What is the integral from 0 to 5 of x^2-3x-4 ?

The integral from 0 to 5 of x^2-3x-4 is -95/6

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integral from 0 to 5 of x^2-3x-4

\int_{0}^{5} x^{2} - 3 x - 4 d x

- \frac{95}{6}

Decimal Notation

- 15.83333 \dots

Solution steps

\int_{0}^{5} 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_{0}^{5} x^{2} d x - \int_{0}^{5} 3 x d x - \int_{0}^{5} 4 d x

\int_{0}^{5} x^{2} d x = \frac{125}{3}

\int_{0}^{5} 3 x d x = \frac{75}{2}

\int_{0}^{5} 4 d x = 20

= \frac{125}{3} - \frac{75}{2} - 20

Simplify

\frac{125}{3} - \frac{75}{2} - 20 : - \frac{95}{6}

= - \frac{95}{6}

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

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

\int_{0}^{0.4} 2 (1 - x) d x

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

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

What is the integral from 0 to 5 of x^2-3x-4 ?
The integral from 0 to 5 of x^2-3x-4 is -95/6