review

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Final Exam Review

Review for the final exam is attached.

Solutions have been posted.

Review Quiz 2

Math 2300 Section 005 – Calculus II – Fall 2010

Quiz – Monday, November 15, 2010

We will be having a review in class every day until the next midterm exam on Wednesday of this week. For this quiz, you are given 10 minutes to decide which problems you think you can solve and which ones you want to see solved during our review session today.

  1. Use the Lagrange Error Bound Formula for P_{n}(x) to find a reasonable bound for the error in approximating the quantity e^{{0.40}} with a third-degree Taylor polynomial for the function f(x)=e^{x} about x=0. Choose the best error estimate.

    Solution: We use the formula

    |E_{3}(x)|\le\frac{M}{4!}|x|^{4}

    where M\ge|f^{{(4)}}(x)| on the interval [0,0.4] (i.e., the interval between where our Taylor polynomial is centered and where we're approximating the value). Since f^{{(4)}}(x)=e^{x} has a maximum value of e^{{0.4}} on the interval [0,0.4], we obtain

    |E_{3}(0.4)|\le\frac{e^{{0.4}}}{4!}0.4^{4}.
  2. Use the Lagrange Error Bound Formula for P_{n}(x) to find a reasonable bound for the error in approximating the quantity 17/\sqrt{3} with a third-degree Taylor polynomial for the function

    g(x)=\frac{17}{\sqrt{4-x}}

    about x=0. Choose the best error estimate.

Midterm 3 Review

The attached file is a review for the third midterm. Many thanks to instructor Anca Radulescu for typing up solutions.

Midterm 2 Review Solutions

Math 2300 – Calculus II – University of Colorado

Fall 2010 – Review for Midterm II

1. Rotating the ellipse \displaystyle\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 about the x-axis generates an ellipsoid. Compute its volume.

Solution: In this situation, the a and b would be numbers. But, in general, we have that

\displaystyle\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \displaystyle=1
\displaystyle\frac{y^{2}}{b^{2}} \displaystyle=1-\frac{x^{2}}{a^{2}}
\displaystyle y^{2} \displaystyle=b^{2}-\frac{b^{2}}{a^{2}}x^{2}
\displaystyle y \displaystyle=\sqrt{b^{2}-\frac{b^{2}}{a^{2}}x^{2}}

This object is symmetric with respect to the x-axis and y-axis, meaning that we only need to find one of the limits of integration. Setting y=0 in the above equation implies

\sqrt{b^{2}-\frac{b^{2}}{a^{2}}x^{2}}=0\qquad\Leftrightarrow\qquad b^{2}-\frac{b^{2}}{a^{2}}x^{2}=0\qquad\Leftrightarrow\qquad b^{2}\left(1-\frac{x^{2}}{a^{2}}\right)=0.

When we create such objects, we never have b=0 and so we must have 1=x^{2}/a^{2} and a^{2}=x^{2}. So, the x-value where the ellipsoid has a zero y-value must be a. (This makes perfect sense from the definition, if you should know it or look it up.)

So, we consider slices in the ellipsoid revolved around the x-axis, where the radius is given as

r=\sqrt{b^{2}-\frac{b^{2}}{a^{2}}x^{2}}

and the width of a Riemann disk is dx. So, we find that the volume is

\displaystyle\int _{{-a}}^{a}\pi r^{2}\, dx \displaystyle=\int _{{-a}}^{a}\pi\left(b^{2}-\frac{b^{2}}{a^{2}}x^{2}\right)\, dx
\displaystyle=\pi\int _{0}^{a}\left(b^{2}-\frac{b^{2}}{a^{2}}x^{2}\right)\, dx
\displaystyle=\pi\left[b^{2}x-\frac{b^{2}}{3a^{2}}x^{3}\right]_{{x=0}}^{{x=a}}
\displaystyle=\pi\left[ab^{2}-\frac{ab^{2}}{3}\right]
\displaystyle=\frac{2\pi ab^{2}}{3}

Series Convergence Notes

Math 2300 Section 005 – Calculus II – Fall 2010

Review on Series Convergence – October 6, 2010

§ Summary of Convergence of Series

§ Some General Notes

  • Pulling a finite number of terms off from a series will not affect convergence / divergence.

