Recitation One
In this recitation, we covered several problems from “Worksheet 0.” We discuss a few of them as well as the concepts behind them below (note that the solutions are freely available on Carmen).
Basic Algebra review
There’s nothing really interesting to write here. We covered a few of the common algebra mistakes people make during exams and homeworks. Some properties we covered are the following.
- For constants \(a,b,c\), we have \((ab)^c = a^c b^c.\)
- When dividing by a single variable, we can distribute this amongst the numerator. One example is
- However, on the other hand, we cannot distribute the numerator amongst the denominator. For example:
- The Freshman’s dream doesn’t hold. For example:
Also:
\[\sqrt{x^2 + 1} = (x^2 + 1)^{1/2} \neq \sqrt{x} + \sqrt{1}.\]- We have to remember that \(x^2\) is not one-to-one, i.e. there isn’t a unique inverse. If we write something like \(x^2 = 9\), we have two answers: \(x = \pm 3.\)
There are a few others on the worksheet, but this covers the major ones.
Limits
While not super important at first, limits will be relevant throughout the course.
- Recall that \(\lim_{x \rightarrow a} f(x) = L\) if, as \(x\) approaches \(a\), we have \(f(x)\) approaches \(L\) (notice that \(f(a)\) is not necessarily \(L\)).
- Recall that a function \(f(x)\) is continuous if \(\lim_{x \rightarrow a} f(x) = f(a)\).
- If \(\lim_{x \rightarrow a} g(x) = L\) and \(f(x)\) is continuous, then the composition rule says \(\lim_{x \rightarrow a} f(g(x)) = f(L).\) In other words, we can bring limits inside of \(f\) as long as the function is continuous and the limit exists:
- We have L’Hopital’s rule: If a limit \(\lim_{x \rightarrow a} f(x)/g(x)\) is indeterminate of the form \(0/0\) or \(\pm \infty/\infty\), then
Using the above facts, we were able to solve the following (notice this is a little different from the one on the worksheet).
Problem: Solve \(\lim_{k \rightarrow \infty} \sqrt[k+1]{k}.\)
Solution: None of the rules for limits from Calculus apply directly, so we need to transform our problem into one where we can use our rules. Let’s rewrite the inside of the limit using the logarithm. This brings the power out, which turns it into a “doable” problem.
\[\sqrt[k+1]{k} = k^{1/k+1} = e^{ \frac{\ln(k)}{k+1}}.\]Now, as long as \(\lim_{k \rightarrow \infty} \frac{\ln(k)}{k+1}\) exists, we can use the composition rule to say
\[\lim_{k \rightarrow \infty} \sqrt[k+1]{k} = e^{\lim_{k \rightarrow \infty} \frac{\ln(k)}{k+1}}.\]Notice that the form of \(\lim_{k \rightarrow \infty} \frac{\ln(k)}{k+1}\) is \(\infty/\infty\), so we can use L’Hopital’s rule:
\[\lim_{k \rightarrow \infty} \frac{\ln(k)}{k+1} = \lim_{k \rightarrow \infty} \frac{\frac{1}{k}}{1} = 0.\]Putting it all together:
\[\lim_{k \rightarrow \infty} \sqrt[k+1]{k} = e^{\lim_{k \rightarrow \infty} \frac{\ln(k)}{k+1}} = e^0 = 1.\]Derivatives
Remember that the derivative of a function is the instantaneous rate of change of that function at a point (equivalently, it is the slope of the tangent line at that point). Formally speaking, we can write the dreadful limit definition of the derivative:
\[f'(a) = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a} = \lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h}.\]In general, we don’t use this definition to solve a derivative problem. We have a bunch of “derivative rules” that we can use to make the whole process easier.
