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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.
Integral of x cosx dx?
The integral of x cos(x) dx is cos(x) + x sin(x) + C
What is the derivative of cos pi x?
By the 'Chain Rule' dy/dx = dy/du X du/dx Y = Cos (pi*x). Let pi*x = u Y = Cos(u) u = pi*x dy/du = -Sin(u) du/dx = pi Hence dy/dx = dy/du X du/dx => ( Chain Rule) dy/dx = -Sin(u) X pi Substitute u for pi*x Hence dy/dx = -Sin(pi*x) X pi Tidying up dy/dx = -piSin(pix) Done!!!!
What is the period of y equals sin5x?
The period of y=sin(x) is 2*pi, so sin(x) repeats every 2*pi units. sin(5x) repeats every 2*pi/5 units. In general, the period of y=sin(n*x) is 2*pi/n.
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.
