Welcome to Calculus 1301B 007 17WI!

Course description

  • General info
    • Instructor name: Pál Zsámboki
    • Instructor email: pzsambok at uwo.ca
    • Office hours: Tuesdays 4pm-6pm at MC 133, or by appointment
    • Section website: follow this link. The notes are here. The md files are the common source of the notes and the slides, that is, the contents of the three are the same
    • Course website: follow this link This is maintained by the course coordinator. It only contains information and announcements relevant to all sections.
    • Course outline and suggested practice problems: on OWL, on the section website, or on the course website. You're expected to know the contents of the course outline. You should at least solve the suggested problems from the textbook sections, but the more the merrier.
    • Math Help Centre: 2:30pm-6:30pm at MC 106, on weekdays, from Monday January 16 until Friday April 8. There will be no Help Centre during Reading Week.

Course description

  • Textbook
    • Single Variable Calculus: Early Transcendentals (8th edition), by James Stewart (Brooks/Cole) with the Student's Solution Manual
    • Practice problems are renumbered in each edition, so it's important to have access to the current edition.
    • Alternatively, you can get an Enhanced Enhanced Web Assign and electronic “You Book” Access Card. It comes with access to the electronic version of the textbook.
  • Optional sources
    • Lecture Notes for Calculus Volume 2 (8th edition), by R. N. Bryan (Custom Course Materials)
    • Midterm Tests and Final Exams for Calculus 1301B, by R. N. Bryan ( Custom Course Materials). I highly recommend getting this. It's important not to just solve problems one by one but to actually time yourself doing an exam. It gives a better evaluation of how prepared you are, and you can practice solving the easier problems first etc.

Course description

  • Tests
    • In-class test 15%. This 45 minutes in-class test will be written before Reading week. Further information will be provided closer to the exam date. No makeup test will be scheduled. If you are unable to write the test due to documented illness or other serious circumstances, your final grade will be calculated as follows: midterm 40%, final 60%.
    • Midterm exam 35%. The midterm will be written Friday March 3, 7pm-9:30pm. The location will vary depending on your section and surname. Specific details on the exam will be posted online closer to the exam date.
    • Final exam 50%. The final exam will be 3 hours long, and it will be cumulative. It will be scheduled by the Registrar’s Office and will be held during the exam period.

Review of derivation

Definition of derivatives derivatives of elementary functions

  • Calculus II builds upon the content of Calculus I. Today, we'll quickly review derivation and integration. If any of the concepts covered feels unfamiliar, refer to the corresponding section of the textbook, and solve enough practice problems that you're comfortable with it.
  • Let \(f\) be a function with domain \(D\), and \(x\) a point in an open interval in \(D\). Then the derivative of \(f\) at \(x\) is \[ f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}, \] provided that the limit exists.
    • If the limit exists, then we say that \(f\) is differentiable at \(x\).
    • Graphical interpretation: \(f'(x_0)\) is the slope of the tangent line to the graph \(y=f(x)\) at \(x_0\).
    • Recall that an equation of the line through the point \(P(x_0,y_0)\) with slope \(k\) is \[ y-y_0=k(x-x_0). \]
    • The tangent line at \(x_0\) is the line through \(P(x_0,f(x_0))\), which approximates the graph \(y=f(x)\) the best.
    • Click here for an illustration

Derivatives of elementary function, rules of derivation

  • Derivatives of elementary functions.
    • For a constant \(c\), we have \((c)'=0\).
    • For a nonzero number \(n\), we have \((x^n)'=nx^{n-1}\).
    • We have \((e^x)'=e^x\).
    • We have \((\sin x)'=\cos x\) and \((\cos x)'=-\sin x\).
  • Let \(f\) and \(g\) be functions differentiable at \(x\), and \(c,d\) constants.
  • Derivation is linear.
    • To be linear means to respect addition and constant multiplication. That is, this is two rules in one:
    • \((f(x)+g(x))'=f'(x)+g'(x)\), and
    • \((cf(x))'=cf'(x)\)
    • Or you can put the two formulas together as \((cf(x)+dg(x))'=cf'(x)+dg'(x)\).
    • Example: let \(f(x)=x^3-3x^2+4x-6\).
    • Then we have \[ f'(x)=(x^3)'-3(x^2)'+4(x)'-6(1)'=3x^2-6x+4. \]

