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AP Calculus AB : Definite integral of the rate of change of a quantity over an interval interpreted as the change of the quantity over the interval

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Example Questions

Example Question #1 : Interpretations And Properties Of Definite Integrals

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:

Correct answer:

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

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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

Find the limit.

lim as n approaches infiniti of ((4n³) – 6n)/((n³) – 2n²+ 6)

Possible Answers:

–6

nonexistent

Correct answer:

Explanation:

lim as n approaches infiniti of ((4n³) – 6n)/((n³) – 2n²+ 6)

Use L'Hopitals rule to find the limit.

lim as n approaches infiniti of ((4n³) – 6n)/((n³) – 2n²+ 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.

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

Possible Answers:

$\frac{1}{6}$

Correct answer:

$\frac{1}{6}$

Explanation:

The answer is $\frac{1}{6}$ seconds.

$p=3t^{2}-t+16$

$v=p'=6t-1$

now set v=0 because that is what the question is asking for.

$v=0=6t-1$

$t=\frac{1}{6}$ seconds

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Example Question #3 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

A particle's movement is represented by $p=-t^{2}+12t+2$

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

Possible Answers:

Correct answer:

Explanation:

The answer is at 6 seconds.

$p=-t^{2}+12t+2$

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

$v=p'=-2t+12$

Now we set v=0 to find the local maximum.

$v=0=-2t+12$

$t=6$ seconds

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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

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

Possible Answers:

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

$(3,+\infty)$

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

$(-5,+\infty)$

$(-\infty ,+\infty )$

Correct answer:

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

Explanation:

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

$x^2-9>0$

$x^2>9$

$\sqrt{x^2}>\sqrt{9}$

$\left | x \right |>3$

$x>3\: or\, x<-3$

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Example Question #4 : Interpretations And Properties Of Definite Integrals

If $y-6x-x^{2}=4$ ,

then at x=1 , what is 's instantaneous rate of change?

Possible Answers:

Correct answer:

Explanation:

The answer is 8.

$y-6x-x^{2}=4$

$y=x^{2}+6x+4$

$y'=2x+6$

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

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Example Question #3 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

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

Possible Answers:

Correct answer:

Explanation:

The average function value is given by the following formula:

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

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

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

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Example Question #4 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

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

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

then find $h^{'}(x)$ .

Possible Answers:

$\frac{-1}{2}$

$\frac{5}{2}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

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

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

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

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

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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

$\begin{align*}&\text{The velocity of a particle is given as:}\\&v(t)=\frac{(26sin(t + 2))}{3}\\&\text{Find its change in position over the time interval }t=4,5.5\end{align*}$

Possible Answers:

1.97

0.54

35.09

5.32

Correct answer:

5.32

Explanation:

$\begin{align*}&\text{If the velocity function of a particle is known, finding the}\\&\text{distance it travels over a given interval of time is only a}\\&\text{matter of integrating the velocity function over this same interval.}\\&\text{Considering the function given:}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{4}^{5.5}(\frac{(26sin(t + 2))}{3})dt=-\frac{(26cos(t + 2))}{3}|_{4}^{5.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{4}^{5.5}(\frac{(26sin(t + 2))}{3})dt=-\frac{(26cos((5.5) + 2))}{3}-(-\frac{(26cos((4) + 2))}{3})\\&\int_{4}^{5.5}(\frac{(26sin(t + 2))}{3})dt=5.32\end{align*}$

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Example Question #7 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

$\begin{align*}&\text{The velocity of a particle is given as:}\\&v(t)=\frac{21}{5}\\&\text{Find its change in position over the time interval }t=4,7\end{align*}$

Possible Answers:

3.50

12.60

1.45

1.85

Correct answer:

12.60

Explanation:

$\begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{4}^{7}(\frac{21}{5})dt=\frac{(21t)}{5}|_{4}^{7}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{4}^{7}(\frac{21}{5})dt=\frac{(21(7))}{5}-(\frac{(21(4))}{5})\\&\int_{4}^{7}(\frac{21}{5})dt=12.60\end{align*}$

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AP Calculus AB : Definite integral of the rate of change of a quantity over an interval interpreted as the change of the quantity over the interval

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1 : Interpretations And Properties Of Definite Integrals

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

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

Example Question #3 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

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

Example Question #4 : Interpretations And Properties Of Definite Integrals

Example Question #3 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

Example Question #4 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

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

Example Question #7 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

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