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

The integral from 0 to 1 of (-x^2-x+2) is 7/6

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

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

\frac{7}{6}

Solution steps

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

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

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} 2 d x = 2

= - \frac{1}{3} - \frac{1}{2} + 2

Simplify

- \frac{1}{3} - \frac{1}{2} + 2 : \frac{7}{6}

= \frac{7}{6}

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What is the integral from 0 to 1 of (-x^2-x+2) ?
The integral from 0 to 1 of (-x^2-x+2) is 7/6