What is the integral from 1 to 6 of |t^2-2t-8| ?

The integral from 1 to 6 of |t^2-2t-8| is 98/3

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integral from 1 to 6 of |t^2-2t-8|

\int_{1}^{6} t^{2} - 2 t - 8 d t

\frac{98}{3}

Decimal

32.66666 \dots

Solution steps

\int_{1}^{6} t^{2} - 2 t - 8 d t

Eliminate Absolutes

= \int_{1}^{4} - t^{2} + 2 t + 8 d t + \int_{4}^{6} t^{2} - 2 t - 8 d t

\int_{1}^{4} - t^{2} + 2 t + 8 d t = 18

\int_{4}^{6} t^{2} - 2 t - 8 d t = \frac{44}{3}

= 18 + \frac{44}{3}

Simplify

18 + \frac{44}{3} : \frac{98}{3}

= \frac{98}{3}

\int_{- 4}^{3} - x^{2} - x + 12 d x

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

\int_{1}^{e^{2}} \frac{ln ( z )}{x z} d z

\int_{0}^{x} ln (t) d t

\int_{0}^{x} \frac{y}{x} d y

What is the integral from 1 to 6 of |t^2-2t-8| ?
The integral from 1 to 6 of |t^2-2t-8| is 98/3