{1} 2 x^{2} d x = 6

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

= - \frac{45}{4} - 6 - \frac{3}{2}

Simplify

- \frac{45}{4} - 6 - \frac{3}{2} : - \frac{75}{4}

= - \frac{75}{4}

\int_{0}^{5 π} x cos (2 x) d x

\int_{0}^{4} r^{2} d r

\int_{0}^{2} x^{3} - 2 x d x

\int_{1}^{5} g (x) d x

\int_{1}^{5} \frac{x ^{2}}{12} d x

What is the integral from-2 to 1 of (3x^3-2x^2+x) ?
The integral from-2 to 1 of (3x^3-2x^2+x) is -75/4