Problem: What is the derivative of
\[\frac{1}{2 x^3}?\]Solution: This is probably best done using the power rule:
\[\frac{d}{dx} x^n = n x^{n-1}.\]We also need to use the fact that we can pull out constants: if \(k\) is a constant, then
\[\frac{d}{dx} (k f(x)) = k f'(x).\]Notice that
\[\frac{1}{2x^3} = \frac{x^{-3}}{2},\]so using the above rules we have
\[\frac{d}{dx} \left( \frac{1}{2x^3}\right) = \frac{1}{2} \frac{d}{dx} (x^{-3}) = \frac{1}{2} (-3 x^{-4}) = - \frac{3}{2x^4}.\]Definite and Indefinite Integrals
Given a function \(f(x)\), we say that \(F(x)\) is an antiderivative if \(F'(x) = f(x)\). Notice that antiderivatives are not unique. For example, we know that \(\frac{d}{dx} (x^2) = 2x,\) but we also have \(\frac{d}{dx}(x^2 + 4) = 2x.\)
The indefinite integral is the family of antiderivatives of a function. Using our previous example, we would write
\[\int 2x dx = x^2 + C.\]On the other hand, the definite integral has nothing to do with the indefinite integral (at least ahead of time). This is just asking about the signed area underneath a curve (signed here just means we keep track of positives and negatives). For example, if one draws the graph of \(f(x) = x\) from \(x=0\) to \(x=1\), one would have a right triangle with base one and height one. The definite integral from \(x=0\) to \(x=1\) will be
\[\int_0^1 x dx = \frac{1}{2}.\]The reason why we use the same notation for the definite and indefinite integral is the fundamental theorem of calculus. This says that if \(f(x)\) is continuous on the interval \([a,b]\), then
\[\int_a^b f(x)dx = F(b) - F(a),\]where \(F'(x) = f(x)\). So we use the indefinite integral to find an antiderivative, and picking the most convenient one we can use it to find the definite integral.
Problem: Find
\[\int_0^2 x^3 dx.\]Solution: It would be tricky to use the definition to find this answer, so instead we’re going to use the fundamental theorem of calculus. Polynomials are continuous, so it satisfies this criteria. Notice that
\[\int x^3 dx = \frac{x^4}{4} + C,\]so let’s choose the antiderivative \(x^4/4\). Using the theorem, we have
\[\int_0^2 x^3dx = \frac{x^4}{4} \Big|_{x=0}^2 = \frac{2^4}{4} - \frac{0^4}{4} = 4.\]Problem: Is it true that we can use the fundamental theorem of calculus to get
\[\int_{-1}^1 \frac{1}{x^2} dx = -2?\]Solution: It is tempting to want to use the fundamental theorem of calculus. We might write
\[\int \frac{1}{x^2}dx = -x^{-1} + C,\]so
\[\int_{-1}^1 \frac{1}{x^2}dx = -x^{-1} \Big|_{x=-1}^1 = -2.\]However, notice that \(1/x^2\) is not continuous at \(x=0\), which is in the interval \([-1,1]\). We cannot use the fundamental theorem of calculus to solve this problem.
Integration Technique: U-Sub
Remember that chain rule says
\[\frac{d}{dx} f(g(x)) = f'(g(x)) g'(x).\]If we take the indefinite integral of both sides, we see that
\[\int f'(g(x))g'(x)dx = f(g(x)) + C.\]The idea behind \(u\)-substitution (or just “\(u\)-sub”) is to formalize this trick. It’s hard to write a good criteria of when to use \(u\)-sub, but it’s generally a good first step for a difficult integral. Let’s test one.
Problem: Find
\[\int_0^2 3x^2 e^{x^3}dx.\]Solution: We’re going to use the fundamental theorem of calculus since the function \(2xe^{x^3}\) is continuous.
None of the usual tricks for integration work, so we need to try a new one. If we take a second, we might notice that \(\frac{d}{dx} x^3 = 3x^2\). Let’s let
\[u = x^3, \ \ du = 3x^2 dx.\]Then
\[\int 3x^2 e^{x^3}dx = \int 3x^2 e^u dx = \int e^u du = e^u + C = e^{x^3} + C.\]By the fundamental theorem of calculus,
\[\int_0^2 3x^2 e^{x^3} dx = e^{x^3} \Big|_{x=0}^2 = e^{2^3} - e^0 = e^8 - 1.\]Problem: Is it true that
\[\int e^{x^2} dx = \frac{e^{x^2}}{2x} + C?\]Solution: It is really tempting to use \(u\)-sub incorrectly here. Let \(u = x^2\), \(du = 2x dx\). The wrong answer would say
\[\int e^{x^2}dx = \int \frac{e^u}{2x} du = \frac{e^u}{2x} + C = \frac{e^{x^2}}{2x} + C.\]If we pause for a moment, we can see a problem with the right hand side. We can use quotient rule to get
\[\frac{d}{dx} \left( \frac{e^{x^2}}{2x} \right) = -\frac{2 e^{x^2} - 4x^2 e^{x^2}}{4x^2} \neq e^{x^2}.\]We must have made a mistake. Notice that we cannot use \(u\)-sub if we did not eliminate all of the \(x\)’s in the problem. So it is not true.
If you’re interested, we actually can’t solve this indefinite integral using what are called “elementary functions.” More details can be found here.