What is the integral from-1 to 4 of |6x^2-x^3-5x| ?

The integral from-1 to 4 of |6x^2-x^3-5x| is 121/4

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integral from-1 to 4 of |6x^2-x^3-5x|

\int_{- 1}^{4} 6 x^{2} - x^{3} - 5 x d x

\frac{121}{4}

Decimal Notation

30.25

Solution steps

\int_{- 1}^{4} 6 x^{2} - x^{3} - 5 x d x

Eliminate Absolutes

= \int_{- 1}^{0} 6 x^{2} - x^{3} - 5 x d x + \int_{0}^{1} - 6 x^{2} + x^{3} + 5 x d x + \int_{1}^{4} 6 x^{2} - x^{3} - 5 x d x

\int_{- 1}^{0} 6 x^{2} - x^{3} - 5 x d x = \frac{19}{4}

\int_{0}^{1} - 6 x^{2} + x^{3} + 5 x d x = \frac{3}{4}

\int_{1}^{4} 6 x^{2} - x^{3} - 5 x d x = \frac{99}{4}

= \frac{19}{4} + \frac{3}{4} + \frac{99}{4}

Simplify

\frac{19}{4} + \frac{3}{4} + \frac{99}{4} : \frac{121}{4}

= \frac{121}{4}

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What is the integral from-1 to 4 of |6x^2-x^3-5x| ?
The integral from-1 to 4 of |6x^2-x^3-5x| is 121/4