Rules of derivation

  • Product rule.
    • We have \((f(x)g(x))'=f'(x)g(x)+f(x)g'(x)\).
    • Example: let \(f(x)=(\sin x)e^x\). Then we have \[ f'(x)=(\sin x)'e^x+(\sin x)(e^x)'=(\cos x)e^x+(\sin x)e^x. \]
  • Quotient rule.
    • We have \[ \left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}. \]
    • Example: let \(f(x)=\tan x\). Then we have \[ f'(x)=(\tan x)'=\left(\frac{\sin x}{\cos x}\right)'=\frac{(\sin x)'(\cos x)-(\sin x)(\cos x)'}{(\cos x)^2}=\frac{(\cos x)^2+(\sin x)^2}{(\cos x)^2}=\frac{1}{(\cos x)^2}=(\sec x)^2. \]
  • Chain rule.
    • Let \(h(x)\) denote the composite function \(f(g(x))\). Then we have \[ h'(x)=f'(g(x))g'(x). \]
    • Example: derivatives of exponential functions. Let \(a\) be a positive number. Then we have \(a^x=e^{(\ln a)x}\), which we can write as the composite function \(h(x)=f(g(x))\) with \(f(x)=e^x\) and \(g(x)=(\ln a)x\). Therefore, we have \[ (a^x)'=f'(g(x))g'(x)=e^{(\ln a)x}\ln a=(\ln a)a^x. \]

Implicit differentiation

  • Using the chain rule, we can derivate equations involving both the variable \(x\), and a function of that variable \(y=f(x)\).
    • The collection of the points \(P(x,y)\) satisfying a given equation is referred to as the implicit curve defined by the equation. Then the derivative \(y'\) gives the slopes of the tangent lines to the curve.
    • Note that the graphs \(y=f(x)\) form a special case of this.
    • The point in implicit derivation is that we can find the tangent lines of curves which cannot be expressed as graphs, that is there's a number \(x_0\) such that the curve intersects the line \(x=x_0\) more than once.
    • For example, consider the ellipse \(E:4x^2+y^2=25\). We can find its tangent line at the point \(P(2,3)\) as follows.
    • First, we implicitly derivate the equation. We get \[ 8x+2yy'=(4x^2+y^2)'=(25)'=0. \]
    • Then we solve for \(y'\). We get \[ y'=-4\frac{x}{y}. \]
    • This will give the slopes of the tangent lines. At \(P(x=2,y=3)\), we have \[ y'(x=2,y=3)=-\frac83. \]
    • Using this, we can write the tangent line as follows \[ y-3=(y-y_0)=y'(x=x_0,y=y_0)(x-x_0)=-\frac83(x-2). \]
    • Click here for an illustration

Application: derivatives of inverse functions.

  • Let \(f\) be a function which is one-to-one, that is if we have \(f(x_1)=f(x_2)\), then we need \(x_1=x_2\).
    • In this case, we can get the inverse function of \(f\), denoted by \(f^{-1}(x)\), using the formula \[ f^{-1}(y)=x\text{ precisely when }y=f(x). \]
    • Note that sometimes it is more convenient to set the variable of the inverse function to \(y\). At other times, it is more convenient to make it have variable \(x\). It is important to be comfortable with both notations.
    • For example, the natural logarithm function \(\ln(x)\) is defined by \[ \ln(x)=y\text{ precisely when }x=e^y. \]
  • We can use implicit derivation to find derivatives of inverse functions.
    • For this, we set \(y=f^{-1}(x)\).
    • Then we implicitly derivate the equation \(f(y)=x\). We get \[ f'(y)y'=1. \]
    • Solving for \(y'\) gives \[ (f^{-1}(x))'=y'=\frac{1}{f'(y)}. \]
    • For example, let \(f(y)=e^y\). Then we get \(f^{-1}(x)=\ln x\).
    • Substituting this to the formula, we get \[ (\ln x)'=\frac{1}{(e^y)'}=\frac{1}{e^y}=\frac{1}{e^{\ln x}}=\frac{1}{x}. \]

Application: derivatives of inverse functions.

