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}

\int_{\frac{1}{2}}^{\frac{3}{2}} (x^{2} + x + 1) d x

\int_{0}^{1} - x d x

\int_{0}^{2} 9 x^{2} x^{3} + 1 d x

\int_{- 2}^{2} \frac{1}{x + 1} d x

\int_{0}^{1} 2 t 1 + 4 t^{2} d t

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