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

The integral from-1 to 4 of (x^2-3x-4) is -125/6

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

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

- \frac{125}{6}

Decimal Notation

- 20.83333 \dots

Solution steps

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

\int_{- 1}^{4} x^{2} d x = \frac{65}{3}

\int_{- 1}^{4} 3 x d x = \frac{45}{2}

\int_{- 1}^{4} 4 d x = 20

= \frac{65}{3} - \frac{45}{2} - 20

Simplify

\frac{65}{3} - \frac{45}{2} - 20 : - \frac{125}{6}

= - \frac{125}{6}

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What is the integral from-1 to 4 of (x^2-3x-4) ?
The integral from-1 to 4 of (x^2-3x-4) is -125/6