What is the integral from-4 to-1 of (x^3+5x^2+4x) ?

The integral from-4 to-1 of (x^3+5x^2+4x) is 45/4

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integral from-4 to-1 of (x^3+5x^2+4x)

\int_{- 4}^{- 1} (x^{3} + 5 x^{2} + 4 x) d x

\frac{45}{4}

Decimal

11.25

Solution steps

\int_{- 4}^{- 1} x^{3} + 5 x^{2} + 4 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_{- 4}^{- 1} x^{3} d x + \int_{- 4}^{- 1} 5 x^{2} d x + \int_{- 4}^{- 1} 4 x d x

\int_{- 4}^{- 1} x^{3} d x = - \frac{255}{4}

\int_{- 4}^{- 1} 5 x^{2} d x = 105

\int_{- 4}^{- 1} 4 x d x = - 30

= - \frac{255}{4} + 105 - 30

Simplify

- \frac{255}{4} + 105 - 30 : \frac{45}{4}

= \frac{45}{4}

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What is the integral from-4 to-1 of (x^3+5x^2+4x) ?
The integral from-4 to-1 of (x^3+5x^2+4x) is 45/4