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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.
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Advanced engineering mathematics by erwin kreyszig 9th edition solution?
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Use the concept of a limit to explain how you could find the exact value for the definite integral value for a section of your graph?
The definite integral value for a section of a graph is the area under the graph. To compute the area, one method is to add up the areas of the rectangles that can fit under the graph. By making the rectangles arbitrarily narrow, creating many of them, you can better and better approximate the area under the graph. The limit of this process is the summation of the areas (height times width, which is delta x) as delta x approaches zero. The deriviative of a function is the slope of the function. If you were to know the slope of a function at any point, you could calculate the value of the function at any arbitrary point by adding up the delta y's between two x's, again, as the limit of delta x approaches zero, and by knowing a starting value for x and y. Conversely, if you know the antideriviative of a function, the you know a function for which its deriviative is the first function, the function in question. This is exactly how integration works. You calculate the integral, or antideriviative, of a function. That, in itself, is called an indefinite integral, because you don't know the starting value, which is why there is always a +C term. To make it into a definite integral, you evaluate it at both x endpoints of the region, and subtract the first from the second. In this process, the +C's cancel out. The integral already contains an implicit dx, or delta x as delta x approaches zero, so this becomes the area under the graph.
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Could Give and explain the two basic classifications of calculus?
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Advanced engineering mathematics by erwin kreyszig 9th edition solution?
A book to introduce engineering and physics students to areas of math that seem to be most important in relation to practical problems. Book was first published in 1962 - so it is a bit out of date - and has had several reprints. Erwin Kreyszig (Jan 6, 1922 - December 12, 2008) was Professor of at Ohio State University, later moved to Carleton University in Ottawa). The book covers: Ordinary Differential Equations; Ordinary Linear Differential Equations; Power Series Solutions of Diff. Equations; Laplace Transform; Vector Analysis; Line and Surface Integrals; Systems of Linear Equations; Fourier Series and Integrals; Partial Differential Equations; Complex analytic Functions; Conformal Mapping; Complex Integrals; and so on. A very useful book when I did my engineering, though it must be out of date now. GSC
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