Section 1 SingleVariable Calculus Review
¶Multivariable calculus is all about trying to generalize the ideas of singlevariable calculus, so it's important to review your singlevariable calculus knowledge. The problems below are not remotely close to a complete review, but they highlight some of the key ideas we'll be using in the next few weeks. Try them on your own before checking your answers.
Subsection 1.1 Derivative Review
 For a single variable function \(f\text{,}\) how do you picture the derivative \(f'(a)\) at a point \(a\) on the graph of \(f\text{?}\) Solution
\(f'(a)\) is the slope of the line tangent to the graph of \(f(x)\) at \(x = a\text{:}\)
 What information do we get about a function from knowing whether its derivative \(f'(a)\) at a point \(a\) is positive or negative? Solution
Knowing whether the tangent line has positive or negative slope at \(a\) tells us whether the function \(f\) is increasing or decreasing at \(a\text{.}\)
 The owner of a cupcake truck has found that the number of cupcakes she sells per day depends on the price she charges. Let \(C(p)\) be the number of cupcakes she sells when the price of a cupcake is \(p\) cents. Interpret in words the statement that \(C'(300) = 2\text{;}\) be sure to explain the units involved in this statement. (Imagine you were trying to explain to someone who's never taken calculus what the practical meaning of \(C'(300) = 2\) is in terms of cupcakes and price.) Solution
In words, the statement \(C'(300) = 2\) says that, when the price of a cupcake is 300 cents, the instantaneous rate of change in the number of cupcakes sold in a day (with respect to price) is \(2\) cupcakes per cent. In less technical language, for each cent the owner increases the price above 300 cents, she can expect to sell 2 fewer cupcakes (and similarly, for each cent she decreases the price, she can expect to sell 2 more cupcakes). But these expectations are only reasonable for small price changes. So, for example, it would be reasonable to use this guideline to estimate the number of cupcakes she'd sell if she changed the price to 299 or 302 cents, but it would not be reasonable to use this to estimate the number of cupcakes she'd sell if she raised the price to 700 cents. (That's what being an “instantaneous” rate of change means.)
 The actual definition of the derivative \(f'(x)\) of a function \(f(x)\) is expressed as a limit; fill in the boxes to finish the definition.\begin{equation*} f'(x) = \lim_{\fcolorbox{blue}{white}{$\vphantom{h}\phantom{h \to 0}$}} \fcolorbox{red}{white}{$\vphantom{\dfrac{1}{1}}\phantom{\dfrac{f(x + h)  f(x)}{h}}$} \end{equation*}Explain the meaning of this definition; what does the expression in the red box represent? How do you visualize it on a graph of \(f\text{?}\) Solution
The definition is
\begin{equation*} f'(x) = \lim_{\fcolorbox{blue}{white}{$h \to 0$}} \fcolorbox{red}{white}{$\dfrac{f(x + h)  f(x)}{h}$} \end{equation*}The expression in the red box, \(\dfrac{f(x + h)  f(x)}{h}\text{,}\) is the slope of the secant line between the points \((x, f(x))\) and \((x + h, f(x + h))\text{:}\)
When we let \(h \to 0\text{,}\) the purple point approaches the green point, and the secant line approaches the tangent line at \(x\text{,}\) so the slope of the secant line, \(\dfrac{f(x + h)  f(x)}{h}\text{,}\) approaches the slope of the tangent line, \(f'(x)\text{.}\)
Subsection 1.2 Integration Review

Here's the graph of a function \(f\text{.}\)
 How do you visualize \(\displaystyle \int_1^7 f(x)\,dx\) on this graph? Solution
It's the signed area between \(y = f(x)\) and the \(x\)axis, between \(x = 1\) and \(x = 7\text{.}\) That is, it's the area of the green region minus the area of the red region:
 Like the derivative, the definite integral \(\displaystyle \int_1^7 f(x)\,dx\) is really defined as a limit. Explain what limit this is. Solution
If we slice the interval \([1, 7]\) into \(n\) pieces of equal width \(\Delta x\) and label the endpoints \(x_0 = 1, x_1, x_2, ..., x_n = 7\text{,}\) then here's one way to approximate the signed area \(\displaystyle \int_1^7 f(x)\,dx\) using rectangles:
Here, the signed area of the \(k\)th rectangle is \(f(x_k) \Delta x\text{,}\) so the total area in the rectangles is \(\displaystyle \sum_{k=1}^n f(x_k) \Delta x\text{.}\)
If we use more rectangles, so that each one is thinner, our approximation gets more and more accurate:
So, \(\displaystyle \int_1^7 f(x)\,dx\) is the limit of these approximations as the number of rectangles increases without bound, or \(\boxed{\lim_{n \to \infty} \sum_{k=1}^n f(x_k) \Delta x}\text{.}\)
 How do you visualize \(\displaystyle \int_1^7 f(x)\,dx\) on this graph? Solution

Below is a list of integrals. Any variables in the integral other than the variable of integration should be considered constants. For example, in the integral \(\displaystyle \int xyz\ dy\text{,}\) the \(dy\) tells us that \(y\) is the variable of integration, so we'll treat \(x\) and \(z\) as constants.
