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

The integral from 0 to 1 of 2x^2-3x+2 is 7/6

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

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

\frac{7}{6}

Decimal Notation

1.16666 \dots

Solution steps

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

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

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

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

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

Simplify

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

= \frac{7}{6}

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What is the integral from 0 to 1 of 2x^2-3x+2 ?
The integral from 0 to 1 of 2x^2-3x+2 is 7/6