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

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

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

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

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

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

\int_{0}^{2} x^{2} d x = \frac{8}{3}

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

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

= \frac{8}{3} - 6 + 4

Simplify

\frac{8}{3} - 6 + 4 : \frac{2}{3}

= \frac{2}{3}

\int_{- 1}^{1} y d y

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

\int_{0}^{π} sin (x) sin (2 x) d x

\int_{1}^{5} 3 x d x

\int_{1}^{4} (3 x^{2} - 2 x) d x

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