You CAN Ace Calculus

Integration by Parts |

Integrals - Trig Substitution |

Integrals - Partial Fractions |

Volume Integrals |

Centroid |

Work |

Parametrics |

Calculus 2 Exams |

Exam B1 - Exam B2 |

Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |

Arc Length (Vector Functions) |

Arc Length Function |

Arc Length Parameter |

Conservative Vector Fields |

Cross Product |

Curl |

Curvature |

Cylindrical Coordinates |

Lagrange Multipliers |

Line Integrals |

Partial Derivatives |

Partial Integrals |

Path Integrals |

Potential Functions |

Principal Unit Normal Vector |

Differential Equations |
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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

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Help Keep 17Calculus Free |
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This is the midterm exam for second semester single variable calculus.

Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.

- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.

- Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.

- Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).

- Use your calculator as little as possible except for graphing and checking your calculations.

- Work the entire exam before checking any solutions.

- After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.

- Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.

IMPORTANT -

Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.

Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

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Time | 1 hour and 40 minutes |

Questions | 11 |

Total Points | 100 |

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Calculator | see instructions |

Formula Sheet(s) | none |

Other Tools | none |

*Instructions:*

- This exam is in two main parts, labeled parts A and B, with different instructions for each part.

- Show all your work.

- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.

- Correct notation counts (i.e. points will be taken off for incorrect notation).

- Give exact, simplified answers.

Evaluate these integrals.

*Instructions for Part A* - - You have 40 minutes to complete this part of the exam. No calculators are allowed. All questions in this section are worth 10 points.

\( \int{ 5t(2 + t^2)^3 ~dt } \)

Problem Statement |
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\( \int{ 5t(2 + t^2)^3 ~dt } \)

Final Answer |
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\(\displaystyle{ \frac{5}{8}(2+t^2)^4 + C }\) |

Problem Statement |
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\( \int{ 5t(2 + t^2)^3 ~dt } \)

Solution |
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\( \int{ 5t(2 + t^2)^3 ~dt } \) |

Use integration by substitution. |

\( u = 2+t^2 \to du=2t~dt \to dt=du/(2t)\) |

\(\displaystyle{ \int{ 5t ~ u^3 \frac{du}{2t} } }\) |

\(\displaystyle{ \frac{5}{2}\int{ u^3 ~ du } }\) |

\(\displaystyle{ \frac{5}{2} \frac{u^4}{4} + C }\) |

\(\displaystyle{ \frac{5}{8}(2+t^2)^4 + C }\) |

Final Answer |
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\(\displaystyle{ \frac{5}{8}(2+t^2)^4 + C }\) |

close solution |

\(\int{ (2x+1)\ln x ~ dx } \)

Problem Statement |
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\(\int{ (2x+1)\ln x ~ dx } \)

Final Answer |
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\((x^2+x)\ln x - (1/2)x^2 - x + C \) |

Problem Statement |
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\(\int{ (2x+1)\ln x ~ dx } \)

Solution |
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Use integration by parts. |

\( u=\ln x \to du=(1/x)dx \) |

\(dv = (2x+1)dx \to v = x^2+x \) |

\(\int{ u ~dv } = u v - \int{ v ~du } \) |

\(\displaystyle{ (x^2+x)\ln x - \int{ (x^2+x)\frac{dx}{x} } }\) |

\( (x^2+x)\ln x - \int{ x+1~dx } \) |

\((x^2+x)\ln x - (1/2)x^2 - x + C \) |

Final Answer |
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\((x^2+x)\ln x - (1/2)x^2 - x + C \) |

close solution |

\( \int{ \sin^3\theta ~ d\theta } \)

Problem Statement |
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\( \int{ \sin^3\theta ~ d\theta } \)

Final Answer |
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\( (1/3)\cos^3\theta - \cos\theta + C \) |

Problem Statement |
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\( \int{ \sin^3\theta ~ d\theta } \)

Solution |
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\( \int{ \sin^3\theta ~ d\theta } \) |

\( \int{ \sin \theta (\sin^2\theta)~d\theta } \) |

\( \int{ \sin \theta ( 1-\cos^2\theta )~d\theta } \) |

\( u=\cos\theta \to du = -\sin\theta d\theta \) |

\( \int{ (1-u^2)(-du) }\) |

\( \int{ u^2-1 ~du } \) |

\( u^3 / 3 - u + C \) |

\( (1/3)\cos^3\theta - \cos\theta + C \) |

Final Answer |
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\( (1/3)\cos^3\theta - \cos\theta + C \) |

close solution |

\(\displaystyle{ \int{ \frac{dx}{(9+x^2)^{3/2}} } }\)

Problem Statement |
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\(\displaystyle{ \int{ \frac{dx}{(9+x^2)^{3/2}} } }\)

