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## Integral Calc – A Look at Calculus Integration

While Calculus I is primarily devoted to differential calculus, or the study of derivatives, most of Calculus 2 and beyond focuses on integral calculus, which revolves around the study of integrals and the process of integration. Integration has entire courses devoted to it because it is such a critical exercise in mathematics, and in integral calculus there are many different methods and techniques that can be used for integration in different situations. Here we will look at 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 itself a function. A definite integral finds the difference between two particular values ​​of the indefinite integral, and usually 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 by 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 will find the area between the x-axis and the parabola from 0 to 3. 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 basic rule for obtaining derivatives. Its counterpart in integral calculus is the process of integration by substitution, also known as u-substitution. In general, when one tries to take the integral of a function of the form f(g(x)) * g'(x), the result is simply f (x). However, there are many variations on this general theme, and it can even be extended to include functions that have multiple variables. For a basic example, suppose you want to find the integral of (x+1)² dx. We will let u = x+1, and du = dx. After substituting u for x+1, and du for dx, we’ll try to take the integral of u² du, which we know from our basic formulas is just u³/3 + C. We substitute x+ for 1 in our last answer back to you, and we quickly have (x+1)³/3 + C.

Integration in accounting is often seen as a strategic process rather than a straightforward 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, let’s call them u and v, then the integral of u dv becomes the combination of uv – the integral of v two. This may seem like another random formula, but its importance is that it often allows us to simplify a function over which we are taking the integral. This strategy requires that we choose u and du in such a way that the derivative of u is less complicated than u. When we divide the integral by parts, our resulting integral is two, but not u, meaning that the function we are taking the integral from has been simplified in the process.

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