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

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

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

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

- \frac{1}{6}

Decimal

- 0.16666 \dots

Solution steps

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

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

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

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

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

Simplify

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

= - \frac{1}{6}

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