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

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Question

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)?

Answer

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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Question

Find the limit.

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

Answer

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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Question

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

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

Answer

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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Question

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

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

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

Answer

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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Question

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

Answer

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

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

$v=p'=6t-1$

now set because that is what the question is asking for.

$v=0=6t-1$

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

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Question

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

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

Answer

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 to find the local maximum.

$v=0=-2t+12$

$t=6$ seconds

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Question

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

Answer

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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Question

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

Answer

${\sqrt{x^2-9}}>0$ 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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Question

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

Answer

$r = 4\cos\theta$

First, mulitply both sides by r.

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

Then, use the identities and $x = r\cos\theta$ .

The answer is .

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Question

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

Answer

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 to .

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

$\begin{align}&\text{The rate of change of the length of an elastic is:}\&\frac{dl(t)}{dt}=\frac{(17sin(t + 2))}{4}\&\text{Determine the change in length over the time interval }t=[5,6.5]\end{align}$

Answer

$\begin{align}&\text{If the function defining a rate of change is known, the total}\&\text{change over a given interval of time is only a matter of integrating}\&\text{this rate function over this same time interval. Considering}\&\text{the function given:}\&\text{Utilizing integral rules:}\&\int[sin(ax)]=-\frac{cos(ax)}{a}\&\int_{5}^{6.5}(\frac{(17sin(t + 2))}{4})dt=-\frac{(17cos(t + 2))}{4}|_{5}^{6.5}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{5}^{6.5}(\frac{(17sin(t + 2))}{4})dt=-\frac{(17cos((6.5) + 2))}{4}-(-\frac{(17cos((5) + 2))}{4})\&\int_{5}^{6.5}(\frac{(17sin(t + 2))}{4})dt=5.76\end{align}$

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Question

$\begin{align}&\text{The flow of water into a tank is given as:}\&\frac{dV(t)}{dt}=\frac{(57cos(3t))}{8}\&\text{Find the change in volume over the time interval }t=[1,6]\end{align}$

Answer

$\begin{align}&\text{If the function defining a rate of change is known, the total}\&\text{change over a given interval of time is only a matter of integrating}\&\text{this rate function over this same time interval. Considering}\&\text{the function given:}\&\text{Utilizing integral rules:}\&\int[cos(ax)]=\frac{sin(ax)}{a}\&\int_{1}^{6}(\frac{(57cos(3t))}{8})dt=\frac{(19sin(3t))}{8}|_{1}^{6}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{1}^{6}(\frac{(57cos(3t))}{8})dt=\frac{(19sin(3(6)))}{8}-(\frac{(19sin(3(1)))}{8})\&\int_{1}^{6}(\frac{(57cos(3t))}{8})dt=-2.12\end{align}$

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Question

$\begin{align}&\text{The velocity of a particle is given as:}\&v(t)=\frac{(12e^{(t)})}{37}\&\text{Find its change in position over the time interval }t=[0.5,3]\end{align}$

Answer

$\begin{align}&\text{If the function defining a rate of change is known, the total}\&\text{change over a given interval of time is only a matter of integrating}\&\text{this rate function over this same time interval. Considering}\&\text{the function given:}\&\text{Utilizing integral rules:}\&\int[e^{ax}]=\frac{e^{ax}}{a}\&\int_{0.5}^{3}(\frac{(12e^{(t)})}{37})dt=\frac{(12e^{(t)})}{37}|_{0.5}^{3}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{0.5}^{3}(\frac{(12e^{(t)})}{37})dt=\frac{(12e^{((3))})}{37}-(\frac{(12e^{((0.5))})}{37})\&\int_{0.5}^{3}(\frac{(12e^{(t)})}{37})dt=5.98\end{align}$

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Question

$\begin{align}&\text{The rate of change of the length of an elastic is:}\&\frac{dl(t)}{dt}=\frac{(73sin(t + 1))}{11}\&\text{Determine the change in length over the time interval }t=[3,5]\end{align}$

Answer

$\begin{align}&\text{If the function defining a rate of change is known, the total}\&\text{change over a given interval of time is only a matter of integrating}\&\text{this rate function over this same time interval. Considering}\&\text{the function given:}\&\text{Utilizing integral rules:}\&\int[sin(ax)]=-\frac{cos(ax)}{a}\&\int_{3}^{5}(\frac{(73sin(t + 1))}{11})dt=-\frac{(73cos(t + 1))}{11}|_{3}^{5}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{3}^{5}(\frac{(73sin(t + 1))}{11})dt=-\frac{(73cos((5) + 1))}{11}-(-\frac{(73cos((3) + 1))}{11})\&\int_{3}^{5}(\frac{(73sin(t + 1))}{11})dt=-10.71\end{align}$

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Question

$\begin{align}&\text{The rate of change of the length of an elastic is:}\&\frac{dl(t)}{dt}=\frac{(71sin(3t))}{7}\&\text{Determine the change in length over the time interval }t=[4,6]\end{align}$

Answer

$\begin{align}&\text{If the function defining a rate of change is known, the total}\&\text{change over a given interval of time is only a matter of integrating}\&\text{this rate function over this same time interval. Considering}\&\text{the function given:}\&\text{Utilizing integral rules:}\&\int[sin(ax)]=-\frac{cos(ax)}{a}\&\int_{4}^{6}(\frac{(71sin(3t))}{7})dt=-\frac{(71cos(3t))}{21}|_{4}^{6}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{4}^{6}(\frac{(71sin(3t))}{7})dt=-\frac{(71cos(3(6)))}{21}-(-\frac{(71cos(3(4)))}{21})\&\int_{4}^{6}(\frac{(71sin(3t))}{7})dt=0.62\end{align}$

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Question

$\begin{align}&\text{The rate of change of the length of an elastic is:}\&\frac{dl(t)}{dt}=\frac{(37t)}{2}\&\text{Determine the change in length over the time interval }t=[3,7]\end{align}$

Answer

$\begin{align}&\text{If the function defining a rate of change is known, the total}\&\text{change over a given interval of time is only a matter of integrating}\&\text{this rate function over this same time interval. Considering}\&\text{the function given:}\&\text{Utilizing integral rules:}\&\int x^{a}=\frac{x^{a+1}}{a+1}\&\int_{3}^{7}(\frac{(37t)}{2})dt=\frac{(37t^{2})}{4}|_{3}^{7}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{3}^{7}(\frac{(37t)}{2})dt=\frac{(37(7)^{2})}{4}-(\frac{(37(3)^{2})}{4})\&\int_{3}^{7}(\frac{(37t)}{2})dt=370.00\end{align}$

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