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

The integral from 1 to 2 of (x^2+x-2) is 11/6

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

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

\frac{11}{6}

Decimal

1.83333 \dots

Solution steps

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

\int_{1}^{2} x^{2} d x = \frac{7}{3}

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

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

= \frac{7}{3} + \frac{3}{2} - 2

Simplify

\frac{7}{3} + \frac{3}{2} - 2 : \frac{11}{6}

= \frac{11}{6}

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What is the integral from 1 to 2 of (x^2+x-2) ?
The integral from 1 to 2 of (x^2+x-2) is 11/6