  • You can pull series apart like integrals (in terms of sums/difference and multiplication by a constant).

  • Series notation is shorthand, and in most cases series can be rewritten in any number of different ways.

    \sum _{{n=1}}^{\infty}\frac{(-1)^{n}}{n^{2}}=\sum _{{n=1}}^{\infty}\frac{(-1)^{{n+2}}}{n^{2}}=\sum _{{n=5}}^{\infty}\frac{(-1)^{{n-2}}}{(n-4)^{2}}=\sum _{{n=5}}^{\infty}\frac{(-1)^{n}}{(-1)^{2}(n-4)^{2}}=\sum _{{n=5}}^{\infty}\frac{\cos(n\pi)}{n^{2}-8n+16}

Exam 2 Review 1

A review sheet for exam 2 is now available, as an attachment to this post.

Midterm 1 Solutions

Math 2300 – Calculus II – University of Colorado

Fall 2010 – Midterm I – Solutions

  1. (26 points) Consider the integral \displaystyle\int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx

    1. (6 points) Draw a picture of the area that this integral represents. Evaluate this integral any way you like.

      Solution: The graph of the function f(x)=\sqrt{2-x^{2}} for x in [0,\sqrt{2}] represents the upper right quarter (first quadrant) of the circle centered at the origin and of radius r=\sqrt{2}. We now want to compute the integral:

      Solution 1: Since the integrand f is positive on [0,\sqrt{2}], the integral \int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx represents the area under the graph of f and above the x-axis; this is exactly the area of a quarter of the disc, which is \frac{1}{4}\pi r^{2}=\frac{2\pi}{4}=\frac{\pi}{2}.

      Solution 2: We make the trig substitution, and write x in [-\sqrt{2},\sqrt{2}] as x=\sqrt{2}\sin t, with t in [\frac{-\pi}{2},\frac{\pi}{2}]. Then the radical becomes: \sqrt{2-x^{2}}=\sqrt{2-2\sin^{2}t}=\sqrt{2\cos^{2}t}=\sqrt{2}\cos t. The integration limits for t are t=\sin^{{-1}}(0)=0 and t=\sin^{{-1}}(1)=\frac{\pi}{2}, and dx=\sqrt{2}\cos t\, dt. So:

      \displaystyle\int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx \displaystyle=\int _{{0}}^{{\pi/2}}{\sqrt{2}\cos t\sqrt{2}\cos t\, dt}
      \displaystyle=\int _{{0}}^{{\pi/2}}{2\cos^{2}t\, dt}
      \displaystyle=\int _{{0}}^{{\pi/2}}{[1+\cos(2t)]\, dt}
      \displaystyle=\left.\left(t+\frac{\sin(2t)}{2}\right)\right\rvert _{0}^{{\pi/2}}
      \displaystyle=\left(\frac{\pi}{2}+\frac{\sin(\pi)}{2}\right)-\left(0+\frac{\sin(0)}{2}\right)
      \displaystyle=\frac{\pi}{2}
      \displaystyle\approx 1.57
    2. (6 points) Compute Left(2) and Mid(2). Show your work. An answer that appears to have come from a calculator will receive 0 points.

      Solution:

      \displaystyle\textsc{Left}(2) \displaystyle=\Delta x\left[f(x_{0})+f(x_{1})\right]
      \displaystyle=\frac{\sqrt{2}}{2}\left[\sqrt{2-0^{2}}+\sqrt{2-(\sqrt{2}/2)^{2}}\right]
      \displaystyle=\frac{\sqrt{2}}{2}\left[\sqrt{2}+\sqrt{3/2}\right]
      \displaystyle\approx 1.86
      \displaystyle\textsc{Mid}(2) \displaystyle=\Delta x\left[f(\frac{x_{0}+x_{1}}{2})+f(\frac{x_{1}+x_{2}}{2})\right]
      \displaystyle=\frac{\sqrt{2}}{2}\left[\sqrt{2-(\sqrt{2}/4)^{2}}+\sqrt{2-(3\sqrt{2}/4)^{2}}\right]
      \displaystyle\approx 1.63
    3. (4 points) The formula for error in an approximation is given by

      \textsc{Error }=\textsc{Actual Value }-\textsc{ Approximation}.

