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How do you integrate periodic functions?
Same as any other function - but in the case of a definite integral, you can take advantage of the periodicity. For example, assuming that a certain function has a period of pi, and the value of the definite integral from zero to pi is 2, then the integral from zero to 2 x pi is 4.
How is definite integral connected to area under the curve?
If the values of the function are all positive, then the integral IS the area under the curve.
What is the relationship of integral and differential calculus?
We say function F is an anti derivative, or indefinite integral of f if F' = f. Also, if f has an anti-derivative and is integrable on interval [a, b], then the definite integral of f from a to b is equal to F(b) - F(a) Thirdly, Let F(x) be the definite integral of integrable function f from a to x for all x in [a, b] of f, then F is an anti-derivative of f on [a,b] The definition of indefinite integral as anti-derivative, and the relation of definite integral with anti-derivative, we can conclude that integration and differentiation can be considered as two opposite operations.
What is the relation between definite integrals and areas?
Consider the integral of sin x over the interval from 0 to 2pi. In this interval the value of sin x rises from 0 to 1 then falls through 0 to -1 and then rises again to 0. In other words the part of the sin x function between 0 and pi is 'above' the axis and the part between pi and 2pi is 'below' the axis. The value of this integral is zero because although the areas enclosed by the parts of the function between 0 and pi and pi and 2pi are the same the integral of the latter part is negative. The point I am trying to make is that a definite integral gives the area between a function and the horizontal axis but areas below the axis are negative. The integral of sin x over the interval from 0 to pi is 2. The integral of six x over the interval from pi to 2pi is -2.
What is the fundamental theorem of calculus?
If df(x)/dx = g(x), then integral [from a to b] g(x) dx = f(b)-f(a). In plain English: the definite integral can be calculated by finding the antiderivative, evaluating it at the endpoints, and subtracting.
