AP Calculus AB : Fundamental Theorem of Calculus

Study concepts, example questions & explanations for AP Calculus AB

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Example Questions

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Example Question #1 : Integrals

Evaluate .

Possible Answers:

Does not exist

Correct answer:

Explanation:

Even though an antideritvative of  does not exist, we can still use the Fundamental Theorem of Calculus to "cancel out" the integral sign in this expression.

 

. Start

. You can "cancel out" the integral sign with the derivative by making sure the lower bound of the integral is a constant, the upper bound is a differentiable function of , , and then substituting  in the integrand. Lastly the Theorem states you must multiply your result by  (similar to the directions in using the chain rule).

.

Example Question #1 : Integrals

The graph of a function  is drawn below. Select the best answers to the following: 

 

Pbstm

 

 What is the best interpretation of the function?

 

 

 Which plot shows the derivative of the function ?

 

 

 

 

Possible Answers:

Wrngan2

Wrong3q10

Wrn4

Question 10 correct answer

Correct answer:

Question 10 correct answer

Explanation:

   

 

The function  represents the area under the curve  from  to some value of .  

 

Do not be confused by the use of  in the integrand. The reason we use  is because are writing the area as a function of , which requires that we treat the upper limit of integration as a variable . So we replace the independent variable of  with a dummy index  when we write down the integral. It does not change the fundamental behavior of the function  or 

 

  The graph of the derivative of  is the same as the graph for . This follows directly from the Second Fundamental Theorem of Calculus.

If the function  is continuous on an interval  containing , then the function defined by: 

 

has for its' derivative 

 

 

Example Question #1 : Fundamental Theorem Of Calculus

Evaluate 

Possible Answers:

Correct answer:

Explanation:

Here we could use the Fundamental Theorem of Calculus to evaluate the definite integral; however, that might be difficult and messy.

Instead, we make a clever observation of the graph of

Namely, that

This means that the values of the graph when comparing x and -x are equal but opposite. Then we can conclude that

 

 

Example Question #1 : Fundamental Theorem Of Calculus

Wolframalpha--plot_piecewisexxlt03-x0ltxlt213-xxgt2--2013-09-10_1008

Find 

Possible Answers:

Does not exist

Correct answer:

Does not exist

Explanation:

The one side limits are not equal: left is 0 and right is 3

Example Question #1 : Fundamental Theorem Of Calculus

Wolframalpha--plot_piecewisexxlt03-x0ltxlt213-xxgt2--2013-09-10_1008

Which of the following is a vertical asymptote?

Possible Answers:

Correct answer:

Explanation:

When  approaches 3,  approaches .

Vertical asymptotes occur at values.  The horizontal asymptote occurs at

.

Example Question #1 : Fundamental Theorem Of Calculus

 

What are the horizontal asymptotes of ?

Possible Answers:

Correct answer:

Explanation:

Compute the limits of  as  approaches infinity.

Example Question #1 : Fundamental Theorem Of Calculus

Write the domain of the function.

f(x)=\frac{(x^{4}-81)^{1/2}}{x-4}

Possible Answers:

Correct answer:

Explanation:

The answer is 

The denominator must not equal zero and anything under a radical must be a nonnegative number.

Example Question #5 : Fundamental Theorem Of Calculus

What is the value of the derivative of  at x=1?

Possible Answers:

Correct answer:

Explanation:

First, find the derivative of the function, which is:

 

Then, plug in 1 for x:

 

The result is .

Example Question #121 : Integrals

Evaluate the following limit:

\lim_{x\rightarrow \infty }\frac{1-x^4}{x^2-4x^4}

Possible Answers:

Does not exist.

\frac{1}{4}

\infty

\frac{1}{2}

Correct answer:

\frac{1}{4}

Explanation:

First, let's multiply the numerator and denominator of the fraction in the limit by \frac{1}{x^{4}}.

\lim_{x\rightarrow \infty }\frac{1-x^4}{x^2-4x^4}=\lim_{x\rightarrow \infty }\frac{(\frac{1}{x^4})(1-x^4)}{\frac{1}{x^{4}}(x^2-4x^4)}

=\lim_{x\rightarrow \infty }\frac{\frac{1}{x^4}-1}{\frac{1}{x^2}-4}

As  becomes increasingly large the \frac{1}{x^{4}} and\frac{1}{x^{2}} ^{} terms will tend to zero. This leaves us with the limit of .

\lim_{x\rightarrow \infty }\frac{-1}{-4}=\frac{1}{4}.

 

The answer is \frac{1}{4}.

Example Question #7 : Fundamental Theorem Of Calculus

Let  and  be inverse functions, and let

.

What is the value of ?

Possible Answers:

Correct answer:

Explanation:

Since  and  are inverse functions, . We can differentiate both sides of the equation  with respect to to obtain the following:

g'(f(x))\cdot f'(x)=1

We are asked to find , which means that we will need to find  such that . The given information tells us that , which means that . Thus, we will substitute 3 into the equation.

g'(f(3))\cdot f'(3)=1

The given information tells us that.

The equation then becomes g'(4)\cdot (-2)=1.

We can now solve for .

g'(4)=-\frac{1}{2}.

The answer is -\frac{1}{2}.

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