Calculus 2  Exam B2
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This is the second exam for second semester single variable calculus.
Instructions:
 This exam is in four main parts, labeled sections 14, with different instructions for each section.
 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 answers.
Evaluate each of the limits or explain why it doesn't exist. Each question in this section is worth 5 points.
Question 1 
\(\displaystyle{ \lim_{x \to \infty}{\frac{\ln(x)}{x}} }\)



\(\displaystyle{ \lim_{x \to \infty}{\frac{\ln(x)}{x}} = 0 }\)

Direct substitution yields \( \infty/\infty \) which is indeterminate.
So use L'Hôpital's Rule.
\(\displaystyle{ \lim_{x \to \infty}{\frac{\ln(x)}{x}} = }\) \(\displaystyle{
\lim_{x \to \infty}{\frac{1/x}{1}} = }\) \(\displaystyle{ \lim_{x \to \infty}{\frac{1}{x}} = 0 }\)
Question 1 Final Answer 
\(\displaystyle{ \lim_{x \to \infty}{\frac{\ln(x)}{x}} = 0 }\)

Question 2 
\(\displaystyle{ \lim_{x \to 0}{\frac{\sec(x)1}{x^2}} }\)



\(\displaystyle{ \lim_{x \to 0}{\frac{\sec(x)1}{x^2}} = \frac{1}{2} }\)

Direct substitution yields \( 0/0 \) which is indeterminate.
So use L'Hôpital's Rule.
\(\displaystyle{ \lim_{x \to 0}{\frac{\sec(x)}{x^2}} = \lim_{x \to 0}{\frac{\sec(x)\tan(x)}{2x}} = \frac{0}{0}}\)
Still indeterminate, so use L'Hôpital's Rule again.
\(\displaystyle{ \lim_{x \to 0}{\frac{\sec(x)\tan(x)}{2x}} =
\lim_{x \to 0}{\frac{\sec(x)\tan^2(x) + \sec(x)\sec^2(x)}{2}} = \frac{1}{2}
}\)
Question 2 Final Answer 
\(\displaystyle{ \lim_{x \to 0}{\frac{\sec(x)1}{x^2}} = \frac{1}{2} }\)

Question 3 
\(\displaystyle{ \lim_{x \to 0^+}{\left[ \frac{1}{x}  \frac{1}{\sin(x)} \right]} }\)



\(\displaystyle{ \lim_{x \to 0^+}{\left[ \frac{1}{x}  \frac{1}{\sin(x)} \right]} = 0 }\)

Direct substitution yields \( \infty  \infty \) which is indeterminate.
So use L'Hôpital's Rule. Before we can use this rule, we need the limit in fraction form.
\(\displaystyle{
\lim_{x \to 0^+}{\left[ \frac{1}{x}  \frac{1}{\sin(x)}\right]} = }\) \(\displaystyle{
\lim_{x \to 0^+}{\frac{\sin(x)x}{x\sin(x)}} = }\) \(\displaystyle{
\lim_{x \to 0^+}{\frac{\cos(x)1}{x\cos(x)+\sin(x)}}
}\)
Direct subsitution gives us \( 0/0 \) which is also indeterminate. So we need to use L'Hôpital's Rule again.
\(\displaystyle{
\lim_{x \to 0^+}{\frac{\cos(x)1}{x\cos(x)+\sin(x)}} = \lim_{x \to 0^+}{\frac{\sin(x)}{x(\sin(x))+\cos(x)+\cos(x)}} = \frac{0}{2} = 0
}\)
Question 3 Final Answer 
\(\displaystyle{ \lim_{x \to 0^+}{\left[ \frac{1}{x}  \frac{1}{\sin(x)} \right]} = 0 }\)

Evaluate each improper integral or show that it diverges. Each question in this section is worth 5 points.
Question 4 
\(\displaystyle{ \int_{0}^{\infty}{\frac{3x}{(x^2+9)^2}dx} }\)



\(\displaystyle{ \int_{0}^{\infty}{\frac{3x}{(x^2+9)^2}dx} = \frac{1}{6} }\)

