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## Integral Calc – A Look at Calculus Integration

While Calculus I is primarily concerned with differential calculus, or the study of derivatives, most of Calculus 2 and beyond focuses on integral calculus, which is based on the study of integrals and the process of integration. Integration has entire courses devoted to it because it is such a critical operation in mathematics, and there are many different methods and techniques in integral calculus that are used for integration in different situations. Here we will 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 antiderivative of a function and is a function itself. A definite integral finds the difference between two specific values of the indefinite integral, and usually produces a numerical answer rather than a function. Definite integrals can be used to find areas and volumes of irregular figures that cannot be found with basic geometry, as long as the sides of the figure being measured follow 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 parabola curve for a hypotenuse, and it is a large example to quickly find 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 the 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 has 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 for x+1, and du for dx, we’re left trying to take the integral of u² du, which we know from our basic patterns is just u³/3 + C. Substitute x+ Substitute ua back into our final answer and quickly have (x+1)³/3 + C.

Integration in computation is often considered a strategic process rather than a simple mechanical process because of the large number of tools at your disposal for integrating functions. A very important tool is integration by parts, which is a game of the product rule for differentiation. In short, if there are two functions, call them uiv, then the integral of u dv is equal to uv – the integral of v du. This may seem like just another random formula, but its importance is that it often allows us to simplify a function that we are taking the integral of. This strategy requires us to choose ui du so that the derivative of u is less complicated than u. Once we divide the integral by parts, our resulting integral contains du, but not u, meaning that the function we’re taking the integral from has been simplified in the process.

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