AP Calculus AB : Interpretations and properties of definite integrals

Study concepts, example questions & explanations for AP Calculus AB

varsity tutors app store varsity tutors android store

Example Questions

Example Question #1 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

If f(1) = 12, f' is continuous, and the integral from 1 to 4 of f'(x)dx = 16, what is the value of f(4)?

Possible Answers:

27

12

16

4

28

Correct answer:

28

Explanation:

You are provided f(1) and are told to find the value of f(4). By the FTC, the following follows:

(integral from 1 to 4 of f'(x)dx) + f(1) = f(4)

16 + 12 = 28

Example Question #503 : Ap Calculus Ab

Find the limit. 

lim as n approaches infiniti of ((4n3) – 6n)/((n3) – 2n+ 6)

Possible Answers:

0

1

nonexistent

–6

4

Correct answer:

4

Explanation:

lim as n approaches infiniti of ((4n3) – 6n)/((n3) – 2n+ 6)

Use L'Hopitals rule to find the limit. 

lim as n approaches infiniti of ((4n3) – 6n)/((n3) – 2n+ 6)

lim as n approaches infiniti of ((12n2) – 6)/((3n2) – 4n + 6)

lim as n approaches infiniti of 24n/(6n – 4)

lim as n approaches infiniti of 24/6

The limit approaches 4. 

Example Question #1 : Interpretations And Properties Of Definite Integrals

If a particle's movement is represented by p=3t^{2}-t+16\(\displaystyle p=3t^{2}-t+16\), then when is the velocity equal to zero?

Possible Answers:

\(\displaystyle 0\)

\(\displaystyle 1\)

\(\displaystyle 16\)

\(\displaystyle 6\)

\(\displaystyle \frac{1}{6}\)

Correct answer:

\(\displaystyle \frac{1}{6}\)

Explanation:

The answer is \(\displaystyle \frac{1}{6}\) seconds.

 

p=3t^{2}-t+16\(\displaystyle p=3t^{2}-t+16\)

v=p'=6t-1\(\displaystyle v=p'=6t-1\)

now set \(\displaystyle v=0\) because that is what the question is asking for. 

v=0=6t-1\(\displaystyle v=0=6t-1\)

t=\frac{1}{6}\(\displaystyle t=\frac{1}{6}\) seconds

Example Question #44 : Integrals

A particle's movement is represented by p=-t^{2}+12t+2\(\displaystyle p=-t^{2}+12t+2\)

 At what time is the velocity at it's greatest?

Possible Answers:

\(\displaystyle 6\)

\(\displaystyle 0\)

\(\displaystyle 12\)

\(\displaystyle -2\)

\(\displaystyle 2\)

Correct answer:

\(\displaystyle 6\)

Explanation:

The answer is at 6 seconds. 

 

p=-t^{2}+12t+2\(\displaystyle p=-t^{2}+12t+2\) 

We can see that this equation will look like a upside down parabola so we know there will be only one maximum.

v=p'=-2t+12\(\displaystyle v=p'=-2t+12\)

Now we set \(\displaystyle v=0\) to find the local maximum. 

v=0=-2t+12\(\displaystyle v=0=-2t+12\)

t=6\(\displaystyle t=6\) seconds

Example Question #1 : Interpretations And Properties Of Definite Integrals

What is the domain of f(x)=\frac{x+5}{\sqrt{x^2-9}}\(\displaystyle f(x)=\frac{x+5}{\sqrt{x^2-9}}\)?

Possible Answers:

(-\infty,-3)\cup (3,+\infty)\(\displaystyle (-\infty,-3)\cup (3,+\infty)\)

(-5,+\infty)\(\displaystyle (-5,+\infty)\)

(3,+\infty)\(\displaystyle (3,+\infty)\)

(-\infty,-3]\cup [3,+\infty)\(\displaystyle (-\infty,-3]\cup [3,+\infty)\)

\(\displaystyle (-\infty ,+\infty )\)

Correct answer:

(-\infty,-3)\cup (3,+\infty)\(\displaystyle (-\infty,-3)\cup (3,+\infty)\)

Explanation:

{\sqrt{x^2-9}}>0\(\displaystyle {\sqrt{x^2-9}}>0\) because the denominator cannot be zero and square roots cannot be taken of negative numbers

x^2-9>0\(\displaystyle x^2-9>0\)

x^2>9\(\displaystyle x^2>9\)

\sqrt{x^2}>\sqrt{9}\(\displaystyle \sqrt{x^2}>\sqrt{9}\)

\left | x \right |>3\(\displaystyle \left | x \right |>3\)

x>3\: or\, x<-3\(\displaystyle x>3\: or\, x< -3\)

Example Question #2 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

If y-6x-x^{2}=4\(\displaystyle y-6x-x^{2}=4\),

then at \(\displaystyle x=1\), what is \(\displaystyle y\)'s instantaneous rate of change?

Possible Answers:

\(\displaystyle 8\)

\(\displaystyle 0\)

\(\displaystyle -6\)

\(\displaystyle 1\)

\(\displaystyle 6\)

Correct answer:

\(\displaystyle 8\)

Explanation:

The answer is 8.

 

y-6x-x^{2}=4\(\displaystyle y-6x-x^{2}=4\)

y=x^{2}+6x+4\(\displaystyle y=x^{2}+6x+4\)

y'=2x+6\(\displaystyle y'=2x+6\)

\(\displaystyle y^{'}=2(1)+6=8\)

Example Question #5 : Calculus 3

Which of the following represents the graph of the polar function \(\displaystyle r = 4\cos\theta\) in Cartestian coordinates?

