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AP Calculus AB : Integrals

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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

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

Possible Answers:

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

$(-\infty ,+\infty )$

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

$(-5,+\infty)$

$(3,+\infty)$

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 #1 : 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 : Calculus 3

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

Possible Answers:

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

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

x^2 +y^2 =4

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

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

Correct answer:

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

Explanation:

$r = 4\cos\theta$

First, mulitply both sides by r.

$r^2 =4r\cos\theta$

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

x^2+y^2=4x

x^2-4x+y^2=0

x^2 -4x + 4 +y^2 =4

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

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

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Example Question #2 : Interpretations And Properties Of Definite Integrals

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 #3 : 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:

$\frac{5}{2}$

$\frac{-1}{2}$

$\frac{1}{2}$

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 : Interpretations And Properties Of Definite Integrals

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

Possible Answers:

$2xf(x^{2})$

$f(x^{2})$

$g(f(x^{2}))$

$f(x^{2})-f(0)$

$f(2x)$

Correct answer:

$2xf(x^{2})$

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

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

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

$=2xf(x^{2})$

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

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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

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

Possible Answers:

0.54

35.09

1.97

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 #2 : 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:

1.45

1.85

12.60

3.50

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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Example Question #11 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

$\begin{align*}&\text{The flow of water into a tank is given as:}\\&\frac{dV(t)}{dt}=\frac{(79e^{(-t)})}{9}\\&\text{Find the change in volume over the time interval }t=[0.5,1.5]\end{align*}$

Possible Answers:

0.50

10.10

31.97

3.37

Correct answer:

3.37

Explanation:

$\begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int_{0.5}^{1.5}(\frac{(79e^{(-t)})}{9})dt=-\frac{(79e^{(-t)})}{9}|_{0.5}^{1.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{0.5}^{1.5}(\frac{(79e^{(-t)})}{9})dt=-\frac{(79e^{(-(1.5))})}{9}-(-\frac{(79e^{(-(0.5))})}{9})\\&\int_{0.5}^{1.5}(\frac{(79e^{(-t)})}{9})dt=3.37\end{align*}$

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Example Question #11 : Interpretations And Properties Of Definite Integrals

$\begin{align*}&\text{The flow of water into a tank is given as:}\\&\frac{dV(t)}{dt}=\frac{(7\cdot 3^{(\frac{t}{3})})}{2}\\&\text{Find the change in volume over the time interval }t=[0.5,2]\end{align*}$

Possible Answers:

68.06

8.40

1.25

19.33

Correct answer:

8.40

Explanation:

$\begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int_{0.5}^{2}(\frac{(7\cdot 3^{(\frac{t}{3})})}{2})dt=\frac{(21\cdot 3^{(\frac{t}{3})})}{(2ln(3))}|_{0.5}^{2}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{0.5}^{2}(\frac{(7\cdot 3^{(\frac{t}{3})})}{2})dt=\frac{(21\cdot 3^{(\frac{(2)}{3})})}{(2ln(3))}-(\frac{(21\cdot 3^{(\frac{(0.5)}{3})})}{(2ln(3))})\\&\int_{0.5}^{2}(\frac{(7\cdot 3^{(\frac{t}{3})})}{2})dt=8.40\end{align*}$

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AP Calculus AB : Integrals

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

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

Example Question #1 : Interpretations And Properties Of Definite Integrals

Example Question #3 : Calculus 3

Example Question #2 : 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 #1 : Interpretations And Properties Of Definite Integrals

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 #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 #11 : 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 #11 : Interpretations And Properties Of Definite Integrals

All AP Calculus AB Resources

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