Final Answer |
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\(\displaystyle{ \frac{x}{27\sqrt{9+x^2}} + C }\) |

Problem Statement |
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\(\displaystyle{ \int{ \frac{dx}{(9+x^2)^{3/2}} } }\)

Solution |
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Use trigonometric substitution. |

\(x=3\tan t \to dx = 3\sec^2 t ~dt \) |

Draw a triangle that matches \(\tan t = x/3\) including the hypotenuse \(\sqrt{9+x^2}\). |

\(\displaystyle{ \int{ \frac{dx}{(9+x^2)^{3/2}} } }\) |

\(\displaystyle{ \int{ \frac{\sec^2t~dt}{(9+9\tan^2 t)^{3/2}} } }\) |

\(\displaystyle{ \int{ \frac{\sec^2t~dt}{27 (\sec^2 t)^{3/2}} } }\) |

\(\displaystyle{ \frac{1}{27}\int{ \frac{\sec^2 t}{\sec^3 t}~dt } }\) |

\(\displaystyle{ \frac{1}{27}\int{ \frac{1}{\sec t}~dt } }\) |

\(\displaystyle{ \frac{1}{27}\int{ \cos t ~dt } }\) |

\(\displaystyle{ \frac{1}{27}\sin t + C }\) |

Use the triangle that you drew above to write \(\sin t\) in terms of |

\(\displaystyle{ \frac{1}{27} \frac{x}{\sqrt{9+x^2}} + C }\) |

Final Answer |
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\(\displaystyle{ \frac{x}{27\sqrt{9+x^2}} + C }\) |

close solution |

\(\displaystyle{ \int{ \frac{2x^2+x-4}{x^3-x^2-2x} ~dx } }\)

Problem Statement |
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\(\displaystyle{ \int{ \frac{2x^2+x-4}{x^3-x^2-2x} ~dx } }\)

Final Answer |
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\(\displaystyle{ \ln\left| \frac{x^2(x-2)}{x+1} \right| + C }\) |

Problem Statement |
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\(\displaystyle{ \int{ \frac{2x^2+x-4}{x^3-x^2-2x} ~dx } }\)

Solution |
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Check to see if the denominator factors. If so, then try partial fractions.

\(x^3-x^2-2x = x(x^2-x-2) = x(x-2)(x+1)\) |

\(\displaystyle{ \frac{2x^2+x-4}{x^3-x^2-2x} }\) |

\(\displaystyle{ \frac{A}{x} + \frac{B}{x-2} + \frac{D}{x+1} }\) |

Solving for A, B and C, we get \(A=2\), \(B=1\) and \(D=-1\). Now the integral looks like this. |

\(\displaystyle{ \int{ \frac{2}{x} + \frac{1}{x-2} - \frac{1}{x+1} ~ dx } }\) |

\(2\ln|x| + \ln|x-2| - \ln|x+1| + C \) |

\(\displaystyle{ \ln\left| \frac{x^2(x-2)}{x+1} \right| + C }\) |

Notes:

1. The absolute values on each term are required since we are not guaranteed that the factors are positive.

2. Some instructors consider the 2nd to the last expression to be simplified, others require the terms to be combined as in the last expression. Check with your instructor to see what they require.

Final Answer |
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\(\displaystyle{ \ln\left| \frac{x^2(x-2)}{x+1} \right| + C }\) |

close solution |

*Instructions for Part B* - - You may use your calculator. You have one hour to complete this part of the exam. Show all your work. Correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions. Give exact, simplified answers.

Section 1 |
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For questions 6-8, consider the region \(\mathcal{R}\) bounded by the graph of \(y=4-x^2\), the *y*-axis and the horizontal line \(y=1\).

[5 points] Sketch the region \(\mathcal{R}\) and calculate it's area.

Problem Statement |
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[5 points] Sketch the region \(\mathcal{R}\) and calculate it's area.

Final Answer |
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\(A = 2\sqrt{3} \) |

Problem Statement |
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[5 points] Sketch the region \(\mathcal{R}\) and calculate it's area.

Solution |
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built with geoGeobra |
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The region is show in the plot.

For the area, we will set up and evaluate both integrals, i.e. with respect to x and y. The general equation for area is \(A = \int_a^b{f(x)-g(x)~dx}\). A similar equation holds for y.

First, with respect to x.

The upper function is \(y=4-x^2\), the lower function is \(y=1\). We need to find the x-value where these intersect on the right end of the area.

\(4-x^2=1 \to x^2 = 3 \to x=\pm \sqrt{3}\)

Since we are in the first quadrant, x is positive, so \(x=\sqrt{3}\).

