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

The integral from 0 to 2 of (2x^2-3x+2) is 10/3

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integral from 0 to 2 of (2x^2-3x+2)

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

\frac{10}{3}

Decimal

3.33333 \dots

Solution steps

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

\int_{0}^{2} 2 x^{2} d x = \frac{16}{3}

\int_{0}^{2} 3 x d x = 6

\int_{0}^{2} 2 d x = 4

= \frac{16}{3} - 6 + 4

Simplify

\frac{16}{3} - 6 + 4 : \frac{10}{3}

= \frac{10}{3}

What is the integral from 0 to 2 of (2x^2-3x+2) ?
The integral from 0 to 2 of (2x^2-3x+2) is 10/3