Possible Answers:

\(\displaystyle x^2 +y^2 =4\)

\(\displaystyle x^2+(y-2)^2 =4\)

\(\displaystyle (x-1)^2 +(y-2)^2 =4\)

\(\displaystyle (x-2)^2 +y^2 =4\)

\(\displaystyle (x-2)^2 +(y-2)^2 =4\)

Correct answer:

\(\displaystyle (x-2)^2 +y^2 =4\)

Explanation:

\(\displaystyle r = 4\cos\theta\)

First, mulitply both sides by r. 

\(\displaystyle r^2 =4r\cos\theta\)

Then, use the identities \(\displaystyle x^2+y^2 =r^2\) and \(\displaystyle x = r\cos\theta\).

\(\displaystyle x^2+y^2=4x\)

\(\displaystyle x^2-4x+y^2=0\)

\(\displaystyle x^2 -4x + 4 +y^2 =4\)

\(\displaystyle (x-2)^2 +y^2 =4\)

The answer is \(\displaystyle (x-2)^2 +y^2 =4\).

Example Question #3 : Interpretations And Properties Of Definite Integrals

What is the average value of the function \(\displaystyle f(x) = 3x^{2} - 5x +7\) from \(\displaystyle x = 1\) to \(\displaystyle x=5\)?

Possible Answers:

\(\displaystyle 92\)

\(\displaystyle 23\)

\(\displaystyle 13\)

\(\displaystyle 65\)

\(\displaystyle 52\)

Correct answer:

\(\displaystyle 23\)

Explanation:

The average function value is given by the following formula:

\(\displaystyle Ave = \frac{1}{5-1}\int_{1}^{5} f(x) dx = \frac{1}{5-1}\int_{1}^{5} (3x^{2} - 5x + 7)dx\)

\(\displaystyle Ave = \frac{1}{4} [x^{3}-\frac{5}{2}x^{2} + 7x]\), evaluated from \(\displaystyle x = 1\) to \(\displaystyle x = 5\).

\(\displaystyle Ave = \frac{1}{4}[(125-\frac{125}{2} + 35) - (1-\frac{5}{2}+7)] = 23\)

Example Question #2 : Interpretations And Properties Of Definite Integrals

h(x)=\frac{g(x)}{(1+f(x))}\(\displaystyle h(x)=\frac{g(x)}{(1+f(x))}\)

If \(\displaystyle f(x)=1,\ f^{'}(x)=2,\ g(x)=3,\ and\ g^{'}(x)=4\)

then find \(\displaystyle h^{'}(x)\).

Possible Answers:

\(\displaystyle 2\)

\frac{-1}{2}\(\displaystyle \frac{-1}{2}\)

\(\displaystyle 3\)

\(\displaystyle \frac{1}{2}\)

\frac{5}{2}\(\displaystyle \frac{5}{2}\)

Correct answer:

\(\displaystyle \frac{1}{2}\)

Explanation:

We see the answer is 0 after we do the quotient rule. 

 

h(x)=\frac{g(x)}{(1+f(x))}\(\displaystyle h(x)=\frac{g(x)}{(1+f(x))}\)

h'(x)=\frac{g'(x)(1+f(x))-g(x)(f'(x))}{(1+f(x))^{2}}\(\displaystyle h'(x)=\frac{g'(x)(1+f(x))-g(x)(f'(x))}{(1+f(x))^{2}}\)

\(\displaystyle {h}'(x)=\frac{4(1+1)-3(2)}{(1+1)^{2}}=\frac{8-6}{4}=\frac{1}{2}\)

Example Question #2 : Interpretations And Properties Of Definite Integrals

If g(x)=\int_{0}^{x^2}f(t)dt\(\displaystyle g(x)=\int_{0}^{x^2}f(t)dt\), then which of the following is equal to g'(x)\(\displaystyle g'(x)\)?

Possible Answers:

f(2x)\(\displaystyle f(2x)\)

f(x^{2})-f(0)\(\displaystyle f(x^{2})-f(0)\)

2xf(x^{2})\(\displaystyle 2xf(x^{2})\)

f(x^{2})\(\displaystyle f(x^{2})\)

g(f(x^{2}))\(\displaystyle g(f(x^{2}))\)

Correct answer:

2xf(x^{2})\(\displaystyle 2xf(x^{2})\)

Explanation:

According to the Fundamental Theorem of Calculus, if we take the derivative of the integral of a function, the result is the original function. This is because differentiation and integration are inverse operations.

For example, if h(x)=\int_{a}^{x}f(u)du\(\displaystyle h(x)=\int_{a}^{x}f(u)du\), where \(\displaystyle a\) is a constant, then h'(x)=f(x)\(\displaystyle h'(x)=f(x)\).

We will apply the same principle to this problem. Because the integral is evaluated from 0 to x^{2}\(\displaystyle x^{2}\), we must apply the chain rule.

g'(x)=\frac{d}{dx}\int_{0}^{x^{2}}f(t)dt=f(x^{2})\cdot \frac{d}{dx}(x^{2})\(\displaystyle g'(x)=\frac{d}{dx}\int_{0}^{x^{2}}f(t)dt=f(x^{2})\cdot \frac{d}{dx}(x^{2})\)

=2xf(x^{2})\(\displaystyle =2xf(x^{2})\)

The answer is 2xf(x^{2})\(\displaystyle 2xf(x^{2})\).

Learning Tools by Varsity Tutors