\(A = \int_a^b{f(x)-g(x)~dx}\) |

\(A = \int_0^{\sqrt{3}}{ (4-x^2) - 1 ~ dx }\) |

\(A = \int_0^{\sqrt{3}}{ 3-x^2 ~ dx }\) |

\(A = [ 3x - x^3/3 ]_0^{\sqrt{3}} \) |

\(A = 2\sqrt{3} \) |

For the y-direction, the right function is \(y=4-x^2\) which we need to solve for y.

\(y=4-x^2 \to x^2=4-y \to x=\pm\sqrt{4-y} \)

Since the part of the function that we need is in the first quadrant, x is positive, so our right function is \( x=\sqrt{4-y} \).

The left function is \(x=0\). We integrate from 1 to 4.

\(A = \int_c^d{f(y)-g(y)~dx}\) |

\(A = \int_1^4{\sqrt{4-y}-0~dx}\) |

Use integration by substitution. |

\(u = 4-y \to du = -dy\) |

Convert the limits of integration in terms of u. |

lower limit: \(y=1 \to u=4-1=3 \) |

upper limit: \(y=4 \to u=4-4= 0\) |

\(-\int_3^0{ u^{1/2}~du } \) |

\( \int_0^3{ u^{1/2}~du } \) |

\(\displaystyle{ \left. \frac{u^{3/2}}{3/2} \right|_0^3 }\) |

\(\displaystyle{ \frac{2}{3}3^{3/2} - 0 }\) |

\(A = 2\sqrt{3}\) |

Final Answer |
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\(A = 2\sqrt{3} \) |

close solution |

[5 points] A solid is generated by revolving the region \(\mathcal{R}\) about the x-axis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of disc-washers.

Problem Statement |
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[5 points] A solid is generated by revolving the region \(\mathcal{R}\) about the x-axis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of disc-washers.

Final Answer |
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\( V = \pi \int_0^{\sqrt{3}}{ (4-x^2)^2 - 1 ~ dx } \) |

Problem Statement |
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Solution |
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\(V = \pi \int_a^b{ R^2 - r^2 ~dx } \) |

\(R=y=4-x^2\); \(r=1\) |

x ranges from 0 to \(\sqrt{3}\) |

\( V = \pi \int_0^{\sqrt{3}}{ (4-x^2)^2 - 1 ~ dx } \) |

Final Answer |
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\( V = \pi \int_0^{\sqrt{3}}{ (4-x^2)^2 - 1 ~ dx } \) |

close solution |

[5 points] A solid is generated by revolving the region \(\mathcal{R}\) about the x-axis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of shell-cylinders.

Problem Statement |
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[5 points] A solid is generated by revolving the region \(\mathcal{R}\) about the x-axis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of shell-cylinders.

Final Answer |
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\( V = 2\pi\int_1^4{ y\sqrt{4-y}~dy } \) |

Problem Statement |
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Solution |
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\(V = 2\pi\int_c^d{ ph~ dy }\) |

\(p=y; h=x = \sqrt{4-y} \) |

y ranges from 1 to 4 |

\( V = 2\pi\int_1^4{ y\sqrt{4-y}~dy } \) |

Final Answer |
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\( V = 2\pi\int_1^4{ y\sqrt{4-y}~dy } \) |

close solution |

Section 2 |
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[15 points] Find the centroid of the region bounded by the graph of \(y=\cos x\) and the x-axis, between \(x=-\pi/2\) and \(x=\pi/2\). Make a sketch and use symmetry where possible.

Problem Statement |
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[15 points] Find the centroid of the region bounded by the graph of \(y=\cos x\) and the x-axis, between \(x=-\pi/2\) and \(x=\pi/2\). Make a sketch and use symmetry where possible.

Final Answer |
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centroid \((0,\pi/8)\) |

Problem Statement |
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Solution |
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built with geoGeobra |
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The plot is shown here. Using symmetry, \(\bar{x}=0\). The centroid is moment/area with the equation

\(\displaystyle{ \bar{y} = \frac{\int_c^d{y(x_2-x_1)dy}}{\int_c^d{(x_2-x_1~dy)}} }\)

First, calculate the area.

area = \( \int_{-\pi/2}^{\pi/2}{ \cos x ~dx } \) |

area = \(\sin x |_{-\pi/2}^{\pi/2} = 2 \) |

Now, let's calculate the moment.

