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Focus Area: Antidifferentiation Created by Mr. Lajoie and Mr. Herron Finding the Constant of Integration

Focus Area: Antidifferentiation Created by Mr. Lajoie and Mr. Herron

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Page 1: Focus Area: Antidifferentiation Created by Mr. Lajoie and Mr. Herron

Focus Area: Antidifferentiation

Created by Mr. Lajoie and Mr. Herron

Finding the Constant of Integration

Page 2: Focus Area: Antidifferentiation Created by Mr. Lajoie and Mr. Herron

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Initial-Value Problems One of the richest topics in calculus is

differential equations. In fact, whole college courses are devoted to the study of differential equations.

Quite simply, any equation of the form: is a differential equation.

Recall that the solution to the equation above is a family of functions whose derivative is .

If we are also told the value of the antiderivative at a certain value of , then we are able to solve for the unique antiderivative that is a solution to the differential equation.

This is called an initial-value problem.

)(xfdx

dy

Page 3: Focus Area: Antidifferentiation Created by Mr. Lajoie and Mr. Herron

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Example 1

If

Then what is f(9)?

14,)(' 2/3 fandttf

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Solution Step 1: Find the Antiderivative

I used the “reverse Power Rule” here. If you are confused on this part, you need to review the first section in this Playlist.

Here, I simplified (-3/2 + 1 = -1/2)

I simplified because 1 divided by -1/2 equals -2

It made more sense for me to write the square root of t in the denominator than to leave it as “t to the negative one-half power”

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Solution Step 2: Find the value of “C” (constant)

Here, I know that f(4)=-1. So, when my “input” for t is 4, my output, or answer, is -1. Here, I have substituted in 4 to my answer from the previous slide and set the problem equal to -1. The goal of this is to see what the value of C must be.

I can now see that the value of C will be zero. This time, there was no Constant of Integration, so the equation will not have a value for C at the end.

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Solution Step 3: Use antiderivative and C to find f(9).

Now, I am ready to answer the final question:

What is f(9)?

Here, I substituted in a 9 for t in the denominator of the function. Remember, C = 0, so there is no “constant” that I must add or subtract at the end.

The square root of 9 is 3, and there is no more simplification possible, so my final answer is that f(9) = -2/3.

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Example 2 Solve the initial-value problem:

What is the full integral, including the exact constant of integration (C) for this problem?

01,)34( 2 ytdt

dy

The answer key to these example problems is on the last slide. If you do not know why you got a problem wrong, ask your teacher during Office Hours!

Page 8: Focus Area: Antidifferentiation Created by Mr. Lajoie and Mr. Herron

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Example 3 Solve the initial-value problem:

What is the full integral, including the exact constant of integration (C) for this problem?

24

),3(sec2

yxdx

dy

The answer key to these example problems is on the last slide. If you do not know why you got a problem wrong, ask your teacher during Office Hours!

Page 9: Focus Area: Antidifferentiation Created by Mr. Lajoie and Mr. Herron

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Example 4 Solve the initial-value problem:

What is the constant of integration (“C”) for this problem?

11,4 yedx

dy x

The answer key to these example problems is on the last slide. If you do not know why you got a problem wrong, ask your teacher during Office Hours!

Page 10: Focus Area: Antidifferentiation Created by Mr. Lajoie and Mr. Herron

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Example 5 Find a function that satisfies the following

conditions:

What is the constant of integration (“C”) for this problem?

𝑓 ′ ′ (𝑥 )=𝑥3−2𝑥+1 , 𝑓 ′ (1 )=0 , 𝑓 (1 )=4

The answer key to these example problems is on the last slide. If you do not know why you got a problem wrong, ask your teacher during Office Hours!

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Example 6 Find a function that satisfies the following

conditions:

What is the constant of integration (“C”) for this problem?

𝑓 ′ ′ (𝜃 )=cos𝜃 , 𝑓 ′( 𝜋2 )=1 , 𝑓 ( 𝜋2 )=6

The answer key to these example problems is on the last slide. If you do not know why you got a problem wrong, ask your teacher during Office Hours!

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Answer Key