      Compute the errors in Left(2) and Mid(2).

      Solution:

      \textsc{Error Left}(2)\approx 1.57-1.86=-0.29
      \textsc{Error Mid}(2)\approx 1.57-1.62=-0.05
    4. (6 points) Estimate the errors in Left(20) and Mid(20) without actually computing these values. Again, an answer that appears to have come from a calculator will receive 0 points.

      Solution:

      \textsc{Error Left}(20)\approx-0.29/10=-0.029
      \textsc{Error Mid}(20)\approx-0.05/100=-0.0005
    5. (4 points) Order the following approximations from least to greatest for any integer n\ge 2.

      \textsc{Left}(n),\qquad\textsc{Right}(n),\qquad\textsc{Mid}(n)\qquad\text{and}\qquad\textsc{Trap}(n)

      Solution:

      The function f is decreasing, so \textsc{Right}(n)<\displaystyle\int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx<\textsc{Left}(n).

      The function f is concave down, so \textsc{Trap}(n)<\displaystyle\int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx<\textsc{Mid}(n).

      Since Mid(n) and Trap(n) are generally closer estimates than Left(n) and Right(n), we can order:

      \textsc{Right}(n)<\textsc{Trap}(n)<\displaystyle\int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx<\textsc{Mid}(n)<\textsc{Left}(n)

Midterm 1 Review #2

Math 2300 – Calculus II – University of Colorado

Fall 2010 – Review for Midterm I

This review is from instructor Anca Radulescu. I've included my own solutions up through the 'table of integrals' problems. (There may exist better, or more correct solutions.)

§ Substitution

Understand the concept of substitution (as a rewrite of the chain rule in terms of integrals), when to use it and how to carry it through (for both indefinite and definite integrals).

Some practice exercises: Using an appropriate substitution, show that the two integrals are identical:

  1. A\displaystyle{\int _{0}^{{1}}{f(Ax)dx}=\int _{0}^{{A}}{f(x)dx}}

    Solution: If w=Ax and dw=A\, dx then

    \displaystyle A\int _{0}^{1}f(Ax)\, dx \displaystyle=A\int _{{w=0}}^{{w=A}}f(w)\frac{dw}{A}
    \displaystyle=\int _{0}^{A}f(w)\, dw
    \displaystyle=\int _{0}^{A}f(x)\, dx

Midterm 1 Review

Math 2300 – Calculus II – University of Colorado

Fall 2010 – Review for Midterm I

One of the main areas that will be emphasized during this midterm, compared to previous calculus II exams that you may find, is abstract and conceptual understanding. From what we have done in class and on homework/webwork, you have all of the tools to solve these problems. But, they may initially look a bit unfamiliar. Some examples of such problems are given here to help prepare you for the sorts of questions that will be asked on the midterm.

Some Things to Make Sure You Know

  • Substitution of indefinite and definite integrals.

  • Integration by parts of indefinite and definite integrals. It is also always helpful conceptually to know how to derive the formula for integration by parts.

  • Know how to use the table of integrals. More importantly, know how to derive some of the “basic” formulas on and related to that table: \int e^{x}\cos x\, dx and \int\cos^{n}x\, dx are classic examples, and you did one of them as a homework problem.

  • Know how to use the method of partial fractions, including how to apply long division to simplify an integral.

  • Know which trig substitutions to use, and how to use completing the square to rewrite quadratics when needed.

  • Understand how the Left, Right, Mid and Trap approximation methods work and how they relate (when a function is increasing/decreasing, concave up/concave down). Know how Simp relates. Of course, you should be able to calculate any of these and produce the relevant formulae for their calculation (even if you are using a calculator). Know what increase in the factor of subdivisions results in one more decimal of accuracy for each method.

  • Know how to set up the calculations for improper integrals, be able to follow through with the corresponding limit calculations, and be able to explain why this weird limit process is needed.

© 2011 Jason B. Hill. All Rights Reserved.