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

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

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

- \frac{125}{6}

Decimal

- 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_{0}^{4} (1 - e^{- x}) d x

\int_{0}^{1} x ln (x) d x

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

\int_{0}^{\infty} 2 e^{- x - 2 y} d y

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

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