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

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

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

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

- \frac{2}{3}

Decimal

- 0.66666 \dots

Solution steps

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

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

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

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

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

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

Simplify

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

= - \frac{2}{3}

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