Have you had an annual check-up?
During an exercise stress test, a patient's cardiac output (C.O.) was monitored throughout the 15-minute exercise period while the resting C.O. was 5 L/min.
As shown in the diagram, for t<3, the C.O. correlated to the function:
f(t) = - 8.
From t=3 to t=12, the C.O. reached a plateau with a function:
f(t) = 18.
When t>12, the patient slowed his pace down, and the function during recovery phase was:
f(t) = + 31.
What is the average C.O. during this exercise period in L/min.?
Hint : You may not need a calculator for this.
The answer is 15.4.
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1 solution
Notice that the functions when t<3 and t>12 have opposite +/- sign of the main variables to each other and have the equal domain of 3. In other words, if we add the areas under curve of these 2 domains, the area from x=0 to x=3 will equal to
∫[ 1 + e − 2 t 2 6 - 8 + ( 1 + e − 2 ( t ) − 2 6 + 31)] dt = ∫ 23 dt = 23t.
That is the total area has been shifted to form a function of a constant, and so the total area under curve has become area of simple rectangles, as shown in figure 2.
As a result, we can calculate the new version of area under curve as:
3 × 2 3 + 1 8 × ( 1 2 − 3 ) = 69+162 = 231 L.
Finally, the average C.O. over the 15-minute period = 231/15 = 15.4 L./min.