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Calculus 2 : Integral Applications

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Example Question #51 : Applications In Physics

If v(t)=4t+1 and x(2)=3 , what is the original position function?

Possible Answers:

x(t)=2t^2+t-17

x(t)=2t^2+t+7

x(t)=2t^2+t-7

x(t)=2t^2+3t-7

x(t)=t^2+t-7

Correct answer:

x(t)=2t^2+t-7

Explanation:

Recall that the integral of the velocity function is the position function:

$\int 4t+1dt$

Now, integrate:
$\frac{4t^2}{2}+t=2t^2+t+C$

Now plug in your initial conditions:

2(2^2)+2+C=3; C=-7

Plug your C back into your position function:

x(t)=2t^2+t-7

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Example Question #52 : Applications In Physics

If v(t)=4t-1 and x(2)=3 , what is the original position function?

Possible Answers:

x(t)=2t^2-t

x(t)=2t^2-4t-3

x(t)=2t^2-t+3

x(t)=2t^2-t-3

x(t)=t^2-t-3

Correct answer:

x(t)=2t^2-t-3

Explanation:

Recall that the integral of velocity is the position function:

$\int 4t-1dt$

Now, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{4t^2}{2}-t=2t^2-t+C$

Now plug in your initial conditions to solve for C:

2(2^2)-2+C=3

C=-3

Now plug your C back in to your position function:

x(t)=2t^2-t-3

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Example Question #53 : Applications In Physics

If v(t)=6t+2 and x(1)=4 , what is the original position function?

Possible Answers:

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

x(t)=3t^2+2t-8

x(t)=3t^2+2t-1

x(t)=3t^2+6t-1

x(t)=3t^2+2t+1

Correct answer:

x(t)=3t^2+2t-1

Explanation:

Recall that the integral of the velocity function is the position function:

$\int 6t+2dt$

Now, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{6t^2}{2}+2t=3t^2+2t+C$

Now, plug in your initial conditions:

3(1^2)+2(1)+C=4

C=-1

Now plug your C back into your position function:

x(t)=3t^2+2t-1

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Example Question #51 : Applications In Physics

In the vacuum of space, a particle is initially moving at a constant speed of only 6 meters/second away from you as you watch in your space capsule stationary relative to the particle. You track the object and then observe an acceleration that varies with time as,

$\small a(t)=2t^2$

After travelling for 20 minutes at this speed the particle is now $4\times 10^{9}$ km away from you. How far was it from you at the instant it began accelerating?

Possible Answers:

$x_o \approx 3.12\times 10^{9} m \: \: \:or\: \: \:3.12\times 10^{6} km$

$x_o \approx 2.12\times 10^{9} m \: \: \:or\: \: \:2.12\times 10^{6} km$

$x_o \approx 3.12\times 10^{12} m \: \: \:or\: \: \:3.12\times 10^{9} km$

$x_o \approx 3.65\times 10^{9} m \: \: \:or\: \: \:3.65\times 10^{6} km$

$x_o \approx 3.65\times 10^{12} m \: \: \:or\: \: \:3.65\times 10^{9} km$

Correct answer:

$x_o \approx 3.65\times 10^{12} m \: \: \:or\: \: \:3.65\times 10^{9} km$

Explanation:

$\small a(t)=2t^2$

To solve the problem we first need to find the position function by successively integrating. First find the velocity function by integrating the acceleration:

$\small v(t) = \int a(t)dt =\int 2t^2 dt =\frac{2}{3}t^3 +v_o$

$\small v(t) = \frac{2}{3}t^3 +v_o$

The constant of integration $\small v_o$ is the initial velocity, which we are given as 6 meters-per-second. For convenience, we will omit the units in the equation.

$\small v(t)=\frac{2}{3}t^3 +6$

Now we integrate the velocity function to find the position function:

$\small \small x(t)=\int v(t) dt = \int \left(\frac{2}{3}t^3 +6 \right )dt = \frac{2}{12}t^4+6t+x_o$

$\small \small \small x(t)= \frac{1}{6}t^4+6t+x_o$

The constant of integration x_o represents the initial position when t=0 . After $\small t= 20 minutes = 1200 seconds$ the particle is $4\times 10^{9}$ km, or $4\times 10^{12}$ meters, away. Using standard S.I. units (meters and seconds), solve for x_o at t=1200 seconds where $x(t=1200)=4\times 10^{12}$ .