First, we need to rewrite the improper integral as a limit of a proper integral. This is important since definite integration is only defined on a continuous, closed interval. The integrand itself is continuous on the entire interval from \(0\) to \(\infty\), the only part of the integral that makes it improper is the upper limit.
\(\displaystyle{ \int_{0}^{\infty}{\frac{3x}{(x^2+9)^2}dx} = \lim_{b \to \infty}{\int_{0}^{b}{\frac{3x}{(x^2+9)^2}dx}} }\)
We will use integration by substitution and evaluate the indefinite integral first. Then go back, substitute the limits of integration and take the limit.
\( u = x^2+9 ~~~ \to ~~~ du = 2x~dx ~~~ \to ~~~ du/(2x) = dx \)
\(\displaystyle{
\int{\frac{3x}{(x^2+9)^2}dx} = }\) \(\displaystyle{
\int{\frac{3x}{u^2}\frac{du}{2x}} = }\) \(\displaystyle{
\frac{3}{2}\int{u^{2}du} = }\) \(\displaystyle{ \frac{3}{2}\frac{u^{1}}{1} + C }\) \(\displaystyle{
= \frac{3}{2(x^2+9)} + C
}\)
\(\displaystyle{
\begin{array}{rcl}
\lim_{b \to \infty}{\int_{0}^{b}{\frac{3x}{(x^2+9)^2}dx}} & = & \lim_{b \to \infty}{\left[ \frac{3}{2}\frac{1}{x^2+9} \right]_0^b} \\
& = & \frac{3}{2}\left[ \lim_{b \to \infty}{\frac{1}{b^2+9}}  \frac{1}{9} \right] \\
& = & \frac{3}{2} \left[ 0  \frac{1}{9} \right] = \frac{1}{6}
\end{array}
}\)
Question 4 Final Answer 
\(\displaystyle{ \int_{0}^{\infty}{\frac{3x}{(x^2+9)^2}dx} = \frac{1}{6} }\)

Question 5 
\(\displaystyle{ \int_0^2{\frac{x}{\sqrt{4x^2}}dx} }\)



\(\displaystyle{ \int_0^2{\frac{x}{\sqrt{4x^2}}dx} = 2 }\)

There is a discontinuity at \( x=2 \) since the denominator is zero. (The denominator is also zero at \(x=2\) but this value is not within the limits of integration. So it does not come into play in this problem.) We need to rewrite the improper integral as a limit and a proper integral.
\(\displaystyle{
\lim_{b \to 2^}{\int_0^b{\frac{x}{\sqrt{4x^2}}dx}}
}\)
As with the previous problem, we will evaluate the indefinite integral and then substitute the limits of integration and evaluate the limit later.
\( u = 4x^2 ~~ \to ~~ du = 2x~dx ~~ \to ~~ du/(2x) = dx \)
\(\displaystyle{
\begin{array}{rcl}
\int{\frac{x}{\sqrt{4x^2}}dx} & = & \int{ \frac{x}{u^{1/2}} \frac{du}{2x} } \\
& = & \frac{1}{2} \int{u^{1/2}du} \\
& = & \frac{1}{2} \frac{u^{1/2}}{1/2} + C
\end{array}
}\)
\(\displaystyle{
\begin{array}{rcl}
\lim_{b \to 2^}{\int_0^b{\frac{x}{\sqrt{4x^2}}dx}} & = & \lim_{b \to 2^}{\left[ (4x^2)^{1/2} \right]_0^b } \\
& = & \lim_{b \to 2^}{\left[ (4b^2)^{1/2}  2 \right] } \\
& = &  \left[ (44)^{1/2}  2 \right] = 2
\end{array}
}\)
Note: There may be a tendency to write the limit as \( b \to 2 \), rather than \( b \to 2^ \). However, we need to restrict the limit to the left of \( 2 \) since the limits of integration are from \(0\) to \(2\). What is going on to the right of \( 2\) is not part of the integral. So it is important that the limit say \( b \to 2^ \).
Question 5 Final Answer 
\(\displaystyle{ \int_0^2{\frac{x}{\sqrt{4x^2}}dx} = 2 }\)

Question 6 
\(\displaystyle{ \int_0^{\infty}{xe^{5x}dx} }\)



