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

The integral from-1 to 2 of (x^2+x-4) is -15/2

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

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

- \frac{15}{2}

Solution steps

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

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

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

\int_{- 1}^{2} 4 d x = 12

= 3 + \frac{3}{2} - 12

Simplify

3 + \frac{3}{2} - 12 : - \frac{15}{2}

= - \frac{15}{2}

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What is the integral from-1 to 2 of (x^2+x-4) ?
The integral from-1 to 2 of (x^2+x-4) is -15/2