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## Integral Calc – A Look at Calculus Integration

While Calculus I is primarily devoted to the study of differential calculus, or derivatives, much of Calculus 2 and beyond is focused on integral calculus, which is based around the study of integrals and the process of integration. Integration has entire courses devoted to it because it is such an important operation in mathematics, and there are many different methods and techniques in integral calc used for integration in different situations. Here we are going to take an overview of some of these techniques and the types of integrals that can be taken.

First, there are definite integers and indefinite integers. An indefinite integral is simply the antiderivative of a function, and is a function. A definite integral finds the difference between two specific values of an indefinite integral, and usually produces a numerical answer rather than a function. Certain integers can be used to find areas and volumes of irregular shapes that cannot be found with basic geometry, as long as the sides of the shape being measured follow some function that can be integrated. For example, x², a fixed integer from 0 to 3, will 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 to a hypotenuse, and is a great example. Fast Finding Areas of Irregular Two-Dimensional Shapes Using Definite Integrals.

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 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 also be extended to handle functions with multiple variables. For a basic example, suppose you want to find the indefinite 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 are left trying to take the integral of u² du, which we know from our basic formulas is u³/3 + C. We substitute x+ 1 back for you in our final answer, and quickly (x+1)³/3 + C.

Integration in calculus is viewed as a strategic process rather than a straight-forward 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 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 the integral of uv – v du. This may seem like just another random formula, but its importance is that it allows us to simplify a function that we are taking an integral. For this strategy we need to choose u and du in such a way that the derivative of u is less complex than u. Once we break the integral down by parts, our resulting integral contains du, but not u, which means that the process by which we take the integral is simplified.

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