\(\displaystyle{ \int_0^{\infty}{xe^{5x}dx} = \frac{1}{25} }\)

The function is continuous on the entire interval of integration, so the only part of the limit that causes this to be an improper integral is the upper limit. So we need to rewrite this as a limit of a proper integral before we can integrate.
\(\displaystyle{ \int_0^{\infty}{xe^{5x}dx} = \lim_{b \to \infty}{ \int_0^{b}{xe^{5x}dx} } }\)
As in the previous two problems, we will evaluate the indefinite integral \(\displaystyle{ \int{xe^{5x}dx} }\) then substitute the limits of integration and take the limit.
Use integration by parts.
\(u = x\)  \(\to\)  \( du = dx \) 
\(dv = e^{5x}dx\)  \(\to\)  \(v = e^{5x}/(5)\) 
\(\displaystyle{
\int{xe^{5x}dx} = }\) \(\displaystyle{
\frac{xe^{5x}}{5}  \int{\frac{e^{5x}}{5}dx} = }\) \(\displaystyle{
\frac{xe^{5x}}{5} + \frac{1}{5}\frac{e^{5x}}{5} + C
}\)
\(\displaystyle{
\lim_{b \to \infty}{ \int_0^{b}{xe^{5x}dx} } = }\) \(\displaystyle{
\frac{1}{5} \lim_{b \to \infty}{\left[ xe^{5x} + \frac{1}{5}e^{5x} \right]_0^b } = }\) \(\displaystyle{
\frac{1}{5} \lim_{b \to \infty}{ \left[ be^{5b} + \frac{1}{5}e^{5b}  0  \frac{1}{5} \right] }
}\)
Let's look at the first two factors. The second one is
\(\displaystyle{ \lim_{b \to \infty}{ \frac{1}{5e^{5b}} } = \frac{1}{\infty} = 0 }\)
The first one is
\(\displaystyle{ \lim_{b \to \infty}{ \frac{b}{e^{5b}} } = \frac{\infty}{\infty} }\) which is
indeterminate. So we will use
L'Hôpital's Rule.
\(\displaystyle{
\lim_{b \to \infty}{ \frac{b}{e^{5b}} } = }\) \(\displaystyle{
\lim_{b \to \infty}{\frac{1}{5e^{5b}}} = }\) \(\displaystyle{ \frac{1}{\infty} = 0
}\)
So that leaves us with
\(\displaystyle{ \frac{1}{5} \left[ 0 + 0  0  \frac{1}{5} \right] = \frac{1}{25}}\)
Question 6 Final Answer 
\(\displaystyle{ \int_0^{\infty}{xe^{5x}dx} = \frac{1}{25} }\)

Determine whether these series converge or diverge. At the end of each question, state your result and which test you used to determine your answer. Each question in this section is worth 5 points.
Question 7 
\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{n}{n+100}} }\)



\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{n}{n+100}} }\) diverges by the nth Term Test.

Let's try the nth Term Test.
\(\displaystyle{
\lim_{n \to \infty}{\left[ \frac{n}{n+100} \frac{1/n}{1/n}\right]} = }\) \(\displaystyle{
\lim_{n \to \infty}{\left[ \frac{1}{1+100/n} \right]} = }\) \(\displaystyle{ \frac{1}{1+0} = 1
}\)
Since the limit is not zero, the series diverges by the nth Term Test.
Question 7 Final Answer 
\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{n}{n+100}} }\) diverges by the nth Term Test.

Question 8 
\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{n\ln(n)}} }\)



\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{n\ln(n)}} }\) diverges by the integral test

