While Calculus I is primarily dedicated to differential calc, or the study of derivatives, most of Calculus 2 and beyond focuses on integral calc, which is based around the study of integrals and the process of integration. Integration has entire courses dedicated to it because it’s such a critical operation in mathematics, and there are many different methods and techniques in integral calc that are used for integration in different situations. Here we’re going to look at an overview of some of these techniques and the types of integrals that can be taken.

First, there are definite integrals and indefinite integrals. An indefinite integral is just the anti-derivative of a function, and is a function itself. A definite integral finds the difference between two specific values of the indefinite integral, and typically produces a numerical answer instead of a function. Definite integrals can be used to find areas and volumes of irregular figures that cannot be found with basic geometry, so long as the sides of the figure being measured follows some function that can be integrated. For example, the definite integral from 0 to 3 of x² would find the area between the x-axis and the parabola from 0 to 3. This shape is like a triangle with a curve from a parabola for a hypotenuse, and is a great example of quickly finding the area of an irregular two-dimensional shape using a definite integral.

In differential calculus, you learn that the chain rule is a key rule for taking derivatives. Its counterpart in integral calculus is the process of integration by substitution, also known as u-substitution. In general, when trying to take the integral of a function that is of the form f(g(x)) * g'(x), the result is simply f(x). However, there are a number of variations on this general theme, and it can even be extended to handle functions that have multiple variables. For a basic example, suppose you want to find the indefinite integral of (x+1)² dx. We would let u = x+1, and du = dx. After substituting u in place of x+1, and du in place of dx, we’re left with trying to take the integral of u² du, which we know from our basic patterns is just u³/3 + C. We substitute x+1 back in for u in our final answer, and quickly have (x+1)³/3 + C.

Integration in calculus is often seen a strategic process instead of a straight-forward mechanical process because of the large number of tools at your disposal for integrating functions. One very important tool is integration by parts, which is a play on the product rule for differentiation. In short, if there are two functions, call them u and v, then the integral of u dv equals uv – the integral of v du. This may seem like just another random formula, but it’s significance is that it often allows us to simplify a function that we’re taking the integral of. This strategy requires that we pick u and du in a way that the derivative of u is less complicated than u. Once we break the integral up by parts, our resulting integral contains du, but not u, meaning that the function we are taking the integral of has become simplified in the process.