Use this cheat sheet as a reference for the most important formulas, rules, equations, and so on that you need for calculus. The main calculus topics are covered (limits, differentiation, integration, and infinite series), as are the critical pre-calculus topics (algebra, geometry, and trigonometry).Some Essential AlgebraAlgebra is the language of calculus.

A surface of revolution is a three-dimensional surface with circular cross sections, like a vase or a bell, or a wine bottle. This article, and the video, show you how to find its area. For these problems, you divide the surface into narrow circular bands, figure the surface area of a representative band, and then just add up the areas of all the bands to get the total surface area.

The nice thing about finding the area of a surface of revolution is that there’s a formula you can use. Memorize it and you’re halfway done. To find the area of a surface of revolution between a and b, watch this video tutorial or follow the steps below: This formula looks long and complicated, but it makes more sense when you spend a minute thinking about it.

By its nature, calculus can be intimidating. But you can take some of the fear of studying calculus away by understanding its basic principles, such as derivatives and antiderivatives, integration, and solving compound functions. Also discover a few basic rules applied to calculus like Cramer's Rule, the Constant Multiple Rule, and a few others, and you'll be on your way to acing the course.

When g'(x) = f(x), you can use the substitution u = g(x) to integrate expressions of the form f(x) multiplied by h(g(x)), provided that h is a function that you already know how to integrate.Variable substitution helps to fill the gaps left by the absence of a Product Rule and a Chain Rule for integration.Here’s a hairy-looking integral that actually responds well to substitution:The key insight here is that the numerator of this fraction is the derivative of the inner function in the denominator.

You can find the average value of a function over a closed interval by using the mean value theorem for integrals. The best way to understand the mean value theorem for integrals is with a diagram — look at the following figure.The graph on the left shows a rectangle whose area is clearly less than the area under the curve between 2 and 5.

Your first step in any problem that involves partial fractions is to recognize which case you’re dealing with so that you can solve the problem. One case where you can use partial fractions is when the denominator is the product of distinct quadratic factors — that is, quadratic factors that are nonrepeating. For each distinct quadratic factor in the denominator, add a partial fraction of the following form:For example, suppose that you want to integrate this function:The first factor in the denominator is linear, but the second is quadratic and can’t be decomposed to linear factors.