What is the integral from 0 to 3 of (-x^2+x-1) ?

The integral from 0 to 3 of (-x^2+x-1) is -15/2

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integral from 0 to 3 of (-x^2+x-1)

\int_{0}^{3} (- x^{2} + x - 1) d x

- \frac{15}{2}

Decimal

- 7.5

Solution steps

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

\int_{0}^{3} x^{2} d x = 9

\int_{0}^{3} x d x = \frac{9}{2}

\int_{0}^{3} 1 d x = 3

= - 9 + \frac{9}{2} - 3

Simplify

- 9 + \frac{9}{2} - 3 : - \frac{15}{2}

= - \frac{15}{2}

What is the integral from 0 to 3 of (-x^2+x-1) ?
The integral from 0 to 3 of (-x^2+x-1) is -15/2