Try the integral test. We need to evaluate \(\displaystyle{ \int_{k}^{\infty}{f(x)~dx} }\) where
\(\displaystyle{ f(x) = \frac{1}{x\ln(x)} }\)
We need to choose k where \(f(x)\) is decreasing for \( x > k\). This occurs for \( k > 1 \), so we can choose any k greater than \(1\). We choose \(k=2\).
So we have \(\displaystyle{ \int_{2}^{\infty}{\frac{1}{x\ln(x)}dx} }\).
If this integral diverges, so does the series; if the integral converges, so does the series. We need to write the improper integral as a limit of a proper integral.
\(\displaystyle{ \int_{2}^{\infty}{\frac{1}{x\ln(x)}dx} = \lim_{b \to \infty}{ \int_{2}^{b}{\frac{1}{x\ln(x)}dx} } }\)
We will evaluate the indefinite integral and then go back, evaluate the result using the limits of integration and take the limit.
Use integration by substitution and let \( u = \ln(x) ~~ \to ~~ du = dx/x ~~ \to ~~ x~du = dx\)
\(\displaystyle{ \int{\frac{1}{x\ln(x)}} = }\) \(\displaystyle{
\int{\frac{1}{x}\frac{1}{u}x~du} = \int{\frac{1}{u}du} = }\) \(\displaystyle{
\ln(u)+C = \ln(\ln(x))+C }\)
\(\displaystyle{ \lim_{b \to \infty}{\left[ \ln(\ln(x)) \right]_2^b} = }\) \(\displaystyle{
\lim_{b \to \infty}{[\ln(\ln(b))  \ln(\ln(2))]} = }\) \(\displaystyle{ \infty }\)
Since the integral diverges, the series also diverges by the integral test.
Question 8 Final Answer 
\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{n\ln(n)}} }\) diverges by the integral test

Question 9 
\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{5n^2}{11n^3+4n^2+6n+3}} }\)



\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{5n^2}{11n^3+4n^2+6n+3}} }\) diverges by the limit comparison test

Let's try the limit comparison test. We will choose the test series to be \(\sum{1/n}\). We chose this because, if we look at the nth term of the original series for very large n, the highest powers dominate. That leaves \(\displaystyle{ \frac{5n^2}{11n^3} = \frac{5}{11n} = \frac{5}{11}\cdot\frac{1}{n}}\). Multiplication by a constant does not affect convergence or divergence, so to simplify the comparison series, we ignore \( 5/11 \).
The limit comparison test says that we choose a test series and evaluate the limit \(\displaystyle{ \lim_{n \to \infty}{\left[ \frac{a_n}{t_n} \right]} }\). If this limit turns out finite and positive (and nonzero), then the original series will converge or diverge depending on the test series. In this question, \(\displaystyle{ a_n = \frac{5n^2}{11n^3+4n^2+6n+3} }\) and \(\displaystyle{ t_n = \frac{1}{n} }\). Let's evaluate the limit.
\(\displaystyle{
\begin{array}{rcl}
\lim_{n \to \infty}{\left[ \frac{a_n}{t_n} \right]} & = & \lim_{n \to \infty}{ \left[ \frac{5n^2}{11n^3+4n^2+6n+3} \frac{n}{1} \right]} \\
& = & \lim_{n \to \infty}{ \left[ \frac{5n^3}{11n^3+4n^2+6n+3} \frac{1/n^3}{1/n^3} \right]} \\
& = & \lim_{n \to \infty}{ \left[ \frac{5}{11+4/n+6/n^2+3/n^3} \right]} \\
& = & \frac{5}{11+0+0+0} = \frac{5}{11}
\end{array}
}\)
The limit \(\displaystyle{ \lim_{n \to \infty}{\left[ \frac{a_n}{t_n} \right]} = \frac{5}{11} }\) and \( 5/11 \) and is finite and positive. The test series is \(\sum{1/n}\), which is a pseries with \(p=1\) and, so, it diverges. Putting all this together tells us that the original series also diverges.
Question 9 Final Answer 
\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{5n^2}{11n^3+4n^2+6n+3}} }\) diverges by the limit comparison test

Question 10 
\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{(1)^{n+1}}{n+1}} }\)



\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{(1)^{n+1}}{n+1}} }\) converges by the alternating series test

