Midterm 1 Review #2

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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
  2. \displaystyle{\int _{0}^{{\pi/3}}{3\sin^{2}(3x)dx}=\int _{0}^{{\pi}}{\sin^{2}(x)dx}}

    Solution: If w=3x and dw=3\, dx then

    \displaystyle\int _{0}^{{\pi/3}}3\sin^{2}(3x)\, dx \displaystyle=\int _{{w=0}}^{{w=\pi}}3\sin^{2}(w)\frac{dw}{3}
    \displaystyle=\int _{0}^{\pi}\sin^{2}(w)\, dw
    \displaystyle=\int _{0}^{\pi}\sin^{2}(x)\, dx
  3. \displaystyle{\int _{0}^{{B}}{f(x+A)dx}=\int _{A}^{{A+B}}{f(x)dx}}

    Solution: If w=a+A and dw=dx then

    \displaystyle\int _{0}^{B}f(x+A)\, dx \displaystyle=\int _{{w=A}}^{{w=A+B}}f(w)\, dw
    \displaystyle=\int _{A}^{{A+B}}f(x)\, dx
  4. \displaystyle{\int _{0}^{{\pi}}{(\pi-x)\cos(x)dx}=\int _{0}^{{\pi}}{x\cos(\pi-x)dx}}

    Solution: Use the substitution w=\pi-x and dw=-dx. Notice that this implies x=\pi-w.

    \displaystyle\int _{0}^{\pi}(\pi-x)\cos(x)\, dx \displaystyle=\int _{{w=\pi}}^{{w=0}}-w\cos(\pi-w)\, dw
    \displaystyle=-\int _{\pi}^{0}w\cos(\pi-w)\, dw
    \displaystyle=\int _{0}^{\pi}w\cos(\pi-w)\, dw
    \displaystyle=\int _{0}^{x}x\cos(\pi-x)\, dx
  5. I_{{m,n}}=I_{{n,m}}, where \displaystyle{I_{{m,n}}=\int _{0}^{1}{x^{m}(1-x)^{n}dx}}

    Solution: Using the substitution w=1-x and dw=-dx we can solve for x=1-w and then

    \displaystyle I_{{m,n}} \displaystyle=\int _{0}^{1}x^{m}(1-x)^{n}\, dx
    \displaystyle=\int _{{w=1}}^{{w=0}}-(1-w)^{m}w^{n}\, dw
    \displaystyle=-\int _{1}^{0}(1-w)^{m}w^{n}\, dw
    \displaystyle=\int _{0}^{1}(1-w)^{m}w^{n}\, dw
    \displaystyle=\int _{0}^{1}w^{n}(1-w)^{m}\, dw
    \displaystyle=\int _{0}^{1}x^{n}(1-x)^{m}\, dx
    \displaystyle=I_{{n,m}}
  6. Show that: \displaystyle{\int _{{-\pi}}^{{\pi}}{\cos(m\theta)\sin(n\theta)d\theta}=0}

    Solution: I think this is supposed to be

    \displaystyle{\int _{{-\pi}}^{{\pi}}{\cos(m\theta)\sin(m\theta)d\theta}=0}

    If we let w=\sin(m\theta) and dw=m\cos(m\theta)\, d\theta then we have

    \displaystyle\int _{{-\pi}}^{\pi}\cos(m\theta)\sin(m\theta)\, d\theta \displaystyle=\frac{1}{m}\int _{{w=\sin(-m\pi)}}^{{w=\sin(m\pi)}}w\, dw
    \displaystyle=\frac{1}{m}\left[\frac{1}{2}w^{2}\right]_{{\sin(-m\pi)}}^{{\sin(m\pi)}}
    \displaystyle=\frac{1}{2m}\left[\sin^{2}(m\pi)-\sin^{2}(-m\pi)\right]=0

§ Integration by parts

Understand the concept (as a rewrite of the product rule in terms of integrals), when it is convenient to use it. Know how to set up and complete and integration by parts for indefinite and definite integrals (including improper integrals). Know how to derive a recurrence formula (as shown in the Table of Integrals) using integration by parts, such as in \int _{0}^{1}{(-\ln x)^{n}dx} (the improper integral on your last worksheet), or \int{\cos^{n}(t)dt} (in your 7.3 homework) or \int{x^{n}e^{x}dx} (in your last review sheet).