  • We can improve the result \((\ln x)'=x^{-1}\).
    • Recall that the absolute value function is the piecewise linear function \[ |x|=\begin{cases} x & x\ge0,\\ -x & x\le0. \end{cases} \]
    • Let \(f(x)=\ln|x|\). Note that the domain of this function is \((-\infty,0)\cup(0,\infty)\).
    • If \(x>0\), then we have \[ f'(x)=(\ln|x|)'=(\ln(x))'=x^{-1} \] as we have just seen.
    • If \(x<0\), then we have \[ f'(x)=(\ln|x|)'=(\ln(-x))'=(-x)^{-1}(-1)=x^{-1} \]
    • Therefore, for any \(x\ne0\), we have \((\ln|x|)'=x^{-1}\).
  • Similarly, we can find the derivatives of the inverse trigonometric functions. We have \[ (\sin^{-1}(x))'=\frac{1}{\sqrt{1-x^2}},\,(\cos^{-1}(x))'=-\frac{1}{\sqrt{1-x^2}},\,(\tan^{-1}(x))'=\frac{1}{1+x^2},\,\text{etc.} \]

Application: logarithmic differentiation.

  • Another application of implicit differentiation is to derivate functions of the form \(f(x)^{g(x)}\), where both \(f\) and \(g\) are nonconstant functions.
    • Note that if \(g\) is constant, then using the chain rule we get \[ ((f(x))^g)'=gf(x)^{g-1}f'(x), \] and if \(f\) is constant, we get \[ (f^{g(x)})'=(\ln f)f^{g(x)}g'(x), \] but we do need a new method in case both \(f\) and \(g\) are nonconstant.
    • What we do is to take the equation \(y=f(x)^{g(x)}\), and take natural logarithms: \[ \ln y=\ln(f(x)^{g(x)})=g(x)\ln(f(x)). \]
    • Then we use implicit differentiation: \[ \frac{y'}{y}=g'(x)\ln(f(x))+g(x)\frac{f'(x)}{f(x)}. \]
    • Solving for \(y'\), we get \[ (f(x)^{g(x)})'=y'=\left(g'(x)\ln(f(x))+g(x)\frac{f'(x)}{f(x)}\right)f(x)^{g(x)} \]
    • Example: in the prototypical case \(f(x)=g(x)=x\), we get \[ (x^x)'=(\ln x+1)x^x \]

Review of integration


  • Let \(f\) be a function with domain \(I\).
    • Let \(F\) be a function with domain \(I\). If we have \(F'(x)=f(x)\) for all \(x\) in \(I\), then we say that \(F\) is an antiderivative of \(f\).
    • Suppose that \(F_1\) and \(F_2\) are antiderivatives of \(f\). Then we have \[ (F_1-F_2)'=F_1'-F_2'=f-f=0. \]
    • Suppose that \(I\) is an open interval. In that case, \(F_1-F_2=0\) implies that \(F_1(x)-F_2(x)=C\), a constant.
    • In general, \(F_1-F_2\) is a function which is constant on any interval in \(I\). We denote that by \(C\) too.
    • We will call a function which is contant on any interval a locally constant function.
    • What we have seen that if \(F\) is an antiderivative of \(f\), then any other antiderivative is of the form \(F(x)+C\) for some locally constant function \(C\).
    • Therefore, we call \(F(x)+C\) the most general antiderivative of \(f\).
    • We also call \(F(x)+C\) the indefinite integral of \(f\), and we write \(\int f(x)\,\mathrm dx=F(x)+C\)
    • In most cases, the domain \(I\) will be an interval, and thus \(C\) will be a constant. But there are important cases where \(I\) is not an interval. We will soon see one.