This is an alternating series (if you are not sure, write out the first few terms and notice how the sign alternates between positive and negative; that's the function of the \( (1)^{n+1} \) term). So we will use the alternating series test.
Condition 1: \(\displaystyle{ \lim_{n \to \infty}{\frac{1}{n+1}} = 0 }\)
So condition 1 holds.
Condition 2: We need to show that \( 0 < a_{n+1} \le a_n \)
\(\displaystyle{ a_n = \frac{1}{n+1} ~~~ a_{n+1} = \frac{1}{n+2} }\)
So let's test \( a_{n+1} \le a_n \). (We can see that \( 0 < a_{n+1} \) already holds since n starts at 1 and increases.) We will set up the inequality and perform valid operations. If we get an equality that is true for all values of n, then the original inequality holds also.
\(\displaystyle{
\begin{array}{rcl}
\frac{1}{n+2} & \le & \frac{1}{n+1} \\
n+1 & \le & n+2 \\
1 & \le & 2
\end{array}
}\)
Since \( 1 \le 2 \) is true for all \( n > 0 \) then condition 2 holds.
So the series converges by the alternating series test.
Question 10 Final Answer 
\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{(1)^{n+1}}{n+1}} }\) converges by the alternating series test

Solve the following problems. Each problem in this section is worth 10 points.
Question 11 
Determine the convergence set of the power series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{x^n}{n}} }\)



The convergence set of \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{x^n}{n}} }\) is \( 1 \leq x < 1 \).

To determine the convergence set, we need to determine the convergence interval and then evaluate the endpoints individually. For the convergence interval, we use the ratio test.
\(\displaystyle{ \lim_{n \to \infty}{\left \frac{a_{n+1}}{a_n} \right} }\)
For convergence, this limit must be less than one. For our question, \(\displaystyle{ a_n = \frac{x^n}{n} ~~~~ a_{n+1} = \frac{x^{n+1}}{n+1} }\)
\(\displaystyle{
\begin{array}{rcl}
\lim_{n \to \infty}{\left \frac{x^{n+1}}{n+1} \frac{n}{x^n} \right} & < & 1 \\
\lim_{n \to \infty}{\left \frac{n}{n+1} \frac{x^{n+1}}{x^n} \right} & < & 1 \\
\lim_{n \to \infty}{\left \frac{n}{n+1} \frac{1/n}{1/n} x \right} & < & 1 \\
\lim_{n \to \infty}{\left \frac{1}{1+1/n} x \right} & < & 1 \\
x & < & 1 ~~~ \to ~~~ 1 < x < 1
\end{array}
}\)
So the interval of convergence is \( 1 < x < 1 \). Now we need to test the endpoints.
At \( x = 1 \) the series is \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{(1)^n}{n}} }\), which converges by the alternating series test.
At \( x = 1 \), the series is \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{n}} }\), which is a pseries with \(p=1\) and therefore diverges.
So the convergence set of the series \( \sum{a_n} \) is \( [1,1) \) (this can also be written \( 1 \leq x < 1 \) ).
Question 11 Final Answer 
The convergence set of \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{x^n}{n}} }\) is \( 1 \leq x < 1 \).

Question 12 
Find the power series for the function \(\displaystyle{ f(x) = \frac{1}{(13x)^2} }\)
Hint: \(\displaystyle{ \sum_{n=0}^{\infty}{3^n x^n} }\) is the geometric series with ratio \( r=3x \).



\(\displaystyle{ f(x) = \frac{1}{(13x)^2} = \sum_{n=0}^{\infty}{n(3x)^{n1}} }\) for \( 0 < x < 1/3 \)

Let's start with the hint and see what it converges to.
A geometric series with ratio r looks like \(\displaystyle{ \sum_{n=0}^{\infty}{ar^n} = \frac{a}{1r} }\) if \( 0 < r <1 \).
So in the hint, \( r=3x \) giving us \(\displaystyle{ g(x) = \sum_{n=0}^{\infty}{(3x)^n} = \frac{1}{13x} }\) for \( 0 < 3x < 1 ~~ \to ~~ 0 < x < 1/3 \).
We can get \( f(x) \) from \( g(x) \) by taking the derivative of \( g(x) \).
\(\displaystyle{
\begin{array}{rcl}
g'(x) & = & (1)(13x)^{2}(3) = \frac{3}{(13x)^2} \\
\frac{g'(x)}{3} & = & \frac{1}{(13x)^2} = f(x) \\
f(x) & = & \frac{g'(x)}{3} = \frac{1}{3} \sum_{n=0}^{\infty}{3^n (nx^{x1})} \\
& = & \sum_{n=0}^{\infty}{3^{n1}nx^{n1}} \\
& = & \sum_{n=0}^{\infty}{n(3x)^{n1}}
\end{array}
}\)
Since differentiation does not affect the convergence interval, this series converges for \( 0 < x < 1/3 \).
Question 12 Final Answer 
\(\displaystyle{ f(x) = \frac{1}{(13x)^2} = \sum_{n=0}^{\infty}{n(3x)^{n1}} }\) for \( 0 < x < 1/3 \)