Some practice exercises:

  1. Show that, for any continuous function f: \displaystyle{f(x)-f(0)=f^{{\prime}}(0)x+\int _{0}^{x}{f^{{\prime\prime}}(t)(x-t)dt}}

    Solution: By the Fundamental Theorem of Calculus we can write

    f(x)-f(0)=\int _{0}^{x}f^{{\prime}}(t)\, dt.

    We now evaluate this using integration by parts, in a very strange way. We let u=f^{{\prime}}(t) and v^{{\prime}}=1. We must have u^{{\prime}}=f^{{\prime\prime}}(t), but with v we can actually use any antiderivative of v^{{\prime}}, and so we pick v=t-x. Remember that these are functions of t. Here, x is simply the upper limit of the integral, and is therefore a number. So, v=t-x actually satisfies v^{{\prime}}=1 when we differentiate with respect to t. Then we get

    \displaystyle f(x)-f(0) \displaystyle=\int _{0}^{x}f^{{\prime}}(t)\, dt
    \displaystyle=f^{{\prime}}(t)(t-x)\Big|_{0}^{x}-\int _{0}^{x}f^{{\prime\prime}}(t)(t-x)\, dt
    \displaystyle=f^{{\prime}}(x)(x-x)-f^{{\prime}}(0)(0-x)-\int _{0}^{x}f^{{\prime\prime}}(t)(t-x)\, dt
    \displaystyle=-f^{{\prime}}(0)(-x)-\int _{0}^{x}f^{{\prime\prime}}(t)(t-x)\, dt
    \displaystyle=f^{{\prime}}(0)(x)-\int _{0}^{x}f^{{\prime\prime}}(t)(t-x)\, dt
    \displaystyle=f^{{\prime}}(0)(x)+\int _{0}^{x}f^{{\prime\prime}}(t)(-1)(t-x)\, dt
    \displaystyle=f^{{\prime}}(0)(x)+\int _{0}^{x}f^{{\prime\prime}}(t)(x-t)\, dt
  2. Show that: \displaystyle{\int _{{-\pi}}^{{\pi}}{\cos(m\theta)\cos(n\theta)d\theta}=0} if m and n are integers that
    are not equal.

    Solution: I think we may also need m and n to be nonzero. In any case, this requires integration by parts twice and we end up with something strange. Let

    \begin{array}[]{cc}u=\cos(m\theta)&v^{{\prime}}=\cos(n\theta)\\<br />
u^{{\prime}}=-m\sin(m\theta)&v=\frac{1}{n}\sin(n\theta)\end{array}

    and you get

    \displaystyle\int _{{-\pi}}^{\pi}\cos(m\theta)\sin(n\theta)\, d\theta \displaystyle=\frac{1}{n}\cos(m\theta)\sin(n\theta)\Big|_{{-\pi}}^{\pi}+\frac{m}{n}\int _{{-\pi}}^{\pi}\sin(m\theta)\sin(n\theta)\, d\theta

    Since \sin(n\theta)=0 for any integer n this is the same as

    \frac{m}{n}\int _{{-\pi}}^{\pi}\sin(m\theta)\sin(n\theta)\, d\theta.

