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

The integral from-6 to 5 of |x^2-3x-4| is 767/6

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

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

\frac{767}{6}

Decimal Notation

127.83333 \dots

Solution steps

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

Eliminate Absolutes

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

\int_{- 6}^{- 1} x^{2} - 3 x - 4 d x = \frac{625}{6}

\int_{- 1}^{4} - x^{2} + 3 x + 4 d x = \frac{125}{6}

\int_{4}^{5} x^{2} - 3 x - 4 d x = \frac{17}{6}

= \frac{625}{6} + \frac{125}{6} + \frac{17}{6}

Simplify

\frac{625}{6} + \frac{125}{6} + \frac{17}{6} : \frac{767}{6}

= \frac{767}{6}

cos^{2} (x)

\int \frac{x ^{2}}{( x ^{2} + 16 ) ^{\frac{3}{2}}} d x

\frac{\partial}{\partial x} (\frac{6 x}{y ^{2}})

\frac{1}{10 \cdot s}

\int \frac{y}{3} d y

What is the integral from-6 to 5 of |x^2-3x-4| ?
The integral from-6 to 5 of |x^2-3x-4| is 767/6