Answer
The total thermic energy of a meal for the next six hours after a
meal by integrating the thermic effect function between $t=0$
and $t=6$ is given by:
$$
\begin{aligned}
\int_{0}^{6} &\left(-10.28+175.9 t e^{-t / 1.3}\right) d t \approx 219.07
\end{aligned}
$$
Therefore, the total thermic energy of a meal for the next six hours after a meal is $219$ kJ.
Work Step by Step
the total thermic energy of a meal for the next six hours after a
meal by integrating the thermic effect function between $t=0$
and $t=6$ is given by:
$$
\begin{aligned} \int_{0}^{6}\left(-10.28+175.9 t e^{-t / 1.3}\right) d t \end{aligned}
$$
use integration by parts with
$$
\quad\quad\quad \left[\begin{array}{c}{u=t, \quad\quad dv= e^{-t / 1.3} d t \\ {d u= dt , \quad\quad v=-1.3 e^{-t / 1.3}} }\end{array}\right] , \text { then }\\
$$
First , we find indefinite integral
$$
\begin{aligned} \int t e^{-t / 1.3} d t&=(t)\left(-1.3 e^{-t / 1.3}\right)-\int\left(-1.3 e^{-t / 1.3}\right) d t \\
&=-1.3 t e^{-t / 1.3}-1.69 e^{-t / 1.3}+C
\end{aligned}
$$
Now, we obtain the define integral
$$
\begin{aligned}
\int_{0}^{6} &\left(-10.28+175.9 t e^{-t / 1.3}\right) d t= \\
&=-10.28 t+\left.175.9\left(-1.3 t e^{-t / 1.3}-1.69 e^{-t / 1.3}\right)\right|_{0} ^{6} \\
&=\left(-61.68-1669.291 e^{-6 / 1.3}\right)-(-297.271)\\
&\approx 219.07
\end{aligned}
$$
Therefore, the total thermic energy of a meal for the next six hours after a meal is $219$ kJ.