$\small x (t=1200)\: \: =\: \: \frac{1}{6}(1200)^4+6(1200)+x_o\: \: \: =\: \: \: 4\times 10^{12}$

Now solve for x_o

$x_o = 4\times 10^{12} -\frac{1}{6}(1200)^4-6(1200)$

$x_o \approx 3.65\times 10^{12} m \: \: \:or\: \: \:3.65\times 10^{9} km$

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Example Question #51 : Integral Applications

Give the area of the region bounded by the curves of the functions

f(x) =x^2-4x+7

and

g(x) = 2 x^2 -8x+10 .

Possible Answers:

$\frac{7}{6}$

$\frac{2}{3}$

$\frac{5}{6}$

$\frac{4}{3}$

Correct answer:

$\frac{4}{3}$

Explanation:

The two curves intersect when g(x) = f(x) .

2 x^2 -8x+10 = x^2-4x+7

2 x^2 - x^2 -8x+ 4x+10 - 7 = 0

x^2 -4x+ 3 = 0

(x-1)(x-3) = 0

x= 1,x=3

Therefore, the area between the curves is

$\left |\int_{1}^{3} \left [g(x) - f(x) \right ] \mathrm{d}x \right |$

$=\left |\int_{1}^{3} \left (x^2 -4x+ 3 \right) \mathrm{d}x \right |$

$=\left | \left.\begin{matrix} \frac{x^3}{3} -4\cdot \frac{x^{2}}{2}+ 3x \\ \\ \end{matrix}\right| \begin{matrix} 3\\ \\ 1 \end{matrix} \right |$

$=\left | \left.\begin{matrix} \frac{x^3}{3} -2 x^{2} + 3x \\ \\ \end{matrix}\right| \begin{matrix} 3\\ \\ 1 \end{matrix} \right |$

$=\left | \left (\frac{3^3}{3} -2 \cdot 3^{2} + 3 \cdot 3 \right ) - \left (\frac{1^3}{3} -2 \cdot 1^{2} + 3 \cdot 1 \right ) \right |$

$=\left | \left (9 -18 + 9 \right ) - \left (\frac{1}{3} -2 + 3 \right ) \right |$

$=\left | 0 - \frac{4}{3} \right | = \frac{4}{3}$

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Example Question #231 : Integrals

Give the area of the region bounded by the curves of the functions

$f(x) = 4e^{x}$

and

$g(x) = e^{2x} + 3$ .

Possible Answers:

$4 - \ln 27$

$4 - \ln 3$

$12 \ln 3$

$4 + \ln 27$

$4 + \ln 3$

Correct answer:

$4 - \ln 27$

Explanation:

The two curves intersect when

g(x) = f(x) .

$e^{2x} + 3 = 4e^{x}$

$e^{2x} - 4e^{x} + 3 = 0$

$\left (e^{x} \right )^{2} - 4e^{x} + 3 = 0$

$\left ( e^{x} - 1 \right ) \left ( e^{x} - 3 \right ) = 0$

$e^{x} = 1$

x = 0

$e^{x} = 3$

$x = \ln 3$

Therefore, the area between the curves is

$\left |\int_{0}^{\ln 3} \left [g(x) - f(x) \right ] \mathrm{d}x \right |$

$= \left |\int_{0}^{\ln 3} \left (e^{2x} - 4e^{x} + 3 \right ) \mathrm{d}x \right |$

$=\left | \left.\begin{matrix} \frac{e^{2x}}{2} -4e^{x}+ 3x \\ \\ \end{matrix}\right| \begin{matrix} \ln 3\\ \\ 0 \end{matrix} \right |$

$=\left |\left ( \frac{e^{2 \ln 3}}{2} -4e^{\ln 3}+ 3 \ln 3 \right )- \left ( \frac{e^{2 \cdot 0}}{2} -4e^{0}+ 3 \cdot 0 \right ) \right |$

