AP Calculus AB : Basic properties of definite integrals (additivity and linearity)

Study concepts, example questions & explanations for AP Calculus AB

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Example Questions

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Example Question #41 : Interpretations And Properties Of Definite Integrals

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Example Question #42 : Interpretations And Properties Of Definite Integrals

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Example Question #43 : Interpretations And Properties Of Definite Integrals

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Example Question #44 : Interpretations And Properties Of Definite Integrals

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Example Question #23 : Basic Properties Of Definite Integrals (Additivity And Linearity)

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Example Question #52 : Interpretations And Properties Of Definite Integrals

Given that  and , find the value of the following expression:

 

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First, simplifying the given's gives us  

And

Our goal is to get the given expression into terms of these two integrals. Our first step will be to try and get a  from our expression.

 

First note,

And for the third term,

Putting these facts together, we can rewrite the original expression as

Rearranging,

The three terms in parentheses can all be brought together, as the top limit of the previous integral matches the bottom limit of the next integral. Thus, we now have

 

Substituting in our given's, this simplifies to 

Example Question #21 : Basic Properties Of Definite Integrals (Additivity And Linearity)

Evaluate the definite integral 

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Here we are using several basic properties of definite integrals as well as the fundamental theorem of calculus. 

First, you can pull coefficients out to the front of integrals.

Second, we notice that our lower bound is bigger than our upper bound. You can switch the upper and lower bounds if you also switch the sign.

Lastly, our integral "distributes" over addition and subtraction. That means you can split the integral by each term and integrate each term separately. 

Now we integrate and calculate using the Fundamental Theorem of Calculus.

Example Question #21 : Basic Properties Of Definite Integrals (Additivity And Linearity)

Solve:

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Rather than solve each integral independently, we can use the property of linearity to add the integrals together:

The integrals were solved using the following rule:

Finally, we evaluate the definite integral by evaluating the answer at the upper bound and subtracting the answer evaluated at the lower bound:

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