Given F(x) = the integral from 2 to x of (t^2) dt. Find F'(x).

There you got a function in the upper bound of integration, so you just write the exact same integrand, just change "t" for "x" and that´s it.

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Course: AP®︎/College Calculus AB > Unit 8

Lesson 3: Using accumulation functions and definite integrals in applied contexts

Analyzing problems involving definite integrals

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The interpretation of definite integrals as accumulation of quantities can be used to solve various real-world word problems.

Accumulation (or net change) problems are word problems where the rate of change of a quantity is given and we are asked to calculate the value the quantity accumulated over time. These problems are solved using definite integrals. Let's see how it's done.

Accumulation problems are solved using definite integrals

Imagine we are given the following information:

The temperature of a soup is increasing at a rate of $r (t) = 30 e^{- 0.3 t}$ ‍ degrees Celsius per minute (where $t$ ‍ is the time in minutes). At time $t = 0$ ‍, the temperature of the soup is $23$ ‍ degrees Celsius.

And imagine we are asked to find the amount by which the temperature increased between $t = 0$ ‍ and $t = 5$ ‍ minutes. This is an accumulation (or net change) word problem. We can tell it is so because we are given a function that models the rate of change of a quantity, and are asked about the change in that quantity over an interval of time.

For any quantity whose rate is given by the function

r

, the definite integral

\int_{a}^{b} r (t) d t

describes the amount by which the quantity changed between

t = a

and

t = b

So in our case, the amount by which the temperature increased between

t = 0

and

t = 5

minutes is given by

\int_{0}^{5} r (t) d t

\int_{0}^{5} r (t) d t \approx 77.7

degrees Celsius

Now imagine we were asked a different question: what is the soup's temperature at $t = 5$ ‍ minutes? Notice we aren't dealing with a change anymore, we're dealing with an actual value. But have no fear, because definite integrals can help us with this one too! All we need is to add the initial condition.

Recall we were given that the temperature of the soup at time

t = 0

was

23

degrees Celsius. Adding that to the change in temperature between

t = 0

and

t = 5

gives us the temperature at

t = 5

\underset{temp at t = 0}{\underset{⏟}{23}} + \overset{change in temp from t = 0 to t = 5}{\overset{⏞}{\int_{0}^{5} r (t) d t}}

Since we already calculated

\int_{0}^{5} r (t) d t

, we can tell that at

t = 5

minutes, the temperature was

23 + 77.7 = 100.7

degrees Celsius. That's boiling hot!

Problem 1

Consider the following problem:

The population of a town grows at a rate of $r (t) = 300 \cdot e^{0.3 t}$ ‍ people per year (where $t$ ‍ is time in years). At time $t = 2$ ‍, the town's population is $1200$ ‍ people. What is the town's population at $t = 7$ ‍?

Which expression can we use to solve the problem?

Common mistake: Misusing initial conditions

Some accumulation problems ask about a net change, and some ask about an actual value. The difference is that when looking for an actual value we must use the initial conditions.

A common mistake would be using initial conditions when asked about a net change, or not using initial conditions when asked about an actual value.

Common mistake: Using differentiation instead of integration

Applied word problems are common throughout both differential and integral calculus. When given a word problem, we must decide whether the solution involves derivatives or integrals. Making a wrong decision will of course result in a wrong answer.

Derivatives are useful when we are given a quantity and asked about its rate, while integrals are useful when we are given a rate and asked about the quantity.

	What's given?	What's missing?	What to use?
Differential calculus	Quantity	Rate	Derivative
Integral calculus	Rate	Quantity (or change in quantity)	Integral

Problem 2

Consider the following problem:

The depth of the water in a tank is changing at a rate of $r (t) = 0.3 t$ ‍ centimeters per minute (where $t$ ‍ is the time in minutes). At time $t = 0$ ‍, the depth of the water is $35$ ‍ centimeters. What is the change in the water's depth during the fourth minute?

Which expression can we use to solve the problem?

Common mistake: Wrong choice of integration interval

As you just saw, picking the correct integration interval is crucial in order to get to the correct answer. Make sure you are not picking the wrong endpoints, especially for the initial point which is usually ignored.

Want more practice? Try this exercise.

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Rachel
Posted 3 years ago. Direct link to Rachel's post “Given F(x) = the integral...”
Given F(x) = the integral from 2 to x of (t^2) dt. Find F'(x).
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(3 votes)
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- Jeffrey
  Posted 3 years ago. Direct link to Jeffrey's post “There you got a function ...”
  There you got a function in the upper bound of integration, so you just write the exact same integrand, just change "t" for "x" and that´s it.
  Comment on Jeffrey's post “There you got a function ...”
  (3 votes)
Ray2017
Posted 4 months ago. Direct link to Ray2017's post “I find the wording of som...”
I find the wording of something, let's say temperature, "is increasing at a rate r(t) per minute" confusing. I know that r(t) is supposed to be the derivative of the actual temperature, but the phrase "per minute" indicates that it is over the interval of one minute, which contradicts the notion of derivative being the instantaneous rate of change right?
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(1 vote)
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