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

The integral from 1 to 4 of (x^2-x-6) is -9/2

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integral from 1 to 4 of (x^2-x-6)

\int_{1}^{4} (x^{2} - x - 6) d x

- \frac{9}{2}

Decimal Notation

- 4.5

Solution steps

\int_{1}^{4} x^{2} - x - 6 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} x d x - \int_{1}^{4} 6 d x

\int_{1}^{4} x^{2} d x = 21

\int_{1}^{4} x d x = \frac{15}{2}

\int_{1}^{4} 6 d x = 18

= 21 - \frac{15}{2} - 18

Simplify

21 - \frac{15}{2} - 18 : - \frac{9}{2}

= - \frac{9}{2}

\int_{6.3}^{6.6} 12 x^{2} - 6 x + \frac{4}{x} d x

\int_{- 4}^{- 3} \frac{x}{( x ^{2} + 5 ) ^{2}} d x

\int_{0}^{4} (x^{2} - 1) d x

\int_{0}^{2} 12 x^{3} d x

\int_{1}^{6} 8 x^{3} - 6 x^{- 2} d x

What is the integral from 1 to 4 of (x^2-x-6) ?
The integral from 1 to 4 of (x^2-x-6) is -9/2