One of the integrals below cannot be evaluated by hand; identify that one, and evaluate the rest. (Can you check your answers by differentiating them?)
If you need to review your techniques of integration, check out this review material.
 \(\displaystyle \int x \sin(x)\,dx\) Solution
We'll integrate by parts. Let \(u = x\) and \(dv = \sin x\,dx\text{.}\) Then, \(du = dx\) and \(v =  \cos x\text{,}\) so \(\displaystyle \int x \sin x\,dx =  x \cos x  \int  \cos x\,dx = \boxed{ x \cos x + \sin x + C}\text{.}\)
 \(\displaystyle \int x \sin(x^2)\,dx\) Solution
You might be able to do this in your head, but if not, you can use substitution. If we substitute \(u = x^2\text{,}\) then \(du = 2x\,dx\text{,}\) so \(x\,dx = \frac{du}{2}\text{.}\) Then,
\begin{align*} \int x \sin(x^2)\,dx \amp = \int \sin(u) \frac{du}{2} \\ \amp =  \frac{1}{2} \cos u + C \\ \amp = \boxed{\frac{1}{2} \cos (x^2) + C} \end{align*}  \(\displaystyle \int x \sin(xy)\,dy\) Solution
Again, you might be able to do this in your head, or you can use substitution. If we substitute \(u = xy\text{,}\) then (remembering that \(x\) is a constant and \(y\) is the variable), \(du = x\,dy\text{,}\) so
\begin{align*} \int x \sin(xy)\,dy \amp = \int \sin(u)\,du \\ \amp =  \cos u + C \\ \amp = \boxed{\cos(xy) + C} \end{align*}  \(\displaystyle \int e^{x^2 y}\,dx\) Solution
This one cannot be evaluated by hand. This isn't because \(e^{x^2 y}\) doesn't have an antiderivative (when we think of it as a function of \(x\text{,}\) with \(y\) a constant); after all, the Fundamental Theorem of Calculus guarantees that every continuous function has a continuous antiderivative. Rather, mathematicians have proved that the antiderivatives of \(e^{x^2 y}\) can't be expressed in terms of familiar functions like exponentials and trigonometric functions.
 \(\displaystyle \int e^{x^2 y}\,dy\) Solution
We'll substitute \(u = x^2 y\text{.}\) Then (remembering that \(x\) is a constant and \(y\) is the variable), \(du = x^2\,dy\text{,}\) so \(dy = \frac{du}{x^2}\text{.}\) So,
\begin{align*} \int e^{x^2 y}\,dy \amp = \int e^u \, \frac{du}{x^2} \\ \amp = \frac{1}{x^2} \int e^u\,du \\ \amp = \frac{1}{x^2} e^u + C \\ \amp = \boxed{\frac{1}{x^2} e^{x^2 y} + C} \end{align*}
 \(\displaystyle \int x \sin(x)\,dx\) Solution