$=\left |\left ( \frac{e^{ \ln 9}}{2} -4e^{\ln 3}+ 3 \ln 3 \right )- \left ( \frac{e^{ 0}}{2} -4e^{0}+ 0 \right ) \right |$

$\; =\left | \left ( \frac{9}{2} - 4 \cdot 3 + \ln 27 \right) - \left ( \frac{1}{2} -4 \cdot 1 + 0 \right ) \right |$ $\; =\left | \left ( \frac{9}{2} - 4 \cdot 3 + \ln 27 \right) - \left ( \frac{1}{2} -4 \cdot 1 + 0 \right ) \right |$

$\; = \left |\left ( \frac{9}{2} -12 + \ln 27 \right) - \left ( \frac{1}{2} -4 \right ) \right |$

$=\left | \frac{9}{2} -12 + \ln 27 - \frac{1}{2} + 4 \right |$

$=\left | -4 + \ln 27 \right | = 4 - \ln 27$

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Example Question #231 : Integrals

Give the area of the region bounded by the curves of the functions

f(x) = x^3 + x - 5

and

g(x) = 17x - 5 .

Possible Answers:

256

128

Correct answer:

128

Explanation:

The two curves intersect when

f(x) = g(x) .

x^3 + x - 5 = 17x - 5

x^3 + x - 5 - 17x +5 =0

x^3 -16x =0

$x\left ( x^2 -16 \right ) =0$

$x\left ( x+4 \right )\left ( x-4 \right ) =0$

x = -4,0,4

So the curves intersect at three points: x = -4,0,4 .

The area of the region bound by the curves is

$\int_{-4}^{0}\left |f(x) - g(x) \right | \mathrm{d}x+ \int_{0}^{4}\left |f(x) - g(x) \right | \mathrm{d}x$

$=\left |\int_{-4}^{0}\left ( x^3 -16x \right )\mathrm{d}x \right |+\left | \int_{0}^{4} \left ( x^3 -16x \right ) \; \mathrm{d}x \right |$

$=\left | \left.\begin{matrix} \left ( \frac{x^{4}}{4} -\frac{16x^{2}}{2} \right ) \\ \\ \end{matrix}\right|\begin{matrix} 0\\ \\ -4 \end{matrix} \right | + \left | \left.\begin{matrix} \left ( \frac{x^{4}}{4} -\frac{16x^{2}}{2} \right ) \\ \\ \end{matrix}\right|\begin{matrix} 4\\ \\ 0 \end{matrix} \right |$

$=\left | \left.\begin{matrix} \left ( \frac{x^{4}}{4} -8x^{2} \right ) \\ \\ \end{matrix}\right|\begin{matrix} 0\\ \\ -4 \end{matrix} \right | + \left | \left.\begin{matrix} \left \left ( \frac{x^{4}}{4} -8x^{2} \right ) \\ \\ \end{matrix}\right|\begin{matrix} 4\\ \\ 0 \end{matrix} \right |$

$=\left | \left ( \frac{0^{4}}{4} -8 \cdot 0^{2} \right ) - \left ( \frac{(-4)^{4}}{4} -8 \cdot (-4)^{2} \right ) \right | + \left | \left ( \frac{4^{4}}{4} -8 \cdot 4^{2} \right ) - \left ( \frac{0^{4}}{4} -8 \cdot 0^{2} \right ) \right |$

$=\left | 0 - \left ( \frac{256}{4} -8 \cdot 16 \right ) \right | + \left | \left ( \frac{256}{4} -8 \cdot 16 \right ) - 0 \right |$

$=\left | - \left (64 -128\right ) \right | + \left | \left ( 64 -128 \right ) \right |$

= 64 + 64 = 128

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Example Question #52 : Integral Applications

Give the area of the region bounded by the curves of the functions

$f(x) = \frac{16}{x}$

and

g(x) = 10 -x .

Possible Answers:

$16 \ln 4$

$56 - 16 \ln 4$

$30 - 16 \ln 4$

$30 - 16 \ln 16$

$56 - 16 \ln 16$

Correct answer:

$30 - 16 \ln 4$

Explanation:

The two curves intersect when

g(x) = f(x) .

