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# Worked example: problem involving definite integral (algebraic)

AP.CALC:
CHA‑4 (EU)
,
CHA‑4.D (LO)
,
CHA‑4.D.1 (EK)
,
CHA‑4.D.2 (EK)
,
CHA‑4.E (LO)
,
CHA‑4.E.1 (EK)
Using definite integral to solve a word problem about the growth in the population of a town.

## Want to join the conversation?

• but why do we get a fraction for the number of people using calculator
• It may be that in places, the calculator uses decimal approximations of fractions. This is why you should always try to evaluate integrals yourself.
• Why can't you just plug in 5 back into the r(t) function for the second part of the question?
• From - , Sal uses u-substitution to evaluate the indefinite integral of e^1.2t. At , he plugs this value back into the definite integral problem. Why does he not change the bounds of integration after the u-substitution? i.e. from t = [2, 5] to u = [2.4, 6] ?
(1 vote)
• Because he back-substituted to continue working with the t-variable instead.
• Since we are given the population t(2)=1500, why wouldn't that be used as the lower limit for the definite integral? It seems that using it to calculate a C would provide a more accurate solution?
(1 vote)
• We have the rate of change of the population:
𝑃 '(𝑡) = 𝑒^(1.2𝑡) − 2𝑡

What we could do is find the population 𝑃(𝑡) as the indefinite integral
𝑃(𝑡) = ∫𝑃 '(𝑡)𝑑𝑡 = (1∕1.2)𝑒^(1.2𝑡) − 𝑡² + 𝐶

Then, since we know 𝑃(2) = 1500
we can use that as the initial condition and find 𝐶:
𝑃(2) = (1∕1.2)𝑒^2.4 − 4 + 𝐶 = 1500
⇒ 𝐶 = 1504 − (1∕1.2)𝑒^2.4 ≈ 1494.81

Thereby,
𝑃(5) ≈ (1∕1.2)𝑒^6 − 25 + 1494.81 ≈ 1806.00

And the change in population between 𝑡 = 2 and 𝑡 = 5 is
𝑃(5) − 𝑃(2) ≈ 1806.00 − 1500 = 306

– – –

Sal's solution goes back to the fundamental theorem of calculus,
which says:
∫[𝑎, 𝑏] 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎)

Thereby,
∫[2, 5] 𝑃 '(𝑡)𝑑𝑡 = 𝑃(5) − 𝑃(2),
which directly gives us the change in population between 𝑡 = 2 and 𝑡 = 5.