Integrals of elementary functions

  • Integrals of elementary functions.
    • The derivation rules for elementary functions can be turned into integration rules.
    • Let \(n\ne-1\) be a number. Then the derivation rule \((x^{n+1})'=(n+1)x^n\) gives the integration rule \[ \int x^n\,\mathrm dx=\frac{x^{n+1}}{n+1}+C. \]
    • For \(n=-1\), the derivation rule \((\ln|x|)'=x^{-1}\) gives the integration rule \[ \int x^{-1}\,\mathrm dx=\ln|x|+C. \]
    • Note that the domain of \(x^{-1}\) is \((-\infty,0)\cup(0,\infty)\). Therefore, the locally constant function \(C\) can take up an arbitrary constant value on \((-\infty,0)\), and another arbitrary constant value on the interval \((0,\infty)\).
    • We can similarly get \[ \int e^x\,\mathrm dx=e^x+C,\,\int \sin x\,\mathrm dx=-\cos x+C,\,\int\cos x\,\mathrm dx=\sin x+C\text{ and }\int(\sec x)^2\,\mathrm dx=\tan x+C. \]

Linearity of integration

  • Integration is linear.
    • The linearity rule for derivatives \((cf(x)+dg(x))'=cf'(x)+dg'(x)\) turns into the linearity rule for integrals: \[ \int cf(x)+dg(x)\,\mathrm dx=c\int f(x)\,\mathrm dx+d\int g(x)\,\mathrm dx. \]
    • Example: let \(f(x)=x^2-2x+3\).
    • Then we have \[ \int f(x)\,\mathrm dx=\int x^2-2x+3\,\mathrm dx=\int x^2\,\mathrm dx-2\int x\,\mathrm dx+\int3\,\mathrm dx=\frac13x^3-x^2+3x+C. \]

Substitution rule

  • The chain rule \((f(g(x)))'=f'(g(x))g'(x)\) can be turned into an integration rule.
    • For this, let us change notation: \[ (F(u(x)))'=F'(u(x))u'(x). \]
    • Note that this implies \[ \int F'(u(x))u'(x)\,\mathrm dx=F(u(x))+C. \]
    • By writing \(u=u(x)\) we can express that it is possible to think of \(u\) not only as a function but a variable too, into which we substitute the function \(u(x)\). Therefore we have \[ F(u(x))+C=F(u)+C=\int F'(u)\,\mathrm du. \]
    • Putting these together, we get: \[ \int F'(u(x))u'(x)\,\mathrm dx=\int F'(u)\,\mathrm du \]
    • Letting \(f(u)=F'(u)\), we get the Substitution rule: \[ \int f(u(x))u'(x)\,\mathrm dx=\int f(u)\,\mathrm du. \]

Substitution rule.

  • Example: let's calculate \(\int x^2\sqrt{x^3+3}\,\mathrm dx\).
    • We want to write the integrand \(x^2\sqrt{x^3+3}\) in the form \(f(u(x))u'(x)\). Since the composite function in the integrand is \(\sqrt{x^3+3}\), it looks like we could have \(f(u)=k\sqrt u\) for some constant \(k\), and \(u(x)=(x^3+3)\).
    • That is, we need to solve for \(k\) the equation \[ x^2\sqrt{x^3+3}=kf(u(x))u'(x)=k\sqrt{x^3+3}3x^2. \]
    • We get \(k=\frac13\), and thus we can use the substitution rule: \[ \int x^2\sqrt{x^3+3}\,\mathrm dx=\int f(u(x))u'(x)\,\mathrm dx=\int f(u)\,\mathrm du=\int\frac13\sqrt{u}\,\mathrm du=\frac29u^{3/2}+C=\frac29(x^3+3)^{3/2}+C. \]

Definite integrals.

