A person's metabolic rate tends to go up after eating a meal...

A person's metabolic rate tends to go up after eating a meal and then, after some time has passed, it returns to a resting metabolic rate. This phenomenon is known as the thermic effect of food, and the effect (in kJ per hour) for one individual is F(t) = -11.42 + 174.3te^{-t/1.1}, where t is the number of hours that have elapsed since eating a meal. Find the total thermic energy of a meal for the next five hours after a meal by integrating the thermic effect function between t = 0 and t = 5. The total thermic energy is about kJ. (Round to the nearest whole number as needed.)

Transcript

00:01 In this problem we are provided with the thermic effect of food which is given by f of t which equals to negative 11 .42 plus 174 .3 times t times e raised to the power negative t over 1 .1 where the unit is kilojoules per hour and the time t is in r's.

00:33 We are asked to find out the total thermic energy for five hours, that is, from time t equals to zero up to time t equals to five.

00:45 So let us denote this total thermic energy as t.

00:49 So we have t to be equal to integral zero to five f of t which is negative 11 .4 .4 plus 174 .3 times t, times e raised to the power negative t over 1 .1 dt.

01:06 So in order to evaluate this, we make use of the property and write down the integral as negative 11 .42 times integral 0 to 5 d t plus 174 .3 times integral 0 to 5 t times e raised to the power negative t over 1 .1 d t.

01:30 So we begin by evaluating the second integral by integration by parts integral u dv equals to uv minus integral v d u.

01:41 Here we consider u to be equal to t and we consider v to be equal to sorry we consider dv to be equal to e raised to the power negative t over 1 .1.

01:53 So now differentiating u we have d u to be equal to d t and integrating dv we have v equals to e raised to the power negative t over 1 .1 times negative 1 .1.

02:07 So now substituting this in the formula of integration byparts we have integral t times e raised to the power negative t over 1 .1 d t to be equal to u times v which is negative 1 .1 times t times e raised to the power negative t over 1 .1 minus integral negative 1 .1 times e raised to the power negative t over 1 .1 d t.

02:37 So this can be written as negative 1 .1 t e raised to the power negative t over 1 .1 plus 1 .1 times integral of e raised to the power negative t over 1 .1 d t.

02:55 So evaluating the integral we obtain the value of of the integral to be equal to negative 1 .1 times e raised to the power negative t over 1 .1.

03:12 So this can be further simplified and written as negative 1 .1 times e raised to the power negative t over 1 .1 times t plus 1 .1...

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