Question 13 
Taylors Formula with remainder applied to \(\displaystyle{ f(x) = \ln(2+3x) }\) centered about \( a=0\) shows that \(\displaystyle{ \ln(2+3x) = P_2(x) + R_2(x) }\) where \( P_2(x) \) is the Maclaurin polynomial of degree 2 and \( R_2(x) \) is the remainder term. Determine \( P_2(x) \) and \( R_2(x)\).



\(\displaystyle{ P_2(x) = \ln(2) + \frac{3x}{2}  \frac{9}{8}x^2 }\) and \(\displaystyle{ R_2(x) = \frac{27x^2}{8(2+3x)} }\)

Maclaurin Polynomial
The general Taylor polynomial of degree 2 is \(\displaystyle{ f(a) + \frac{f'(a)}{1!}(xa) + \frac{f''(a)}{2!}(xa)^2 }\).
In our case we have a Maclaurin polynomial (a=0), our second degree polynomial is \(\displaystyle{ f(0) + \frac{f'(0)}{1}(x) + \frac{f''(0)}{2}x^2 }\).
\(\displaystyle{
\begin{array}{rclrcl}
f(x) & = & \ln(2+3x) & f(0) & = & \ln(2) \\
f'(x) & = & \frac{3}{2+3x} & f'(0) & = & 3/2 \\
f''(x) & = & \frac{9}{(2+3x)^2} & f''(0) & = & 9/4
\end{array}
}\)
\(\displaystyle{ P_2(x) = \ln(2) + \frac{3x}{2}  \frac{9}{8}x^2 }\)
Remainder Term
The general formula for the remainder term is \(\displaystyle{ R_n(x) = \frac{1}{n!} \int_{a}^{x}{(xt)~f^{(n+1)}(t)~dt} }\)
In our case, \( n=2 \). So we need to calculate \( f^{(3)}(x) \).
\(\displaystyle{ f^{(3)}(x) = \frac{54}{(2+3x)^3} }\)
\(\displaystyle{ R_2(x) = \frac{1}{2} \int_{0}^{x}{(xt)f^{(3)}(t)~dt} = \frac{1}{2}\int_{0}^{x}{(xt)\frac{54}{(2+3t)^3} dt} }\)
We can use integration by substitution with \( u = 2+3t \).
\( u = 2+3t ~~ \to ~~ du = 3dt ~~ \to ~~ du/3 = dt \)
\( u = 2+3t ~~ \to ~~ (u2)/3 = t ~~ \to ~~ (xt) = (xu/3+2/3) \)
Now let's convert the limits of integration.
\( t=0 ~~ \to ~~ u = 2 \) and \( t = x ~~ \to ~~ u = 2+3x \)
\(\displaystyle{
\begin{array}{rcl}
R_2(x) = & = & \frac{1}{2}\int_{2}^{2+3x}{(xu/3+2/3) \frac{54}{u^3} \frac{du}{3} } \\
& = & \frac{54}{6} \int_{2}^{2+3x}{\frac{xu/3+2/3}{u^3} ~du } \\
& = & 9 \int_{2}^{2+3x}{ \frac{x+2/3}{u^3}  \frac{1}{3} \frac{1}{u^2} ~du} \\
& = & 9 \left[ (x+2/3)\frac{u^{2}}{2} \frac{1}{3} \frac{u^{1}}{1} \right]_{2}^{2+3x} \\
& = & 9 \left[ \frac{1}{6}\frac{(2+3x)}{u^2} + \frac{1}{3}\frac{1}{u} \right]_{2}^{2+3x} \\
& = & \frac{9}{6}\left[ \frac{(2+3x)}{(2+3x)^2}  \frac{2}{(2+3x)} \right] +
\frac{9}{6} \left[ \frac{(2+3x)}{4}  \frac{2}{2} \right] \\
& = & \frac{3}{2} \left[ \frac{1}{(2+3x)} + \frac{2}{2+3x} + \frac{2+3x}{4}  1 \right] \\
& = & \frac{3}{2} \left[ \frac{1}{2+3x} + \frac{2+3x}{4}  \frac{4}{4} \right] \\
& = & \frac{3}{2} \left[ \frac{1}{2+3x} \frac{4}{4} + \frac{2+3x}{4}\frac{2+3x}{2+3x} \right] \\
& = & \frac{3}{2} \left[ \frac{4  4 + 9x^2}{4(2+3x)} \right] \\
& = & \frac{27x^2}{8(2+3x)}
\end{array}
}\)
Question 13 Final Answer 
\(\displaystyle{ P_2(x) = \ln(2) + \frac{3x}{2}  \frac{9}{8}x^2 }\) and \(\displaystyle{ R_2(x) = \frac{27x^2}{8(2+3x)} }\)