  • Definite integrals can be used to calculate net areas under graphs of functions.
    • Let \(f\) be a function defined on an interval \([a,b]\). Then the net area or signed area of the graph \(y=f(x)\) along \([a,b]\) is \(A=A_+-A_-\), where \(A_+\) is the area of the region above the \(x\)-axis, and \(A_-\) is the area of the region below the \(x\)-axis.
    • The definite integral of \(f\) along \([a,b]\) is \[ \int_a^bf(x)\,\mathrm dx=\lim_{n\to\infty}\sum_{i=1}^nf(x_i^*)\Delta x, \] provided that the limit exists. In case the limit exists, we say that f(x) is integrable on \([a,b]\).
    • Here, the sum \(R_n=\sum_{i=1}^nf(x_i^*)\Delta x\) is called a partial sum or a Riemann sum. It is formed as follows.
    • The interval \([a,b]\) is split up to \(n\) subintervals of equal length \(\Delta x=\frac{b-a}{n}\).
    • That is, these subintervals are of the form \([x_{i-1},x_i]\), where \(x_i=a+i\Delta x\).
    • Then we pick the sample points \(x_{i-1}\le x_i^*\le x_i\).
    • That is, we're summing the signed areas of the rectangles \([x_{i-1},x_i]\times[0,f(x_i^*)]\) for \(i=1,\dotsc,n\).
    • Click here for an illustration
    • It can be proven that if the limit \[ \lim_{n\to\infty}\sum_{i=1}^nf(x_i^*)\Delta x \] exists, it is independent of the choice of the sample points \(x_i^*\).

Fundamental theorem of calculus

  • The Fundamental theorem of calculus (FTC) is our main tool for calculating definite integrals.
    • Theorem. Let \(f\) and \(F\) be functions defined on an interval \([a,b]\). Suppose that we have \(F'=f\) on \((a,b)\). Then we have \[ \int_a^bf(x)\,\mathrm dx=F(b)-F(a). \]
    • We have the following alternative notations \[ F(x)\Big|_a^b=F(x)\Big]_a^b=\left[F(x)\right]_a^b=F(b)-F(a). \]
    • Example: consider the definite integral \(\int_{-2}^1x^2-x-2\,\mathrm dx\).
    • Since we have \[ \int x^2-x-2\,\mathrm dx=\frac13x^3-\frac12x^2-2x+C, \] we get \[ \int_{-2}^1x^2-x-2\,\mathrm dx=\frac13x^3-\frac12x^2-2x\Big|_{-2}^1=\frac13(1+8)+\frac12(-1+4)-2-4=\frac{18+9-36}{6}=-\frac{3}{2}. \]

Substitution rule for definite integrals

  • Although the substitution rule for indefinite integrals could be applied to calculate definite integrals via the fundamental rule of calculus as is, it's got a version for definite integrals, which makes calculations easier.
    • Let \(u(x)\) be a function on \([a,b]\) which is differentiable on \((a,b)\), and let \(f(u)\) be a function which is integrable on \([u(a),u(b)]\). Then we have \[ \int_{a}^{b}f(u(x))u'(x)\,\mathrm dx=\int_{u(a)}^{u(b)}f(u)\,\mathrm du. \]
    • Note that you have to change the boundaries when you substitute \(u=u(x)\).
    • Example. Let's calculate \(\int_{\pi/6}^{\pi/2}\frac{\cos x}{\sin x}\,\mathrm dx\).
    • We want to write \(\frac{\cos x}{\sin x}=f(u(x))u'(x)\).
    • We see a multiplier of \(\cos x\), so we should have \(u(x)=\sin x\).
    • This means that we should have \(\frac{1}{\sin x}=f(u(x))\), that is \(f(x)=x^{-1}\).
    • Then the substitution rule gives \[ \int_{\pi/6}^{\pi/2}\frac{\cos x}{\sin x}\,\mathrm dx=\int_{1/2}^1u^{-1}\,\mathrm du=\ln|x|\Big|_{1/2}^1=\ln0-\ln\frac12=\ln2. \]