$\frac{16}{x} =10 -x$

$x - 10 + \frac{16}{x} = 0$

Since it can be assumed that $x \neq 0$ ,

$x \left ( x - 10 + \frac{16}{x} \right )=x\cdot 0$

$x ^{2}- 10x + 16 = 0$

(x-2) (x-8)= 0

Therefore, either x-2 = 0 , in which case x = 2 , or x - 8 = 0 , in which case x = 8 .

Therefore, the area between the curves is

$\int_{2}^{8}\left |g(x) - f(x) \right | \mathrm{d}x$

$=\left |\int_{2}^{8} \left ( x - 10 + \frac{16}{x} \right ) \mathrm{d}x \right |$

$=\left | \left.\begin{matrix} \frac{x^2}{2} -10x + 16 \ln x\\ \\ \end{matrix}\right| \begin{matrix} 8\\ \\ 2 \end{matrix} \right |$

$=\left | \left (\frac{8^2}{2} -10 \cdot 8 + 16 \ln 8 \right ) - \left ( \frac{2^2}{2} -10 \cdot 2 + 16 \ln 2 \right ) \right |$

$=\left | \left (32-80 + 16 \ln 8 \right ) - \left ( 2-20 + 16 \ln 2 \right ) \right |$

$=\left | \left (-48 + 16 \ln 8 \right ) - \left ( -18 + 16 \ln 2 \right ) \right |$

$=\left | -48 + 18 + 16 \ln 8 - 16 \ln 2 \right |$

$= \left | -30 + 16( \ln 8 - \ln 2 ) \right | =\left | -30 + 16 \ln 4 \right |= 30 - 16 \ln 4$

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Example Question #234 : Integrals

Find the area under the curve f(x) = e^x from x = -2 to x = 2 .

Possible Answers:

2e^2

$1-2e^{-2}$

$e^2 - e^{-2}$

$e^2 + 2e^{-2}$

2e^2 - 1

Correct answer:

$e^2 - e^{-2}$

Explanation:

The area under a curve f(x) between two x-values $x_{1},x_2$ corresponds to the integral

$\int_{x_1}^{x_2} f(x)dx$ .

In this case, f(x) = e^x, x_1 = -2, x_2 = 2 , and we can find by the fundamental theorem of calculus that...

$\int_{-2}^{2} e^xdx$

= [e^x] evaluated from to

$= e^2 - e^{-2}$

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Example Question #235 : Integrals

Find the area between f(x) and g(x) , such that $f(x) = \sqrt{x}$ and g(x) = x^3

Possible Answers:

$\frac{5}{12}$

$\frac{1}{3}$

$\frac{7}{12}$

$\frac{1}{4}$

$\frac{1}{2}$

Correct answer:

$\frac{5}{12}$

Explanation:

The first step to find the area between two curves is to set them equal to each other and find their intersection points.

$\\f(x) = g(x)\\ \sqrt{x} = x^3\\ x = x^6\\ x = 0, x = 1$

The next step is to find which curve is "on the top" and which is "on the bottom" inside the interval (0,1) . We find that f(x) is on the top and g(x) is on the bottom.

Now, to find the area between the two curves, we take the integral from to of the top curve minus the bottom curve.

$\\\int_{0}^{1}(f(x) - g(x))dx\\ \int_{0}^{1}(\sqrt{x} - x^3)dx\\ \left [ \frac{2}{3}x^{3/2} - \frac{1}{4}x^4 \right ]$ evaluated from to

$\\= \left(\frac{2}{3} - \frac{1}{4}\right) - (0 - 0)\\ = \frac{5}{12}$

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Calculus 2 : Integral Applications

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #51 : Applications In Physics

Example Question #52 : Applications In Physics

Example Question #53 : Applications In Physics

Example Question #51 : Applications In Physics

Example Question #51 : Integral Applications

Example Question #231 : Integrals

Example Question #231 : Integrals

Example Question #52 : Integral Applications

Example Question #234 : Integrals

Example Question #235 : Integrals

All Calculus 2 Resources

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