Question 14 
Find the tangent line to the parametric curve \( x = t^3 + t, ~~~ y = t^2 1 \); \( 0 < t < 2 \) at the point \( (2,0) \).



\( y = x/2  1 \)

First, we need to find the value of t at the point \((2,0)\). We can use either the x or y expression. Using \( y = t^21 \) we have
\( 0 = t^2  1 ~~ \to ~~ t^2 = 1 ~~ \to ~~ t = \pm 1 \)
Since we are given the range on t as \( 0 < t < 2 \), we know that \( t = 1 \).
To find the tangent line, we need the slope and a point. We are given the point \( (2,0) \), so we need to calculate the slope.
\(\displaystyle{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} }\)
\( dy/dt = 2t ~~~~ dx/dt = 3t^2+1 \)
\(\displaystyle{ \frac{dy}{dx} = \frac{2t}{3t^2+1} }\)
At \( t=1 \), the slope is \( m = dy/dx = 2/4 = 1/2 \)
So we have the slope \( m = 1/2\) at the point \( (2,0) \). The equation of the tangent line can now be calculated as
\( y  0 = (1/2)(x2) ~~~~ y = x/2  1 \)
Question 14 Final Answer 
\( y = x/2  1 \)

Question 15 
Calculate the area of one leaf of the 3leaved rose \( r=5\sin(3\theta) \).



Area = \( (25\pi)/4 \)

In order to know how to set up the integral, we need to know what the graph looks like. So it is important to be able to graph polar equations on your calculator quickly. The graph we need is shown on the right. We can calculate the area of any of the leaves and it would be good practice for you to set up the integrals for each leaf. The general integral to calculate area is
Area = \(\displaystyle{ \frac{1}{2} \int_{\theta_1}^{\theta_2}{[ r(\theta) ]^2 ~d\theta} }\)
For this solution, we will calculate the right half of the lower leaf and multiply by \(2\). So for this question and our choice of area to calculate, our integral is
Area = \(\displaystyle{ 2 \left( \frac{1}{2} \int_{3\pi/2}^{2\pi}{ [5\sin(3\theta)]^2 ~d\theta } \right)
= 25 \int_{3\pi/2}^{2\pi}{\sin^2 (3\theta) ~ d\theta} }\)
Recall that \( \sin^2(x) = [1\cos(2x)]/2 \).
\(\displaystyle{
\begin{array}{rcl}
\text{Area} & = & 25 \int_{3\pi/2}^{2\pi}{\frac{1}{2}(1\cos(6\theta)) ~d\theta} \\
& = & \frac{25}{2} \left[ \theta  \frac{\sin(6\theta)}{6} \right]_{3\pi/2}^{2\pi} \\
& = & \frac{25}{2}[ 2\pi  0 ]  \frac{25}{2} [ 3\pi/2  0 ] \\
& = & 25\pi  \frac{75\pi}{4} \\
& = & \frac{25\pi}{4}
\end{array}
}\)
Question 15 Final Answer 
Area = \( (25\pi)/4 \)
