What is the integral from-1 to 2 of (2x^3+x^2-8) ?

The integral from-1 to 2 of (2x^3+x^2-8) is -27/2

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integral from-1 to 2 of (2x^3+x^2-8)

\int_{- 1}^{2} (2 x^{3} + x^{2} - 8) d x

- \frac{27}{2}

Decimal Notation

- 13.5

Solution steps

\int_{- 1}^{2} 2 x^{3} + x^{2} - 8 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}^{2} 2 x^{3} d x + \int_{- 1}^{2} x^{2} d x - \int_{- 1}^{2} 8 d x

\int_{- 1}^{2} 2 x^{3} d x = \frac{15}{2}

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

\int_{- 1}^{2} 8 d x = 24

= \frac{15}{2} + 3 - 24

Simplify

\frac{15}{2} + 3 - 24 : - \frac{27}{2}

= - \frac{27}{2}

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What is the integral from-1 to 2 of (2x^3+x^2-8) ?
The integral from-1 to 2 of (2x^3+x^2-8) is -27/2