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

The integral from 1 to 2 of (x^2+x-1) is 17/6

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

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

\frac{17}{6}

Decimal

2.83333 \dots

Solution steps

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

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

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

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

= \frac{7}{3} + \frac{3}{2} - 1

Simplify

\frac{7}{3} + \frac{3}{2} - 1 : \frac{17}{6}

= \frac{17}{6}

What is the integral from 1 to 2 of (x^2+x-1) ?
The integral from 1 to 2 of (x^2+x-1) is 17/6