    Use integration by parts again with

    \begin{array}[]{cc}u=\sin(m\theta)&v^{{\prime}}=\sin(n\theta)\\<br />
u^{{\prime}}=m\cos(m\theta)&v=-\frac{1}{n}\cos(n\theta)\end{array}

    and we find (using the same reasoning) that

    \displaystyle\frac{m}{n}\int _{{-\pi}}^{\pi}\sin(m\theta)\sin(n\theta)\, d\theta \displaystyle=\frac{m}{n}\left[-\frac{1}{n}\sin(m\theta)\cos(n\theta)\Big|_{{-\pi}}^{\pi}+\frac{m}{n}\int _{{-\pi}}^{\pi}\cos(m\theta)\cos(n\theta)\, d\theta\right]
    \displaystyle=\frac{m}{n}\left[\frac{m}{n}\int _{{-\pi}}^{\pi}\cos(m\theta)\cos(n\theta)\, d\theta\right]
    \displaystyle=\frac{m^{2}}{n^{2}}\int _{{-\pi}}^{\pi}\cos(m\theta)\cos(n\theta)\, d\theta.

    Thus, we have actually shown that

    \int _{{-\pi}}^{\pi}\cos(m\theta)\cos(n\theta)\, d\theta=\frac{m^{2}}{n^{2}}\int _{{-\pi}}^{\pi}\cos(m\theta)\cos(n\theta)\, d\theta.

    Since m\neq n we know that \frac{m^{2}}{n^{2}}\neq 1 and so the integral must be zero by basic algebra.

§ Table of integrals

It is worth remembering classical integrals from the table (such as \int{\frac{1}{x^{2}+1}dx}=\tan^{{-1}}(x)+C, or \int{\sec(t)\tan(t)}=\sec(t)+C). You do not have to memorize all that is in the tables, but rather you should be able to obtain some or these formulas on your own (e.g., any of the ones listed in the previous section). You also have to be able to use a known general formula towards solving a particular case that follows that pattern.

Practice exercise:

  1. Find a formula (that will depend on a and b) for \displaystyle{\int{e^{{ax}}\cos(bx)dx}}.

    Solution: This is an example where you can use integration by parts twice and solve for your original integral. With

    \begin{array}[]{cc}u=e^{{ax}}&v^{{\prime}}=\cos(bx)\\<br />
u^{{\prime}}=ae^{{ax}}&v=\frac{1}{b}\sin(bx)\end{array}

    we find that

    \displaystyle\int e^{{ax}}\cos(bx)\, dx \displaystyle=\frac{1}{b}e^{{ax}}\sin(bx)-\int ae^{{ax}}\frac{1}{b}\sin(bx)\, dx
    \displaystyle=\frac{1}{b}e^{{ax}}\sin(bx)-\frac{a}{b}\int e^{{ax}}\sin(bx)\, dx.

    Using integration by parts again with

    \begin{array}[]{cc}u=e^{{ax}}&v^{{\prime}}=\sin(bx)\\<br />
u^{{\prime}}=ae^{{ax}}&v=-\frac{1}{b}\cos(bx)\end{array}

    gives that this is equal to

    \displaystyle=\frac{1}{b}e^{{ax}}\sin(bx)-\frac{a}{b}\left[-\frac{1}{b}e^{{ax}}\cos(bx)+\frac{a}{b}\int e^{{ax}}\cos(bx)\, dx\right]
    \displaystyle=\frac{1}{b}e^{{ax}}\sin(bx)+\frac{a}{b^{2}}e^{{ax}}\cos(bx)-\frac{a^{2}}{b^{2}}\int e^{{ax}}\cos(bx)\, dx

    Since we now have

    \int e^{{ax}}\cos(bx)\, dx=\frac{1}{b}e^{{ax}}\sin(bx)+\frac{a}{b^{2}}e^{{ax}}\cos(bx)-\frac{a^{2}}{b^{2}}\int e^{{ax}}\cos(bx)\, dx

