What is the integral from 1 to 3 of (t^2-t+6) ?

The integral from 1 to 3 of (t^2-t+6) is 50/3

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integral from 1 to 3 of (t^2-t+6)

\int_{1}^{3} (t^{2} - t + 6) d t

\frac{50}{3}

Decimal

16.66666 \dots

Solution steps

\int_{1}^{3} t^{2} - t + 6 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_{1}^{3} t^{2} d t - \int_{1}^{3} t d t + \int_{1}^{3} 6 d t

\int_{1}^{3} t^{2} d t = \frac{26}{3}

\int_{1}^{3} t d t = 4

\int_{1}^{3} 6 d t = 12

= \frac{26}{3} - 4 + 12

Simplify

\frac{26}{3} - 4 + 12 : \frac{50}{3}

= \frac{50}{3}

What is the integral from 1 to 3 of (t^2-t+6) ?
The integral from 1 to 3 of (t^2-t+6) is 50/3