What is the integral from 1 to 8 of |t^2-4t-12| ?

The integral from 1 to 8 of |t^2-4t-12| is 77

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integral from 1 to 8 of |t^2-4t-12|

\int_{1}^{8} t^{2} - 4 t - 12 d t

77

Solution steps

\int_{1}^{8} t^{2} - 4 t - 12 d t

Eliminate Absolutes

= \int_{1}^{6} - t^{2} + 4 t + 12 d t + \int_{6}^{8} t^{2} - 4 t - 12 d t

\int_{1}^{6} - t^{2} + 4 t + 12 d t = \frac{175}{3}

\int_{6}^{8} t^{2} - 4 t - 12 d t = \frac{56}{3}

= \frac{175}{3} + \frac{56}{3}

Simplify

\frac{175}{3} + \frac{56}{3} : 77

= 77

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What is the integral from 1 to 8 of |t^2-4t-12| ?
The integral from 1 to 8 of |t^2-4t-12| is 77