\(\int_0^1{ y( \arccos(y) - (-\arccos(y)) ) ~ dy } \) |

\(\int_0^1{ 2y \arccos(y) ~ dy } \) |

Use integration by parts. |

\( u=\arccos(y) \to du=-dy/\sqrt{1-y^2} \) |

\( dv = y~dy \to v=y^2/2 \) |

\(\displaystyle{ 2\left[ \left. \frac{y^2}{2} \arccos(y)\right|_0^1 - \int_0^1{ \frac{-1}{2}\frac{y^2}{\sqrt{1-y^2}} ~dy } \right] }\) |

\(\displaystyle{ \int_0^1{ \frac{y^2}{\sqrt{1-y^2}} ~dy } }\) |

Use trig substitution. |

\( y=\sin\theta \to dy=\cos\theta~d\theta\) |

Convert the limits of integration to \(\theta\). |

\(y=0 \to \theta=0\); \(y=1 \to \theta=\pi/2\) |

\(\displaystyle{ \int_{0}^{\pi/2}\frac{\sin^2\theta}{\sqrt{1-\sin^2\theta}}\cos \theta ~d\theta }\) |

\(\displaystyle{ \int_0^{\pi/2}{ \sin^2\theta d\theta } }\) |

\(\displaystyle{ \int_0^{\pi/2}{ (1/2) (1-\cos(2\theta)) d\theta } }\) |

\(\displaystyle{ \frac{1}{2}\left[ \theta - \frac{\sin(2\theta)}{2} \right]_0^{\pi/2} }\) |

\(\displaystyle{ \frac{1}{2}\left[ \frac{\pi}{2} \right] = \frac{\pi}{4} }\) |

\(\displaystyle{ \bar{y}= \frac{\pi/4}{2} = \frac{\pi}{8} }\)

centroid \((0,\pi/8)\)

Final Answer |
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centroid \((0,\pi/8)\) |

close solution |

[10 points] The corn in a grain silo, which as the shape of a right circular cylinder, is to be lifted out over the top edge of the silo. The silo is 20 feet in diameter and is 30 feet high. If the corn in the silo has a uniform depth of 25 feet, how much work is done in lifting all of the corn out of the silo? [Use \(\delta\) lb/ft^{3} for the (weight) density of the corn.]

Problem Statement |
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[10 points] The corn in a grain silo, which as the shape of a right circular cylinder, is to be lifted out over the top edge of the silo. The silo is 20 feet in diameter and is 30 feet high. If the corn in the silo has a uniform depth of 25 feet, how much work is done in lifting all of the corn out of the silo? [Use \(\delta\) lb/ft^{3} for the (weight) density of the corn.]

Final Answer |
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\( 43750\pi\delta \) ft-lb |

Problem Statement |
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^{3} for the (weight) density of the corn.]

Solution |
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Volume of a slice is \(\pi r^2 ~dy \) giving a differential force of \( dF = \pi r^2 \delta ~dy\) lbs.

The distance the slice if moved is \(30-y\) if the bottom of the tank is at height zero.

Work is force times distance giving us the integral

\(\displaystyle{ W = \int_0^{25}{ \pi r^2 \delta (30-y)~dy } }\) |

\(\displaystyle{ W = 100 \pi \delta \int_0^{25}{ 30-y ~dy } }\) |

\(\displaystyle{ W = 100 \pi \delta \left[ 30y - \frac{y^2}{2} \right]_0^{25} }\) |

\( W = 100\pi \delta[ 750 - 625/2 ] = 43750\pi\delta \) ft-lb |

Final Answer |
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\( 43750\pi\delta \) ft-lb |

close solution |

[10 points] Calculate the length of the parametric curve \(x=3t^2, y=2t^3, 0 \leq x \leq 2\).

Problem Statement |
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[10 points] Calculate the length of the parametric curve \(x=3t^2, y=2t^3, 0 \leq x \leq 2\).

Final Answer |
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\( 2[5\sqrt{5}-1] \) |

Problem Statement |
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[10 points] Calculate the length of the parametric curve \(x=3t^2, y=2t^3, 0 \leq x \leq 2\).

Solution |
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The length of a parametric curve is given by the equation

\(\displaystyle{ s = \int_0^2{ \sqrt{ (dx/dt)^2 + (dy/dt)^2 } ~dt } }\) |

\(\displaystyle{ x=3t^2 \to dx/dt = 6t }\) |

\(\displaystyle{ y=2t^3 \to dy/dt = 6t^2 }\) |

\(\displaystyle{ s = \int_0^2{ \sqrt{ 36t^2 + 36t^4 } ~dt } }\) |

\(\displaystyle{ s = \int_0^2{ 6t \sqrt{ 1+t^2 } ~dt } }\) |

Use integration by substitution. |

\( u=1+t^2 \to du=2t~dt \) |

Change the limits of integration. |

\( t=0 \to u=1; t=2 \to u=5 \) |

\(\displaystyle{ s = \int_1^5{ 6t u^{1/2} \frac{du}{2t} } }\) |

\(\displaystyle{ 3\int_1^5{u^{1/2}~du} }\) |

\(\displaystyle{ \left. 3\frac{u^{3/2}}{3/2} \right|_1^5 }\) |

\( 2[5^{3/2}-1] = 2[5\sqrt{5}-1]\) |

Final Answer |
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\( 2[5\sqrt{5}-1] \) |

close solution |