Integration is about chopping things up, and adding the pieces. This “chopping and adding” viewpoint is often helpful in setting up problems involving integration, and will be especially useful later for multiple integrals.

Figure 1: Chopping a line into small pieces.

For example, consider finding the total amount of chocolate on a straight piece of wafer (like a stick of Pocky), given the density of chocolate on the wafer. What does “density” mean? In this case, the amount of chocolate, which could be measured in grams, per unit distance along the wafer, which could be measured in centimeters. We call this quantity the (linear) mass density of chocolate on the wafer, which we will denote by $\lambda$. Using $x$ to measure distance along the wafer, chop the wafer into small pieces of length $dx$, as indicated symbolically in Figure 1. 1) What is the mass of each piece? Clearly, $\lambda \,dx$. The total mass $M$ of the chocolate on the wafer is given by adding up the mass of each piece. Since in general $\lambda$ depends on position, it can be thought of as a function of $x$, that is, $\lambda=\lambda(x)$. Thus, “adding” really means “integrating”, and we obtain \begin{equation} M = \int \lambda(x) \,dx \end{equation}

You may have learned that integration is antidifferentiation, and that integrals are areas. Yes, the total amount of chocolate, thought of as a function of the distance from one end, is indeed an antiderivative of the function $\lambda(x)$. And yes, this integral represents the “area” under the graph of the function $\lambda(x)$, although the dimensions (mass) are not those of geometric area (length squared). But the “infinitesmal mass” $\lambda \,dx$ better represents the relevant physical process, and should therefore be regarded as fundamental. Note the importance of $dx$ (and its units) to this argument!

From this point of view, the fundamental theorem of calculus is easy. If you add up little bits of choclate ($\lambda\,dx$), you get the total amount of chocolate. Similarly, if you add up the small changes in some quantity, say $dq$, the charge on a small piece of wire, you get its total change, so that \begin{equation} \int dq = q \label{intdiff} \end{equation} which might look more familiar if you write $q=f(x)$, so that (\ref{intdiff}) becomes \begin{equation} \int \frac{df}{dx} \,dx = f \end{equation}

Yes, this means that the integral of a derivative is the function you started with. In the chocolate example, that function is the “total chocolate” function, whose derivative is $\lambda(x)$. Nontheless, what you're adding up is chocolate, that is, $\lambda\,dx$, not merely $\lambda$.

The use of differentials has other advantages. In the language of differentials, substitution is easy: If $u=x^2$, then $du=2x\,dx$, so that e.g. \begin{equation} \int \cos(x^2)\, 2x\,dx = \int \cos(u) \,du = \sin(u) \end{equation} And integration by parts is just the product rule: Start with \begin{equation} d(uv) = u\,dv + v\,du \end{equation} and integrate both sides (and rearrange terms as needed).

All of the above integrals are indefinite; both sides of the equality are functions. Definite integrals are obtained simply by evaluating both sides at the endpoints of some interval. For example: \begin{equation} \int\limits_a^b \frac{df}{dx} \,dx = f\Big|_a^b = f(b) - f(a) \end{equation} and both sides of this equality are now numbers.

In practice, ($\ref{intdiff}$) tells us that (single) integration is nothing more than antidifferentation — just run the derivative rules backwards. Here are the basic derivative rules, in differential form: \begin{eqnarray*} d\left(u^n\right) &=& nu^{n-1} \,du \\ d\left(e^u\right) &=& e^u \,du \\ d(\sin u) &=& \cos u \,du \\ d(\cos u) &=& -\sin u \,du \\ d(\ln u) &=& \frac{1}{u} \>du \\ \end{eqnarray*} To obtain the corresponding integration rules, simply integrate both sides, and use ($\ref{intdiff}$). For example, the first rule becomes \begin{equation} \int nu^{n-1} \,du = u^n \end{equation} which is more commonly written in the form \begin{equation} \int u^m \,du = \frac{u^{m+1}}{m+1} \end{equation} where of course $m=n-1\ne-1$. 2)

At this point, you should be able to integrate easily simple expressions involving polynomials, trig, exponentials, and logarithms, including those requiring reasonably obvious substitutions, and you should be comfortable with occasional problems involving more complicated substitutions and/or integration by parts. If you are not comfortable with these skills, including the necessary algebra, you are strongly encouraged to spend as much time as needed on review and practice until you are truly fluent.