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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 8

Lesson 3: Using accumulation functions and definite integrals in applied contexts- Area under rate function gives the net change
- Interpreting definite integral as net change
- Worked examples: interpreting definite integrals in context
- Interpreting definite integrals in context
- Analyzing problems involving definite integrals
- Analyzing problems involving definite integrals
- Analyzing problems involving definite integrals
- Worked example: problem involving definite integral (algebraic)
- Problems involving definite integrals (algebraic)

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# Analyzing problems involving definite integrals

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, left parenthesis, t, right parenthesis, equals, 30, e, start superscript, minus, 0, point, 3, t, end superscript degrees Celsius per minute (where t is the time in minutes). At time t, equals, 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, equals, 0 and t, equals, 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 integral, start subscript, a, end subscript, start superscript, b, end superscript, r, left parenthesis, t, right parenthesis, d, t describes the amount by which the quantity changed between t, equals, a and t, equals, b.

So in our case, the amount by which the temperature increased between t, equals, 0 and t, equals, 5 minutes is given by integral, start subscript, 0, end subscript, start superscript, 5, end superscript, r, left parenthesis, t, right parenthesis, d, t.

integral, start subscript, 0, end subscript, start superscript, 5, end superscript, r, left parenthesis, t, right parenthesis, d, t, approximately equals, 77, point, 7 degrees Celsius

Now imagine we were asked a different question:

**what is the soup's temperature at t, equals, 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, equals, 0 was 23 degrees Celsius. Adding that to the change in temperature between t, equals, 0 and t, equals, 5 gives us the temperature at t, equals, 5:

Since we already calculated integral, start subscript, 0, end subscript, start superscript, 5, end superscript, r, left parenthesis, t, right parenthesis, d, t, we can tell that at t, equals, 5 minutes, the temperature was 23, plus, 77, point, 7, equals, 100, point, 7 degrees Celsius. That's boiling hot!

#### 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 |

#### 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.*

## Want to join the conversation?

- Given F(x) = the integral from 2 to x of (t^2) dt. Find F'(x).(4 votes)
- 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.(4 votes)