    we add the right-most integral to both sides and we get

    \displaystyle\int e^{{ax}}\cos(bx)\, dx+\frac{a^{2}}{b^{2}}\int e^{{ax}}\cos(bx)\, dx \displaystyle=\frac{1}{b}e^{{ax}}\sin(bx)+\frac{a}{b^{2}}e^{{ax}}\cos(bx)
    \displaystyle\left(1+\frac{a^{2}}{b^{2}}\right)\int e^{{ax}}\cos(bx)\, dx \displaystyle=\frac{1}{b}e^{{ax}}\sin(bx)+\frac{a}{b^{2}}e^{{ax}}\cos(bx)
    \displaystyle\left(\frac{a^{2}+b^{2}}{b^{2}}\right)\int e^{{ax}}\cos(bx)\, dx \displaystyle=\frac{1}{b}e^{{ax}}\sin(bx)+\frac{a}{b^{2}}e^{{ax}}\cos(bx)
    \displaystyle\int e^{{ax}}\cos(bx)\, dx \displaystyle=\left(\frac{b^{2}}{a^{2}+b^{2}}\right)\left[\frac{1}{b}e^{{ax}}\sin(bx)+\frac{a}{b^{2}}e^{{ax}}\cos(bx)\right].

    The right hand side will then simplify slightly and we have

    \int e^{{ax}}\cos(bx)\, dx=e^{{ax}}\left(\frac{b}{a^{2}+b^{2}}\sin(bx)+\frac{a}{a^{2}+b^{2}}\cos(bx)\right).
  2. Use this formula to evaluate: \displaystyle{\int _{0}^{{\pi}}{e^{{-x}}\cos(2x)dx}}.

    Solution: Here a=-1 and b=2. So, using the formula we just derived we find

    \displaystyle\int _{0}^{\pi}e^{{-x}}\cos(2x)\, dx \displaystyle=\left[e^{{-x}}\left(\frac{2}{5}\sin(2x)-\frac{1}{5}\cos(2x)\right)\right]_{0}^{\pi}
    \displaystyle=\frac{1}{e^{\pi}}\left(\frac{2}{5}\sin(2\pi)-\frac{1}{5}\cos(2\pi)\right)-e^{0}\left(\frac{2}{5}\sin(0)-\frac{1}{5}\cos(0)\right)
    \displaystyle=-\frac{1}{5e^{\pi}}+\frac{1}{5}
    \displaystyle=\frac{1}{5}\left(1-\frac{1}{e^{\pi}}\right)
    \displaystyle\approx 0.191357...

    This is verified by a calculator.

§ Partial fractions

Know the rules for setting up an integrand as a sum of partial fractions (be able to perform long polynomial division, if necessary).

Practice exercises:

  1. Calculate \displaystyle{\int{\frac{x}{(1-2x)(x-a)}dx}}, for (a) a\neq 1/2 and (b) a=1/2.

  2. Homework problem #66/7.4

§ Trig substitution

Know which substitution is recommended for each case, and be able to carry out the transformation.

Practice exercise: Consider the integral: \displaystyle{\int{\frac{1}{\sqrt{2z-z^{2}}}dz}}

  1. What trig substitution would be appropriate to transform it into \int{\frac{1}{1-w^{2}}dw}?

  2. What method would you use at that point to finish solving the integral?

§ Approximations

Know how to:

  • interpret an integral geometrically in terms of areas;

  • calculate, for some given small n: LEFT(n), RIGHT(n), TRAP(n), MID(n), SIMP(n);

  • tell, when possible, which of the above are under or over-estimates;

  • calculate the error when you know the estimate and the exact value of the integral;

  • estimate (for each approximation scheme) how the error will be affected if n is increased by a multiplicative factor (and conversely, how n needs to change to obtain more error precision).

§ Improper integrals

Know how to:

  • recognize an improper integral and explain why it is improper;

  • set up the integral as a limit and, if necessary, apply substitution, parts etc. to solve the integral inside the limit;

  • evaluate simple limits, possibly using L'Hôpital's rule when necessary;

  • split up an improper integral as a sum of simpler improper integrals, if necessary, and understand that convergence requires convergence of all the components of the sum.

Practice exercise: If g is any arbitrary bounded function, show that:

\int _{{-\infty}}^{{\infty}}{g^{{\prime}}(x)e^{{-x^{2}/2}}dx}=\int _{{-\infty}}^{{\infty}}{xg(x)e^{{-x^{2}/2}}dx}

© 2011 Jason B. Hill. All Rights Reserved.