What is the integral from-2 to 1 of (x^4+x^3-4x^2+6) ?

The integral from-2 to 1 of (x^4+x^3-4x^2+6) is 177/20

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integral from-2 to 1 of (x^4+x^3-4x^2+6)

\int_{- 2}^{1} (x^{4} + x^{3} - 4 x^{2} + 6) d x

\frac{177}{20}

Decimal Notation

8.85

Solution steps

\int_{- 2}^{1} x^{4} + x^{3} - 4 x^{2} + 6 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} x^{4} d x + \int_{- 2}^{1} x^{3} d x - \int_{- 2}^{1} 4 x^{2} d x + \int_{- 2}^{1} 6 d x

\int_{- 2}^{1} x^{4} d x = \frac{33}{5}

\int_{- 2}^{1} x^{3} d x = - \frac{15}{4}

\int_{- 2}^{1} 4 x^{2} d x = 12

\int_{- 2}^{1} 6 d x = 18

= \frac{33}{5} - \frac{15}{4} - 12 + 18

Simplify

\frac{33}{5} - \frac{15}{4} - 12 + 18 : \frac{177}{20}

= \frac{177}{20}

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What is the integral from-2 to 1 of (x^4+x^3-4x^2+6) ?
The integral from-2 to 1 of (x^4+x^3-4x^2+6) is 177/20